Chapter 3 Trigonometric Functions

We are asked to convert the angle 40° 20′ into radian measure.

First, convert the minutes part into degrees. We know that $1^\circ = 60$ minutes (written as 60′).

So, 20 minutes is equal to $\frac{20}{60}$ degrees.

$20' = \left(\frac{20}{60}\right)^\circ = \left(\frac{1}{3}\right)^\circ$

Now, add this to the degree part:

$40^\circ 20' = 40^\circ + \left(\frac{1}{3}\right)^\circ = \left(40 + \frac{1}{3}\right)^\circ$

Combine the terms:

$\left(40 + \frac{1}{3}\right)^\circ = \left(\frac{40 \times 3}{3} + \frac{1}{3}\right)^\circ = \left(\frac{120}{3} + \frac{1}{3}\right)^\circ = \left(\frac{121}{3}\right)^\circ$

Now, we convert degrees to radians. The conversion formula is: $1 \text{ degree} = \frac{\pi}{180} \text{ radians}$.

So, to convert $\left(\frac{121}{3}\right)^\circ$ to radians, we multiply by $\frac{\pi}{180}$.

$\left(\frac{121}{3}\right)^\circ = \frac{121}{3} \times \frac{\pi}{180} \text{ radians}$

Multiply the numerators and the denominators:

$\frac{121 \times \pi}{3 \times 180} \text{ radians} = \frac{121\pi}{540} \text{ radians}$

Thus, 40° 20′ is equal to $\frac{121\pi}{540}$ radians.

The radian measure is $\frac{121\pi}{540}$.

We need to convert 6 radians into degree measure.

The conversion formula from radians to degrees is: $\text{Radians} \times \frac{180}{\pi} = \text{Degrees}$.

Using the approximation $\pi \approx \frac{22}{7}$, we have:

$6 \text{ radians} = 6 \times \frac{180}{\frac{22}{7}} \text{ degrees}$

$= 6 \times 180 \times \frac{7}{22} \text{ degrees}$

$= \frac{6 \times 180 \times 7}{22} \text{ degrees}$

$= \frac{\cancel{6}^{3} \times 180 \times 7}{\cancel{22}_{11}} \text{ degrees}$

$= \frac{3 \times 180 \times 7}{11} \text{ degrees}$

$= \frac{540 \times 7}{11} \text{ degrees}$

$= \frac{3780}{11} \text{ degrees}$

Now, we convert the fractional part of the degree into minutes and seconds.

Divide 3780 by 11:

$\frac{3780}{11} = 343$ with a remainder of $3780 - (11 \times 343) = 3780 - 3773 = 7$.

So, $\frac{3780}{11}\text{ degrees} = 343 \frac{7}{11}\text{ degrees}$. This is $343^\circ + \frac{7}{11}^\circ$.

Convert the fractional part $\frac{7}{11}^\circ$ to minutes. We know $1^\circ = 60'$.

$\frac{7}{11}^\circ = \frac{7}{11} \times 60' = \frac{420}{11}'$

Divide 420 by 11:

$\frac{420}{11} = 38$ with a remainder of $420 - (11 \times 38) = 420 - 418 = 2$.

So, $\frac{420}{11}' = 38 \frac{2}{11}'$. This is $38' + \frac{2}{11}'$.

Convert the fractional part $\frac{2}{11}'$ to seconds. We know $1' = 60''$.

$\frac{2}{11}' = \frac{2}{11} \times 60'' = \frac{120}{11}''$

Divide 120 by 11:

$\frac{120}{11} \approx 10.91''$. We can round this to approximately 11 seconds.

So, $\frac{120}{11}'' \approx 11''$ (rounded to the nearest second).

Combining the degrees, minutes, and seconds, we get:

$343^\circ 38' \frac{120}{11}'' \approx 343^\circ 38' 11''$

Thus, 6 radians is approximately equal to 343 degrees, 38 minutes, and 11 seconds.

The degree measure is approximately $343^\circ 38' 11''$.

Given:

Central angle $\theta = 60^\circ$

Arc length $l = 37.4$ cm

Value of $\pi = \frac{22}{7}$

To Find:

The radius of the circle, $r$.

Solution:

The relationship between the arc length ($l$), the radius ($r$), and the central angle ($\theta$) measured in radians is given by the formula:

$l = r\theta$

For this formula, the angle $\theta$ must be in radians. We are given the angle in degrees, so we first need to convert it to radians.

The conversion formula from degrees to radians is: Degrees $\times \frac{\pi}{180} = $ Radians.

Given angle $\theta = 60^\circ$.

Converting $60^\circ$ to radians:

$\theta \text{ (in radians)} = 60^\circ \times \frac{\pi}{180}$

$\theta \text{ (in radians)} = \frac{60}{180} \pi$

$\theta \text{ (in radians)} = \frac{1}{3} \pi$ radians.

Now, we can use the formula $l = r\theta$. We have $l = 37.4$ cm and $\theta = \frac{\pi}{3}$ radians.

$37.4 = r \times \frac{\pi}{3}$

To find $r$, we rearrange the formula:

$r = \frac{37.4}{\frac{\pi}{3}}$

$r = \frac{37.4 \times 3}{\pi}$

Now, substitute the given value of $\pi = \frac{22}{7}$:

$r = \frac{37.4 \times 3}{\frac{22}{7}}$

$r = \frac{37.4 \times 3 \times 7}{22}$

$r = \frac{37.4 \times 21}{22}$

We can rewrite 37.4 as $\frac{374}{10}$.

$r = \frac{\frac{374}{10} \times 21}{22}$

$r = \frac{374 \times 21}{10 \times 22}$

$r = \frac{374 \times 21}{220}$

We can simplify the fraction $\frac{374}{220}$ by dividing the numerator and denominator by their common factors. Both are divisible by 2.

$\frac{374}{220} = \frac{187}{110}$

Now, 187 is divisible by 11 ($187 = 11 \times 17$) and 110 is divisible by 11 ($110 = 11 \times 10$).

$\frac{\cancel{187}^{17}}{\cancel{110}_{10}} = \frac{17}{10}$

So, $r = \frac{17}{10} \times 21$

$r = \frac{17 \times 21}{10}$

$17 \times 21 = 17 \times (20 + 1) = 17 \times 20 + 17 \times 1 = 340 + 17 = 357$.

$r = \frac{357}{10}$

$r = 35.7$

The radius of the circle is 35.7 cm.

The radius of the circle is 35.7 cm.

Given:

Length of the minute hand (radius), $r = 1.5 \$ cm

Time duration, $t = 40 \$ minutes

Value of $\pi = 3.14$

To Find:

The distance the tip of the minute hand moves in 40 minutes.

Solution:

The minute hand completes one full revolution ($360^\circ$) in 60 minutes.

Angle traced by the minute hand in 60 minutes = $360^\circ$

Angle traced by the minute hand in 1 minute = $\frac{360^\circ}{60} = 6^\circ$

Angle traced by the minute hand in 40 minutes = $40 \times 6^\circ = 240^\circ$

Let $\theta$ be the angle traced in 40 minutes. So, $\theta = 240^\circ$.

To use the arc length formula, we need to convert the angle from degrees to radians.

The relationship between degrees and radians is $180^\circ = \pi$ radians.

So, $1^\circ = \frac{\pi}{180}$ radians.

Therefore, $\theta = 240^\circ = 240 \times \frac{\pi}{180}$ radians.

$\theta = \frac{240}{180}\pi = \frac{24}{18}\pi = \frac{4}{3}\pi$ radians.

The distance moved by the tip of the minute hand is the arc length ($s$) of the sector with radius $r$ and central angle $\theta$ (in radians).

The formula for arc length is $s = r\theta$.

Substituting the values:

$r = 1.5 \$ cm$

$\theta = \frac{4}{3}\pi$ radians

$s = 1.5 \times \frac{4}{3}\pi$

$s = \frac{1.5 \times 4}{3}\pi$

$s = \frac{6}{3}\pi$

$s = 2\pi$

Using the given value of $\pi = 3.14$:

$s = 2 \times 3.14$

$s = 6.28$

The distance moved by the tip of the minute hand is $6.28 \$ cm$.

Alternatively, using the angle in degrees first:

Distance moved = (Fraction of the circle completed) $\times$ (Circumference of the circle)

Fraction of the circle completed in 40 minutes = $\frac{40 \text{ minutes}}{60 \text{ minutes}} = \frac{40}{60} = \frac{2}{3}$

Circumference of the circle = $2\pi r = 2 \times 3.14 \times 1.5$

Circumference = $2 \times 1.5 \times 3.14 = 3 \times 3.14 = 9.42 \$ cm$

Distance moved = $\frac{2}{3} \times 9.42$

Distance moved = $2 \times \frac{9.42}{3} = 2 \times 3.14 = 6.28 \$ cm$

Both methods yield the same result.

Answer:

The tip of the minute hand moves $6.28 \$ cm$ in 40 minutes.

Given:

Let the two circles be Circle 1 and Circle 2.

Let the radius of Circle 1 be $r_1$ and the radius of Circle 2 be $r_2$.

Let the angle subtended at the center of Circle 1 be $\theta_1 = 65^\circ$.

Let the angle subtended at the center of Circle 2 be $\theta_2 = 110^\circ$.

The arcs of the same length are in both circles. Let this length be $s$.

To Find:

The ratio of their radii, i.e., $r_1 : r_2$ or $\frac{r_1}{r_2}$.

Solution:

The formula for the length of an arc ($s$) of a sector with radius ($r$) and central angle ($\theta$) in radians is given by $s = r\theta$.

First, we need to convert the given angles from degrees to radians.

The conversion factor is $1^\circ = \frac{\pi}{180}$ radians.

For Circle 1, the angle in radians is $\theta_1 = 65^\circ \times \frac{\pi}{180} = \frac{65\pi}{180}$ radians.

For Circle 2, the angle in radians is $\theta_2 = 110^\circ \times \frac{\pi}{180} = \frac{110\pi}{180}$ radians.

Since the arc lengths are the same in both circles, we have:

Arc length in Circle 1 = $s_1 = r_1 \theta_1$

Arc length in Circle 2 = $s_2 = r_2 \theta_2$

Given that $s_1 = s_2 = s$, we can write:

$r_1 \theta_1 = r_2 \theta_2$

Substitute the radian values of the angles:

$r_1 \left(\frac{65\pi}{180}\right) = r_2 \left(\frac{110\pi}{180}\right)$

We want to find the ratio $\frac{r_1}{r_2}$. Rearrange the equation:

$\frac{r_1}{r_2} = \frac{\frac{110\pi}{180}}{\frac{65\pi}{180}}$

Cancel out the common term $\frac{\pi}{180}$ from numerator and denominator:

$\frac{r_1}{r_2} = \frac{110}{65}$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 5:

$\frac{110}{5} = 22$ and $\frac{65}{5} = 13$

So, $\frac{r_1}{r_2} = \frac{22}{13}$

The ratio of the radii is $r_1 : r_2 = 22 : 13$.

Answer:

The ratio of the radii of the two circles is $22 : 13$.

To convert a degree measure to radian measure, we use the formula:

Radian measure = Degree measure $\times \frac{\pi}{180}$

(i) $25^\circ$

Radian measure = $25 \times \frac{\pi}{180}$

We can simplify the fraction $\frac{25}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 5.

$\frac{25}{5} = 5$ and $\frac{180}{5} = 36$

So, $\frac{25}{180} = \frac{5}{36}$

Radian measure = $\frac{5\pi}{36}$

Thus, the radian measure corresponding to $25^\circ$ is $\frac{5\pi}{36}$.

(ii) $-47^\circ 30'$

First, convert the minutes part to degrees. We know that $60$ minutes ($60'$) = $1^\circ$.

So, $30' = \frac{30}{60}^\circ = \frac{1}{2}^\circ = 0.5^\circ$.

The given degree measure is $-47^\circ 30' = -(47 + 0.5)^\circ = -47.5^\circ$.

Now, convert $-47.5^\circ$ to radians:

Radian measure = $-47.5 \times \frac{\pi}{180}$

We can write $47.5$ as $\frac{475}{10}$.

Radian measure = $-\frac{475}{10} \times \frac{\pi}{180}$

Radian measure = $-\frac{475\pi}{1800}$

Simplify the fraction $\frac{475}{1800}$. Divide both by 25.

$\frac{475}{25} = 19$ and $\frac{1800}{25} = 72$

So, $\frac{475}{1800} = \frac{19}{72}$

Radian measure = $-\frac{19\pi}{72}$

Thus, the radian measure corresponding to $-47^\circ 30'$ is $-\frac{19\pi}{72}$.

(iii) $240^\circ$

Radian measure = $240 \times \frac{\pi}{180}$

Simplify the fraction $\frac{240}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 60.

$\frac{240}{60} = 4$ and $\frac{180}{60} = 3$

So, $\frac{240}{180} = \frac{4}{3}$

Radian measure = $\frac{4\pi}{3}$

Thus, the radian measure corresponding to $240^\circ$ is $\frac{4\pi}{3}$.

(iv) $520^\circ$

Radian measure = $520 \times \frac{\pi}{180}$

Simplify the fraction $\frac{520}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 20.

$\frac{520}{20} = 26$ and $\frac{180}{20} = 9$

So, $\frac{520}{180} = \frac{26}{9}$

Radian measure = $\frac{26\pi}{9}$

Thus, the radian measure corresponding to $520^\circ$ is $\frac{26\pi}{9}$.

To convert a radian measure to degree measure, we use the formula:

Degree measure = Radian measure $\times \frac{180}{\pi}$

We are given to use $\pi = \frac{22}{7}$.

(i) $\frac{11}{16}$ radians

Degree measure = $\frac{11}{16} \times \frac{180}{\pi}$

Substitute $\pi = \frac{22}{7}$:

Degree measure = $\frac{11}{16} \times \frac{180}{\frac{22}{7}}$

Degree measure = $\frac{11}{16} \times \frac{180 \times 7}{22}$

Degree measure = $\frac{\cancel{11}^1}{16} \times \frac{180 \times 7}{\cancel{22}^2}$

Degree measure = $\frac{180 \times 7}{16 \times 2}$

Degree measure = $\frac{\cancel{180}^{90} \times 7}{16 \times \cancel{2}^1}$

Degree measure = $\frac{\cancel{90}^{45} \times 7}{\cancel{16}^8}$

Degree measure = $\frac{45 \times 7}{8} = \frac{315}{8}$ degrees.

Now, we convert $\frac{315}{8}$ degrees into degrees, minutes, and seconds.

$\frac{315}{8} = 39 \frac{3}{8}$ degrees.

This is $39^\circ$ and $\frac{3}{8}$ of a degree.

Convert the fractional part to minutes: $\frac{3}{8} \times 60 = \frac{180}{8} = \frac{45}{2} = 22.5$ minutes.

This is $22'$ and $0.5$ of a minute.

Convert the fractional part to seconds: $0.5 \times 60 = 30$ seconds.

So, $\frac{315}{8}$ degrees = $39^\circ 22' 30''$.

Thus, the degree measure corresponding to $\frac{11}{16}$ radians is $39^\circ 22' 30''$.

(ii) $-4$ radians

Degree measure = $-4 \times \frac{180}{\pi}$

Substitute $\pi = \frac{22}{7}$:

Degree measure = $-4 \times \frac{180}{\frac{22}{7}}$

Degree measure = $-4 \times \frac{180 \times 7}{22}$

Degree measure = $-\frac{4 \times 180 \times 7}{22}$

Degree measure = $-\frac{\cancel{4}^2 \times 180 \times 7}{\cancel{22}^{11}}$

Degree measure = $-\frac{2 \times 180 \times 7}{11} = -\frac{360 \times 7}{11} = -\frac{2520}{11}$ degrees.

Now, we convert $-\frac{2520}{11}$ degrees into degrees, minutes, and seconds.

$-\frac{2520}{11} = -(229 \frac{1}{11})$ degrees.

This is $-229^\circ$ and $-\frac{1}{11}$ of a degree.

Convert the fractional part to minutes: $\frac{1}{11} \times 60 = \frac{60}{11}$ minutes.

$\frac{60}{11} = 5 \frac{5}{11}$ minutes.

This is $5'$ and $\frac{5}{11}$ of a minute.

Convert the fractional part to seconds: $\frac{5}{11} \times 60 = \frac{300}{11}$ seconds.

$\frac{300}{11} \approx 27.27...$ seconds.

Rounding to the nearest second, we get $27''$.

So, $-\frac{2520}{11}$ degrees $\approx -229^\circ 5' 27''$.

Thus, the degree measure corresponding to $-4$ radians is approximately $-229^\circ 5' 27''$.

(iii) $\frac{5\pi}{3}$ radians

Degree measure = $\frac{5\pi}{3} \times \frac{180}{\pi}$

Cancel out $\pi$ from the numerator and the denominator:

Degree measure = $\frac{5}{\cancel{3}^1} \times \cancel{180}^{60}$

Degree measure = $5 \times 60 = 300^\circ$.

Thus, the degree measure corresponding to $\frac{5\pi}{3}$ radians is $300^\circ$.

(iv) $\frac{7\pi}{6}$ radians

Degree measure = $\frac{7\pi}{6} \times \frac{180}{\pi}$

Cancel out $\pi$ from the numerator and the denominator:

Degree measure = $\frac{7}{\cancel{6}^1} \times \cancel{180}^{30}$

Degree measure = $7 \times 30 = 210^\circ$.

Thus, the degree measure corresponding to $\frac{7\pi}{6}$ radians is $210^\circ$.

Given:

Number of revolutions made by the wheel in one minute = 360

Time duration = 1 minute

To Find:

The angle (in radians) the wheel turns in one second.

Solution:

The wheel makes 360 revolutions in 1 minute.

First, convert the time from minutes to seconds.

$1 \$ minute = 60 \$ seconds$

So, the wheel makes 360 revolutions in 60 seconds.

Number of revolutions made by the wheel in 1 second = $\frac{\text{Total revolutions}}{\text{Total time in seconds}}$

Number of revolutions per second = $\frac{360}{60} = 6 \$ revolutions/second$.

Now, we need to find the angle turned in one revolution.

One complete revolution corresponds to an angle of $360^\circ$.

To convert degrees to radians, we use the relation $180^\circ = \pi$ radians.

So, $360^\circ = 2 \times 180^\circ = 2\pi$ radians.

Angle turned in one revolution = $2\pi$ radians.

The angle turned by the wheel in one second is the product of the number of revolutions per second and the angle turned per revolution.

Total angle turned in 1 second = (Revolutions per second) $\times$ (Angle per revolution)

Total angle turned in 1 second = $6 \times 2\pi$ radians

Total angle turned in 1 second = $12\pi$ radians.

Answer:

The wheel turns through $12\pi$ radians in one second.

Given:

Radius of the circle, $r = 100 \$ cm$

Length of the arc, $s = 22 \$ cm$

Value of $\pi = \frac{22}{7}$

To Find:

The degree measure of the angle subtended at the centre by the arc.

Solution:

The relationship between the arc length ($s$), radius ($r$), and the central angle ($\theta$) in radians is given by the formula:

$s = r\theta$

where $\theta$ is the angle in radians.

Substitute the given values into the formula:

$22 = 100 \times \theta$

Solve for $\theta$:

$\theta = \frac{22}{100} = \frac{11}{50}$ radians.

Now, we need to convert this radian measure into degree measure.

The formula to convert radians to degrees is:

Degree measure = Radian measure $\times \frac{180}{\pi}$

Substitute the value of $\theta$ and the given value of $\pi$ into the formula:

Degree measure = $\frac{11}{50} \times \frac{180}{\frac{22}{7}}$

Degree measure = $\frac{11}{50} \times \frac{180 \times 7}{22}$

Simplify the expression by cancelling out common factors:

Degree measure = $\frac{\cancel{11}^{1}}{50} \times \frac{180 \times 7}{\cancel{22}_{2}}$

Degree measure = $\frac{1}{50} \times \frac{180 \times 7}{2}$

Degree measure = $\frac{180 \times 7}{50 \times 2}$

Degree measure = $\frac{\cancel{180}^{90} \times 7}{50 \times \cancel{2}_{1}}$

Degree measure = $\frac{90 \times 7}{50}$

Degree measure = $\frac{\cancel{90}^{9} \times 7}{\cancel{50}_{5}}$

Degree measure = $\frac{9 \times 7}{5} = \frac{63}{5}$ degrees.

To express this in degrees and minutes, we can convert the fractional part:

$\frac{63}{5}^\circ = 12 \frac{3}{5}^\circ$

$12^\circ + \frac{3}{5}^\circ$

Convert $\frac{3}{5}^\circ$ to minutes: $\frac{3}{5} \times 60' = \frac{180}{5}' = 36'$

So, $\frac{63}{5}^\circ = 12^\circ 36'$.

Answer:

The degree measure of the angle subtended at the centre is $12^\circ 36'$.

Given:

Diameter of the circle = $40 \$ cm$

Length of the chord = $20 \$ cm$

To Find:

The length of the minor arc of the chord.

Solution:

The radius of the circle is half of the diameter.

Radius, $r = \frac{\text{Diameter}}{2} = \frac{40}{2} = 20 \$ cm$.

Let the center of the circle be O, and let the chord be AB. The points A and B are on the circle.

The lengths of the radii OA and OB are equal to the radius of the circle.

$OA = r = 20 \$ cm$

$OB = r = 20 \$ cm$

The length of the chord AB is given as $20 \$ cm$.

Consider the triangle OAB formed by the radii OA, OB, and the chord AB.

In $\triangle OAB$, we have:

$OA = 20 \$ cm$

$OB = 20 \$ cm$

$AB = 20 \$ cm$

Since all three sides of $\triangle OAB$ are equal in length, $\triangle OAB$ is an equilateral triangle.

In an equilateral triangle, each angle is $60^\circ$.

The angle subtended by the chord AB at the center O is the angle $\angle AOB$.

Therefore, $\angle AOB = 60^\circ$.

This central angle corresponds to the minor arc AB.

To find the length of the arc, we need to convert the central angle from degrees to radians.

The conversion formula is: Radian measure = Degree measure $\times \frac{\pi}{180}$.

Central angle in radians, $\theta = 60^\circ \times \frac{\pi}{180} = \frac{60\pi}{180} = \frac{\pi}{3}$ radians.

The length of an arc ($s$) is given by the formula $s = r\theta$, where $r$ is the radius and $\theta$ is the central angle in radians.

Substitute the values of $r$ and $\theta$:

$s = 20 \$ cm \times \frac{\pi}{3}$

$s = \frac{20\pi}{3} \$ cm$.

The length of the minor arc of the chord is $\frac{20\pi}{3} \$ cm$. (Unless a specific value of $\pi$ is required, the answer is usually left in terms of $\pi$).

Answer:

The length of the minor arc of the chord is $\frac{20\pi}{3} \$ cm.

Given:

In two circles, the arcs have the same length. Let this length be $s$.

Let the radii of the two circles be $r_1$ and $r_2$.

The angle subtended at the center of the first circle is $\theta_1 = 60^\circ$.

The angle subtended at the center of the second circle is $\theta_2 = 75^\circ$.

To Find:

The ratio of their radii, i.e., $r_1 : r_2$ or $\frac{r_1}{r_2}$.

Solution:

The relationship between the arc length ($s$), radius ($r$), and the central angle ($\theta$) in radians is given by the formula:

$s = r\theta$

where $\theta$ must be in radians.

First, we need to convert the given angles from degrees to radians.

The conversion formula from degrees to radians is: Radian measure = Degree measure $\times \frac{\pi}{180}$.

For the first circle:

$\theta_1 = 60^\circ = 60 \times \frac{\pi}{180}$ radians

$\theta_1 = \frac{\cancel{60}^{1}\pi}{\cancel{180}_{3}} = \frac{\pi}{3}$ radians.

For the second circle:

$\theta_2 = 75^\circ = 75 \times \frac{\pi}{180}$ radians

$\theta_2 = \frac{\cancel{75}^{5}\pi}{\cancel{180}_{12}} = \frac{5\pi}{12}$ radians.

Since the arc lengths are the same in both circles, we have:

$s_1 = s_2 = s$

Using the formula $s = r\theta$ for each circle:

$r_1 \theta_1 = r_2 \theta_2$

Substitute the radian values of the angles:

$r_1 \left(\frac{\pi}{3}\right) = r_2 \left(\frac{5\pi}{12}\right)$

We want to find the ratio $\frac{r_1}{r_2}$. Rearrange the equation:

$\frac{r_1}{r_2} = \frac{r_2 \left(\frac{5\pi}{12}\right)}{r_2 \left(\frac{\pi}{3}\right)}$

$\frac{r_1}{r_2} = \frac{\frac{5\pi}{12}}{\frac{\pi}{3}}$

$\frac{r_1}{r_2} = \frac{5\pi}{12} \times \frac{3}{\pi}$

Cancel out $\pi$ and simplify the numerical part:

$\frac{r_1}{r_2} = \frac{5 \times \cancel{3}^{1}}{\cancel{12}_{4}}$

$\frac{r_1}{r_2} = \frac{5}{4}$

The ratio of the radii is $r_1 : r_2 = 5 : 4$.

Answer:

The ratio of the radii of the two circles is $5 : 4$.

Given:

Length of the pendulum, which acts as the radius ($r$) = 75 cm.

The arc lengths ($l$) described by the tip are:

(i) 10 cm

(ii) 15 cm

(iii) 21 cm

To Find:

The angle ($\theta$) in radians through which the pendulum swings for each given arc length.

Formula Used:

The relationship between arc length ($l$), radius ($r$), and the angle ($\theta$ in radians) subtended by the arc at the center is given by:

$l = r\theta$

Therefore, $\theta = \frac{l}{r}$

Solution:

We know the radius $r = 75$ cm.

(i) When the arc length $l = 10$ cm:

The angle $\theta$ in radians is:

$\theta = \frac{l}{r} = \frac{10}{75}$

Simplifying the fraction:

$\theta = \frac{2 \times 5}{15 \times 5} = \frac{2}{15}$ radians.

Thus, the angle is $\frac{2}{15}$ radians.

(ii) When the arc length $l = 15$ cm:

The angle $\theta$ in radians is:

$\theta = \frac{l}{r} = \frac{15}{75}$

Simplifying the fraction:

$\theta = \frac{1 \times 15}{5 \times 15} = \frac{1}{5}$ radians.

Thus, the angle is $\frac{1}{5}$ radians.

(iii) When the arc length $l = 21$ cm:

The angle $\theta$ in radians is:

$\theta = \frac{l}{r} = \frac{21}{75}$

Simplifying the fraction (dividing numerator and denominator by 3):

$\theta = \frac{7 \times 3}{25 \times 3} = \frac{7}{25}$ radians.

Thus, the angle is $\frac{7}{25}$ radians.

Given:

$\cos x = -\frac{3}{5}$

$x$ lies in the third quadrant.

To Find:

The values of the other five trigonometric functions: $\sin x$, $\tan x$, $\cot x$, $\sec x$, and $\text{cosec } x$.

Solution:

We are given $\cos x = -\frac{3}{5}$.

Since $x$ lies in the third quadrant, we know that $\sin x$ and $\text{cosec } x$ are negative, while $\tan x$ and $\cot x$ are positive. Also $\sec x$ will be negative.

1. Finding $\sec x$:

Using the reciprocal identity:

$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$

2. Finding $\sin x$:

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$\sin^2 x + \left(-\frac{3}{5}\right)^2 = 1$

$\sin^2 x + \frac{9}{25} = 1$

$\sin^2 x = 1 - \frac{9}{25}$

$\sin^2 x = \frac{25 - 9}{25} = \frac{16}{25}$

$\sin x = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$

Since $x$ lies in the third quadrant, $\sin x$ is negative.

Therefore, $\sin x = -\frac{4}{5}$.

3. Finding $\text{cosec } x$:

Using the reciprocal identity:

$\text{cosec } x = \frac{1}{\sin x} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

4. Finding $\tan x$:

Using the quotient identity:

$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{4}{5}}{-\frac{3}{5}}$

$\tan x = \left(-\frac{4}{5}\right) \times \left(-\frac{5}{3}\right) = \frac{4}{3}$

(Note: $\tan x$ is positive in the third quadrant, which matches our result).

5. Finding $\cot x$:

Using the reciprocal identity:

$\cot x = \frac{1}{\tan x} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$

(Alternatively, $\cot x = \frac{\cos x}{\sin x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$)

(Note: $\cot x$ is positive in the third quadrant, which matches our result).

Summary of Values:

The five trigonometric functions are:

$\sin x = -\frac{4}{5}$

$\sec x = -\frac{5}{3}$

$\text{cosec } x = -\frac{5}{4}$

$\tan x = \frac{4}{3}$

$\cot x = \frac{3}{4}$

Given:

$\cot x = -\frac{5}{12}$

$x$ lies in the second quadrant.

To Find:

The values of the other five trigonometric functions: $\sin x$, $\cos x$, $\tan x$, $\sec x$, and $\text{cosec } x$.

Solution:

We are given $\cot x = -\frac{5}{12}$.

Since $x$ lies in the second quadrant, we know that:

$\sin x$ and $\text{cosec } x$ are positive.

$\cos x$, $\sec x$, and $\tan x$ are negative.

(The given $\cot x$ is negative, which is consistent with the second quadrant).

1. Finding $\tan x$:

Using the reciprocal identity:

$\tan x = \frac{1}{\cot x} = \frac{1}{-\frac{5}{12}} = -\frac{12}{5}$

(This is negative, as expected for the second quadrant).

2. Finding $\text{cosec } x$:

Using the Pythagorean identity $1 + \cot^2 x = \text{cosec}^2 x$:

$\text{cosec}^2 x = 1 + \left(-\frac{5}{12}\right)^2$

$\text{cosec}^2 x = 1 + \frac{25}{144}$

$\text{cosec}^2 x = \frac{144}{144} + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$

$\text{cosec } x = \pm \sqrt{\frac{169}{144}} = \pm \frac{13}{12}$

Since $x$ lies in the second quadrant, $\text{cosec } x$ is positive.

Therefore, $\text{cosec } x = \frac{13}{12}$.

3. Finding $\sin x$:

Using the reciprocal identity:

$\sin x = \frac{1}{\text{cosec } x} = \frac{1}{\frac{13}{12}} = \frac{12}{13}$

(This is positive, as expected for the second quadrant).

4. Finding $\cos x$:

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$\left(\frac{12}{13}\right)^2 + \cos^2 x = 1$

$\frac{144}{169} + \cos^2 x = 1$

$\cos^2 x = 1 - \frac{144}{169}$

$\cos^2 x = \frac{169 - 144}{169} = \frac{25}{169}$

$\cos x = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$

Since $x$ lies in the second quadrant, $\cos x$ is negative.

Therefore, $\cos x = -\frac{5}{13}$.

(Alternatively, we could use $\cot x = \frac{\cos x}{\sin x}$. $\cos x = \cot x \times \sin x = (-\frac{5}{12}) \times (\frac{12}{13}) = -\frac{5}{13}$)

5. Finding $\sec x$:

Using the reciprocal identity:

$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{5}{13}} = -\frac{13}{5}$

(This is negative, as expected for the second quadrant).

Summary of Values:

The other five trigonometric functions are:

$\sin x = \frac{12}{13}$

$\cos x = -\frac{5}{13}$

$\tan x = -\frac{12}{5}$

$\sec x = -\frac{13}{5}$

$\text{cosec } x = \frac{13}{12}$

Given:

The trigonometric expression $\sin \frac{31\pi}{3}$.

To Find:

The value of $\sin \frac{31\pi}{3}$.

Concept Used:

The sine function is periodic with a period of $2\pi$. This means that for any integer $n$:

$\sin(x + 2n\pi) = \sin x$

We need to express the angle $\frac{31\pi}{3}$ in the form $x + 2n\pi$, where $x$ is an angle whose sine value is known, typically between $0$ and $2\pi$.

Solution:

First, let's simplify the angle $\frac{31\pi}{3}$. We can express $\frac{31}{3}$ as a mixed number or find the largest multiple of 3 less than 31.

$\frac{31}{3} = \frac{30 + 1}{3} = \frac{30}{3} + \frac{1}{3} = 10 + \frac{1}{3}$

So, $\frac{31\pi}{3} = \left(10 + \frac{1}{3}\right)\pi = 10\pi + \frac{\pi}{3}$.

Now we can write:

$\sin \frac{31\pi}{3} = \sin \left(10\pi + \frac{\pi}{3}\right)$

Since $10\pi = 5 \times 2\pi$, it represents 5 full rotations. Using the periodicity of the sine function, where $n=5$:

$\sin \left(10\pi + \frac{\pi}{3}\right) = \sin \left(5 \times 2\pi + \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{3}\right)$

We know the standard value for $\sin \left(\frac{\pi}{3}\right)$ (which corresponds to $\sin 60^\circ$):

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Therefore, the value of $\sin \frac{31\pi}{3}$ is $\frac{\sqrt{3}}{2}$.

Final Answer: $\sin \frac{31\pi}{3} = \frac{\sqrt{3}}{2}$

Given:

The trigonometric expression $\cos (–1710°)$.

To Find:

The value of $\cos (–1710°)$.

Concepts Used:

1. Even Function Property of Cosine: The cosine function is an even function, which means $\cos(-\theta) = \cos(\theta)$ for any angle $\theta$.

2. Periodicity of Cosine: The cosine function is periodic with a period of $360^\circ$ (or $2\pi$ radians). This means $\cos(\theta + n \cdot 360^\circ) = \cos(\theta)$ for any integer $n$.

Solution:

We need to find the value of $\cos (–1710°)$.

Step 1: Apply the even function property.

$\cos (–1710°) = \cos (1710°)$

Step 2: Use the periodicity of the cosine function to reduce the angle $1710^\circ$. We need to find an angle between $0^\circ$ and $360^\circ$ that is coterminal with $1710^\circ$. We can do this by finding the remainder when $1710$ is divided by $360$.

$1710 = n \times 360 + \theta$, where $0^\circ \le \theta < 360^\circ$.

Let's find the multiple of 360 closest to 1710:

$360 \times 1 = 360$

$360 \times 2 = 720$

$360 \times 3 = 1080$

$360 \times 4 = 1440$

$360 \times 5 = 1800$

We can write $1710^\circ$ as $1440^\circ + 270^\circ$.

So, $1710^\circ = 4 \times 360^\circ + 270^\circ$.

Using the periodicity property:

$\cos(1710^\circ) = \cos(4 \times 360^\circ + 270^\circ) = \cos(270^\circ)$

Step 3: Evaluate $\cos(270^\circ)$.

We know that $\cos(270^\circ) = 0$.

Therefore, $\cos (–1710°) = 0$.

Alternate Method:

We can also write $1710^\circ$ using the next multiple of $360^\circ$:

$1710^\circ = 1800^\circ - 90^\circ = 5 \times 360^\circ - 90^\circ$.

So, $\cos(1710^\circ) = \cos(5 \times 360^\circ - 90^\circ)$

Using the periodicity property:

$\cos(5 \times 360^\circ - 90^\circ) = \cos(-90^\circ)$

Using the even function property $\cos(-\theta) = \cos(\theta)$:

$\cos(-90^\circ) = \cos(90^\circ)$

We know that $\cos(90^\circ) = 0$.

Thus, $\cos (–1710°) = 0$.

Final Answer: The value of $\cos (–1710°)$ is 0.

Given:

$\cos x = -\frac{1}{2}$

$x$ lies in the third quadrant.

To Find:

The values of the other five trigonometric functions: $\sin x$, $\tan x$, $\cot x$, $\sec x$, and $\text{cosec } x$.

Solution:

We are given $\cos x = -\frac{1}{2}$.

Since $x$ lies in the third quadrant, we know that:

$\sin x$ and $\text{cosec } x$ are negative.

$\tan x$ and $\cot x$ are positive.

$\cos x$ and $\sec x$ are negative. (The given $\cos x$ is negative, consistent with the third quadrant).

1. Finding $\sec x$:

Using the reciprocal identity:

$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{1}{2}} = -2$

(This is negative, as expected for the third quadrant).

2. Finding $\sin x$:

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$\sin^2 x + \left(-\frac{1}{2}\right)^2 = 1$

$\sin^2 x + \frac{1}{4} = 1$

$\sin^2 x = 1 - \frac{1}{4}$

$\sin^2 x = \frac{4 - 1}{4} = \frac{3}{4}$

$\sin x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$

Since $x$ lies in the third quadrant, $\sin x$ is negative.

Therefore, $\sin x = -\frac{\sqrt{3}}{2}$.

3. Finding $\text{cosec } x$:

Using the reciprocal identity:

$\text{cosec } x = \frac{1}{\sin x} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$

Rationalizing the denominator:

$\text{cosec } x = -\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$

(This is negative, as expected for the third quadrant).

4. Finding $\tan x$:

Using the quotient identity:

$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$

$\tan x = \left(-\frac{\sqrt{3}}{2}\right) \times \left(-\frac{2}{1}\right) = \sqrt{3}$

(This is positive, as expected for the third quadrant).

5. Finding $\cot x$:

Using the reciprocal identity:

$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$

Rationalizing the denominator:

$\cot x = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

(Alternatively, $\cot x = \frac{\cos x}{\sin x} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$)

(This is positive, as expected for the third quadrant).

Summary of Values:

The other five trigonometric functions are:

$\sin x = -\frac{\sqrt{3}}{2}$

$\tan x = \sqrt{3}$

$\cot x = \frac{\sqrt{3}}{3}$

$\sec x = -2$

$\text{cosec } x = -\frac{2\sqrt{3}}{3}$

Given:

$\sin x = \frac{3}{5}$

$x$ lies in the second quadrant.

To Find:

The values of the other five trigonometric functions: $\cos x$, $\tan x$, $\cot x$, $\sec x$, and $\text{cosec } x$.

Solution:

We are given $\sin x = \frac{3}{5}$.

Since $x$ lies in the second quadrant, we know that:

$\sin x$ and $\text{cosec } x$ are positive. (The given $\sin x$ is positive, consistent with the second quadrant).

$\cos x$, $\sec x$, $\tan x$, and $\cot x$ are negative.

1. Finding $\text{cosec } x$:

Using the reciprocal identity:

$\text{cosec } x = \frac{1}{\sin x} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$

(This is positive, as expected for the second quadrant).

2. Finding $\cos x$:

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$\left(\frac{3}{5}\right)^2 + \cos^2 x = 1$

$\frac{9}{25} + \cos^2 x = 1$

$\cos^2 x = 1 - \frac{9}{25}$

$\cos^2 x = \frac{25 - 9}{25} = \frac{16}{25}$

$\cos x = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$

Since $x$ lies in the second quadrant, $\cos x$ is negative.

Therefore, $\cos x = -\frac{4}{5}$.

3. Finding $\sec x$:

Using the reciprocal identity:

$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

(This is negative, as expected for the second quadrant).

4. Finding $\tan x$:

Using the quotient identity:

$\tan x = \frac{\sin x}{\cos x} = \frac{\frac{3}{5}}{-\frac{4}{5}}$

$\tan x = \left(\frac{3}{5}\right) \times \left(-\frac{5}{4}\right) = -\frac{3}{4}$

(This is negative, as expected for the second quadrant).

5. Finding $\cot x$:

Using the reciprocal identity:

$\cot x = \frac{1}{\tan x} = \frac{1}{-\frac{3}{4}} = -\frac{4}{3}$

(Alternatively, $\cot x = \frac{\cos x}{\sin x} = \frac{-\frac{4}{5}}{\frac{3}{5}} = -\frac{4}{3}$)

(This is negative, as expected for the second quadrant).

Summary of Values:

The other five trigonometric functions are:

$\cos x = -\frac{4}{5}$

$\tan x = -\frac{3}{4}$

$\cot x = -\frac{4}{3}$

$\sec x = -\frac{5}{4}$

$\text{cosec } x = \frac{5}{3}$

Given:

$\cot x = \frac{3}{4}$

$x$ lies in the third quadrant.

To Find:

The values of the other five trigonometric functions: $\sin x$, $\cos x$, $\tan x$, $\sec x$, and $\text{cosec } x$.

Solution:

We are given $\cot x = \frac{3}{4}$.

Since $x$ lies in the third quadrant, we know that:

$\tan x$ and $\cot x$ are positive. (The given $\cot x$ is positive, consistent with the third quadrant).

$\sin x$, $\text{cosec } x$, $\cos x$, and $\sec x$ are negative.

1. Finding $\tan x$:

Using the reciprocal identity:

$\tan x = \frac{1}{\cot x} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

(This is positive, as expected for the third quadrant).

2. Finding $\text{cosec } x$:

Using the Pythagorean identity $1 + \cot^2 x = \text{cosec}^2 x$:

$\text{cosec}^2 x = 1 + \left(\frac{3}{4}\right)^2$

$\text{cosec}^2 x = 1 + \frac{9}{16}$

$\text{cosec}^2 x = \frac{16}{16} + \frac{9}{16} = \frac{16 + 9}{16} = \frac{25}{16}$

$\text{cosec } x = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}$

Since $x$ lies in the third quadrant, $\text{cosec } x$ is negative.

Therefore, $\text{cosec } x = -\frac{5}{4}$.

3. Finding $\sin x$:

Using the reciprocal identity:

$\sin x = \frac{1}{\text{cosec } x} = \frac{1}{-\frac{5}{4}} = -\frac{4}{5}$

(This is negative, as expected for the third quadrant).

4. Finding $\sec x$:

Using the Pythagorean identity $1 + \tan^2 x = \sec^2 x$:

$\sec^2 x = 1 + \left(\frac{4}{3}\right)^2$

$\sec^2 x = 1 + \frac{16}{9}$

$\sec^2 x = \frac{9}{9} + \frac{16}{9} = \frac{9 + 16}{9} = \frac{25}{9}$

$\sec x = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3}$

Since $x$ lies in the third quadrant, $\sec x$ is negative.

Therefore, $\sec x = -\frac{5}{3}$.

5. Finding $\cos x$:

Using the reciprocal identity:

$\cos x = \frac{1}{\sec x} = \frac{1}{-\frac{5}{3}} = -\frac{3}{5}$

(Alternatively, $\cos x = \cot x \times \sin x = \frac{3}{4} \times \left(-\frac{4}{5}\right) = -\frac{3}{5}$)

(This is negative, as expected for the third quadrant).

Summary of Values:

The other five trigonometric functions are:

$\sin x = -\frac{4}{5}$

$\cos x = -\frac{3}{5}$

$\tan x = \frac{4}{3}$

$\sec x = -\frac{5}{3}$

$\text{cosec } x = -\frac{5}{4}$

Given:

$\sec x = \frac{13}{5}$

$x$ lies in the fourth quadrant.

To Find:

The values of the other five trigonometric functions: $\sin x$, $\cos x$, $\tan x$, $\cot x$, and $\text{cosec } x$.

Solution:

We are given $\sec x = \frac{13}{5}$.

Since $x$ lies in the fourth quadrant, we know that:

$\cos x$ and $\sec x$ are positive. (The given $\sec x$ is positive, consistent with the fourth quadrant).

$\sin x$, $\text{cosec } x$, $\tan x$, and $\cot x$ are negative.

1. Finding $\cos x$:

Using the reciprocal identity:

$\cos x = \frac{1}{\sec x} = \frac{1}{\frac{13}{5}} = \frac{5}{13}$

(This is positive, as expected for the fourth quadrant).

2. Finding $\sin x$:

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$\sin^2 x + \left(\frac{5}{13}\right)^2 = 1$

$\sin^2 x + \frac{25}{169} = 1$

$\sin^2 x = 1 - \frac{25}{169}$

$\sin^2 x = \frac{169 - 25}{169} = \frac{144}{169}$

$\sin x = \pm \sqrt{\frac{144}{169}} = \pm \frac{12}{13}$

Since $x$ lies in the fourth quadrant, $\sin x$ is negative.

Therefore, $\sin x = -\frac{12}{13}$.

3. Finding $\text{cosec } x$:

Using the reciprocal identity:

$\text{cosec } x = \frac{1}{\sin x} = \frac{1}{-\frac{12}{13}} = -\frac{13}{12}$

(This is negative, as expected for the fourth quadrant).

4. Finding $\tan x$:

Using the Pythagorean identity $1 + \tan^2 x = \sec^2 x$:

$1 + \tan^2 x = \left(\frac{13}{5}\right)^2$

$1 + \tan^2 x = \frac{169}{25}$

$\tan^2 x = \frac{169}{25} - 1$

$\tan^2 x = \frac{169 - 25}{25} = \frac{144}{25}$

$\tan x = \pm \sqrt{\frac{144}{25}} = \pm \frac{12}{5}$

Since $x$ lies in the fourth quadrant, $\tan x$ is negative.

Therefore, $\tan x = -\frac{12}{5}$.

(Alternatively, $\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$)

5. Finding $\cot x$:

Using the reciprocal identity:

$\cot x = \frac{1}{\tan x} = \frac{1}{-\frac{12}{5}} = -\frac{5}{12}$

(This is negative, as expected for the fourth quadrant).

Summary of Values:

The other five trigonometric functions are:

$\sin x = -\frac{12}{13}$

$\cos x = \frac{5}{13}$

$\tan x = -\frac{12}{5}$

$\cot x = -\frac{5}{12}$

$\text{cosec } x = -\frac{13}{12}$

Given:

$\tan x = -\frac{5}{12}$

$x$ lies in the second quadrant.

To Find:

The values of the other five trigonometric functions: $\sin x$, $\cos x$, $\cot x$, $\sec x$, and $\text{cosec } x$.

Solution:

We are given $\tan x = -\frac{5}{12}$.

Since $x$ lies in the second quadrant, we know that:

$\sin x$ and $\text{cosec } x$ are positive.

$\cos x$, $\sec x$, $\tan x$, and $\cot x$ are negative. (The given $\tan x$ is negative, consistent with the second quadrant).

1. Finding $\cot x$:

Using the reciprocal identity:

$\cot x = \frac{1}{\tan x} = \frac{1}{-\frac{5}{12}} = -\frac{12}{5}$

(This is negative, as expected for the second quadrant).

2. Finding $\sec x$:

Using the Pythagorean identity $1 + \tan^2 x = \sec^2 x$:

$\sec^2 x = 1 + \left(-\frac{5}{12}\right)^2$

$\sec^2 x = 1 + \frac{25}{144}$

$\sec^2 x = \frac{144}{144} + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$

$\sec x = \pm \sqrt{\frac{169}{144}} = \pm \frac{13}{12}$

Since $x$ lies in the second quadrant, $\sec x$ is negative.

Therefore, $\sec x = -\frac{13}{12}$.

3. Finding $\cos x$:

Using the reciprocal identity:

$\cos x = \frac{1}{\sec x} = \frac{1}{-\frac{13}{12}} = -\frac{12}{13}$

(This is negative, as expected for the second quadrant).

4. Finding $\sin x$:

Using the quotient identity $\tan x = \frac{\sin x}{\cos x}$:

$\sin x = \tan x \times \cos x$

$\sin x = \left(-\frac{5}{12}\right) \times \left(-\frac{12}{13}\right)$

$\sin x = \frac{5}{13}$

(Alternatively, using $\sin^2 x + \cos^2 x = 1$:

$\sin^2 x + \left(-\frac{12}{13}\right)^2 = 1$

$\sin^2 x + \frac{144}{169} = 1$

$\sin^2 x = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$

$\sin x = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$. Since x is in the second quadrant, $\sin x = \frac{5}{13}$)

(This is positive, as expected for the second quadrant).

5. Finding $\text{cosec } x$:

Using the reciprocal identity:

$\text{cosec } x = \frac{1}{\sin x} = \frac{1}{\frac{5}{13}} = \frac{13}{5}$

(This is positive, as expected for the second quadrant).

Summary of Values:

The other five trigonometric functions are:

$\sin x = \frac{5}{13}$

$\cos x = -\frac{12}{13}$

$\cot x = -\frac{12}{5}$

$\sec x = -\frac{13}{12}$

$\text{cosec } x = \frac{13}{5}$

Given:

The trigonometric expression $\sin 765°$.

To Find:

The value of $\sin 765°$.

Concept Used:

Periodicity of Sine: The sine function is periodic with a period of $360^\circ$. This means $\sin(\theta + n \cdot 360^\circ) = \sin(\theta)$ for any integer $n$.

We need to reduce the angle $765^\circ$ to an equivalent angle within the range $[0^\circ, 360^\circ)$.

Solution:

We need to find the value of $\sin 765°$.

To simplify the angle $765°$, we can find how many full rotations ($360^\circ$) are contained within it.

We can divide $765$ by $360$:

$765 = 2 \times 360 + 45$

So, $765^\circ = 2 \times 360^\circ + 45^\circ$.

Using the periodicity property of the sine function:

$\sin(765^\circ) = \sin(2 \times 360^\circ + 45^\circ)$

Since adding or subtracting multiples of $360^\circ$ does not change the value of the sine function, we have:

$\sin(765^\circ) = \sin(45^\circ)$

We know the standard value for $\sin(45^\circ)$:

$\sin(45^\circ) = \frac{1}{\sqrt{2}}$

Rationalizing the denominator:

$\sin(45^\circ) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Therefore, the value of $\sin 765°$ is $\frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$.

Final Answer: $\sin 765° = \frac{1}{\sqrt{2}}$

Given:

The trigonometric expression $\text{cosec} (–1410°)$.

To Find:

The value of $\text{cosec} (–1410°)$.

Concepts Used:

1. Odd Function Property of Cosecant: The cosecant function is an odd function, which means $\text{cosec}(-\theta) = -\text{cosec}(\theta)$ for any angle $\theta$.

2. Periodicity of Cosecant: The cosecant function is periodic with a period of $360^\circ$. This means $\text{cosec}(\theta + n \cdot 360^\circ) = \text{cosec}(\theta)$ for any integer $n$.

3. Relationship with Sine: $\text{cosec}(\theta) = \frac{1}{\sin(\theta)}$.

4. Standard Angle Values: Knowledge of sine/cosecant values for standard angles like $30^\circ$.

Solution:

We need to find the value of $\text{cosec} (–1410°)$.

Step 1: Apply the odd function property.

$\text{cosec} (–1410°) = -\text{cosec} (1410°)$

Step 2: Use the periodicity of the cosecant function to reduce the angle $1410^\circ$. We need to find an angle between $0^\circ$ and $360^\circ$ that is coterminal with $1410^\circ$. We find the remainder when $1410$ is divided by $360$.

$1410 = n \times 360 + \theta$, where $0^\circ \le \theta < 360^\circ$.

Dividing $1410$ by $360$:

$360 \times 3 = 1080$

$1410 - 1080 = 330$

So, $1410^\circ = 3 \times 360^\circ + 330^\circ$.

Using the periodicity property:

$\text{cosec}(1410^\circ) = \text{cosec}(3 \times 360^\circ + 330^\circ) = \text{cosec}(330^\circ)$

Step 3: Evaluate $\text{cosec}(330^\circ)$.

We can write $330^\circ$ as $360^\circ - 30^\circ$.

$\text{cosec}(330^\circ) = \text{cosec}(360^\circ - 30^\circ)$

Since $330^\circ$ lies in the fourth quadrant, where cosecant (and sine) is negative:

$\text{cosec}(360^\circ - 30^\circ) = -\text{cosec}(30^\circ)$

We know that $\sin(30^\circ) = \frac{1}{2}$.

So, $\text{cosec}(30^\circ) = \frac{1}{\sin(30^\circ)} = \frac{1}{1/2} = 2$.

Therefore, $\text{cosec}(330^\circ) = -2$.

Step 4: Combine the results from Step 1 and Step 3.

From Step 1, $\text{cosec} (–1410°) = -\text{cosec} (1410°)$.

From Step 2 and 3, $\text{cosec}(1410^\circ) = \text{cosec}(330^\circ) = -2$.

Substituting this back:

$\text{cosec} (–1410°) = -(-2) = 2$.

Alternate Method:

Step 1: Apply the odd function property.

$\text{cosec} (–1410°) = -\text{cosec} (1410°)$

Step 2: Use a different multiple of $360^\circ$.

$360 \times 4 = 1440$

We can write $1410^\circ = 4 \times 360^\circ - 30^\circ$.

Using the periodicity property:

$\text{cosec}(1410^\circ) = \text{cosec}(4 \times 360^\circ - 30^\circ) = \text{cosec}(-30^\circ)$

Step 3: Apply the odd function property again.

$\text{cosec}(-30^\circ) = -\text{cosec}(30^\circ)$

We know $\text{cosec}(30^\circ) = 2$.

So, $\text{cosec}(-30^\circ) = -2$.

Thus, $\text{cosec}(1410^\circ) = -2$.

Step 4: Combine with Step 1 result.

$\text{cosec} (–1410°) = -\text{cosec} (1410°) = -(-2) = 2$.

Final Answer: The value of $\text{cosec} (–1410°)$ is 2.

Given:

The trigonometric expression $\tan \frac{19\pi}{3}$.

To Find:

The value of $\tan \frac{19\pi}{3}$.

Concept Used:

Periodicity of Tangent: The tangent function is periodic with a period of $\pi$. This means $\tan(x + n\pi) = \tan x$ for any integer $n$.

We need to express the angle $\frac{19\pi}{3}$ in the form $x + n\pi$, where $x$ is an angle whose tangent value is known, typically between $0$ and $\pi$.

Solution:

We need to find the value of $\tan \frac{19\pi}{3}$.

First, let's simplify the angle $\frac{19\pi}{3}$. We can express $\frac{19}{3}$ as a mixed number or find the largest multiple of 3 less than or equal to 19.

Divide 19 by 3:

$19 \div 3 = 6$ with a remainder of $1$.

So, $\frac{19}{3} = 6 + \frac{1}{3}$.

Therefore, $\frac{19\pi}{3} = \left(6 + \frac{1}{3}\right)\pi = 6\pi + \frac{\pi}{3}$.

Now we can write:

$\tan \frac{19\pi}{3} = \tan \left(6\pi + \frac{\pi}{3}\right)$

Since $6\pi$ is an integer multiple of the period $\pi$ (here $n=6$), we can use the periodicity property of the tangent function:

$\tan \left(6\pi + \frac{\pi}{3}\right) = \tan \left(\frac{\pi}{3}\right)$

We know the standard value for $\tan \left(\frac{\pi}{3}\right)$ (which corresponds to $\tan 60^\circ$):

$\tan \left(\frac{\pi}{3}\right) = \sqrt{3}$

Therefore, the value of $\tan \frac{19\pi}{3}$ is $\sqrt{3}$.

Final Answer: $\tan \frac{19\pi}{3} = \sqrt{3}$

Given:

The trigonometric expression $\sin (-\frac{11\pi}{3})$.

To Find:

The value of $\sin (-\frac{11\pi}{3})$.

Concepts Used:

1. Periodicity of Sine: The sine function is periodic with a period of $2\pi$. This means $\sin(x + 2n\pi) = \sin(x)$ for any integer $n$.

2. Odd Function Property of Sine: The sine function is an odd function, which means $\sin(-x) = -\sin(x)$ for any angle $x$. (This can be used as an alternative approach).

3. Standard Angle Values: Knowledge of sine values for standard angles like $\frac{\pi}{3}$.

Solution:

We need to find the value of $\sin (-\frac{11\pi}{3})$.

We can use the periodicity of the sine function, $\sin(x + 2n\pi) = \sin(x)$, to find an equivalent angle in the range $[0, 2\pi)$.

The period is $2\pi$. In terms of thirds, $2\pi = \frac{6\pi}{3}$.

We want to add a multiple of $2\pi$ (or $\frac{6\pi}{3}$) to $-\frac{11\pi}{3}$ to get an angle in the desired range.

Let's add $2 \times 2\pi = 4\pi = \frac{12\pi}{3}$.

$\sin \left(-\frac{11\pi}{3}\right) = \sin \left(-\frac{11\pi}{3} + 4\pi \right)$

$\sin \left(-\frac{11\pi}{3}\right) = \sin \left(-\frac{11\pi}{3} + \frac{12\pi}{3} \right)$

$\sin \left(-\frac{11\pi}{3}\right) = \sin \left(\frac{-11\pi + 12\pi}{3} \right)$

$\sin \left(-\frac{11\pi}{3}\right) = \sin \left(\frac{\pi}{3} \right)$

We know the standard value for $\sin \left(\frac{\pi}{3}\right)$ (which corresponds to $\sin 60^\circ$):

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Alternate Solution:

Step 1: Use the odd function property $\sin(-x) = -\sin(x)$.

$\sin \left(-\frac{11\pi}{3}\right) = -\sin \left(\frac{11\pi}{3}\right)$

Step 2: Simplify the angle $\frac{11\pi}{3}$ using periodicity.

We can write $\frac{11\pi}{3}$ as $\frac{12\pi - \pi}{3} = \frac{12\pi}{3} - \frac{\pi}{3} = 4\pi - \frac{\pi}{3}$.

So, $-\sin \left(\frac{11\pi}{3}\right) = -\sin \left(4\pi - \frac{\pi}{3}\right)$.

Since $4\pi = 2 \times 2\pi$, we use the periodicity $\sin(x + 2n\pi) = \sin(x)$.

$-\sin \left(2 \times 2\pi - \frac{\pi}{3}\right) = -\sin \left(-\frac{\pi}{3}\right)$

Step 3: Apply the odd function property again.

$-\sin \left(-\frac{\pi}{3}\right) = - \left(-\sin \left(\frac{\pi}{3}\right) \right) = \sin \left(\frac{\pi}{3}\right)$

Step 4: Evaluate the standard value.

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Both methods yield the same result.

Final Answer: The value of $\sin (-\frac{11\pi}{3})$ is $\frac{\sqrt{3}}{2}$.

Given:

The trigonometric expression $\cot (-\frac{15\pi}{4})$.

To Find:

The value of $\cot (-\frac{15\pi}{4})$.

Concepts Used:

1. Periodicity of Cotangent: The cotangent function is periodic with a period of $\pi$. This means $\cot(x + n\pi) = \cot(x)$ for any integer $n$.

2. Odd Function Property of Cotangent: The cotangent function is an odd function, which means $\cot(-x) = -\cot(x)$ for any angle $x$. (This can be used as an alternative approach).

3. Standard Angle Values: Knowledge of cotangent values for standard angles like $\frac{\pi}{4}$.

Solution:

We need to find the value of $\cot (-\frac{15\pi}{4})$.

We can use the periodicity of the cotangent function, $\cot(x + n\pi) = \cot(x)$, to find an equivalent angle, preferably in the range $(0, \pi)$.

The period is $\pi$. In terms of fourths, $\pi = \frac{4\pi}{4}$.

We want to add a multiple of $\pi$ (or $\frac{4\pi}{4}$) to $-\frac{15\pi}{4}$ to bring the angle into a more familiar range.

Let's add $4\pi = \frac{16\pi}{4}$ (which is $4 \times \pi$, so $n=4$).

$\cot \left(-\frac{15\pi}{4}\right) = \cot \left(-\frac{15\pi}{4} + 4\pi \right)$

$\cot \left(-\frac{15\pi}{4}\right) = \cot \left(-\frac{15\pi}{4} + \frac{16\pi}{4} \right)$

$\cot \left(-\frac{15\pi}{4}\right) = \cot \left(\frac{-15\pi + 16\pi}{4} \right)$

$\cot \left(-\frac{15\pi}{4}\right) = \cot \left(\frac{\pi}{4} \right)$

We know the standard value for $\cot \left(\frac{\pi}{4}\right)$ (which corresponds to $\cot 45^\circ$):

$\cot \left(\frac{\pi}{4}\right) = 1$

Alternate Solution:

Step 1: Use the odd function property $\cot(-x) = -\cot(x)$.

$\cot \left(-\frac{15\pi}{4}\right) = -\cot \left(\frac{15\pi}{4}\right)$

Step 2: Simplify the angle $\frac{15\pi}{4}$ using periodicity.

We can write $\frac{15\pi}{4}$ as $\frac{16\pi - \pi}{4} = \frac{16\pi}{4} - \frac{\pi}{4} = 4\pi - \frac{\pi}{4}$.

So, $-\cot \left(\frac{15\pi}{4}\right) = -\cot \left(4\pi - \frac{\pi}{4}\right)$.

Since $4\pi$ is an integer multiple of the period $\pi$ ($n=4$), we use the periodicity $\cot(x + n\pi) = \cot(x)$.

$-\cot \left(4\pi - \frac{\pi}{4}\right) = -\cot \left(-\frac{\pi}{4}\right)$

Step 3: Apply the odd function property again.

$-\cot \left(-\frac{\pi}{4}\right) = - \left(-\cot \left(\frac{\pi}{4}\right) \right) = \cot \left(\frac{\pi}{4}\right)$

Step 4: Evaluate the standard value.

$\cot \left(\frac{\pi}{4}\right) = 1$

Both methods yield the same result.

Final Answer: The value of $\cot (-\frac{15\pi}{4})$ is 1.

To Prove:

$3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}=1$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= 3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}$

We know the values of the trigonometric ratios for standard angles:

$\sin\frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$

$\sec\frac{\pi}{3} = \sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2$

$\sin\frac{5\pi}{6} = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$

$\cot\frac{\pi}{4} = \cot 45^\circ = 1$

Substitute these values into the LHS expression:

LHS $= 3 \times \left(\frac{1}{2}\right) \times (2) - 4 \times \left(\frac{1}{2}\right) \times (1)$

LHS $= 3 \times 1 - 4 \times \frac{1}{2}$

LHS $= 3 - 2$

LHS $= 1$

The Right Hand Side (RHS) of the equation is given as $1$.

Since LHS $=$ RHS, the equation is proven.

To Find:

The value of $\sin 15^\circ$.

Solution:

We can express $15^\circ$ as the difference of two standard angles for which the trigonometric values are known. For example, we can use $45^\circ$ and $30^\circ$ because $45^\circ - 30^\circ = 15^\circ$.

We use the trigonometric identity for the sine of the difference of two angles:

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

Let $A = 45^\circ$ and $B = 30^\circ$. Substituting these values into the identity:

$\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

We know the values of the trigonometric ratios for $45^\circ$ and $30^\circ$:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

Substitute these values into the equation:

$\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{2}\right)$

$\sin 15^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} - \frac{1 \times 1}{\sqrt{2} \times 2}$

$\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$

Combine the fractions:

$\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\sin 15^\circ = \frac{(\sqrt{3} - 1) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}}$

$\sin 15^\circ = \frac{\sqrt{3} \times \sqrt{2} - 1 \times \sqrt{2}}{2 \times (\sqrt{2})^2}$

$\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{2 \times 2}$

$\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$

Alternate Solution:

We can also express $15^\circ$ as the difference $60^\circ - 45^\circ$.

$15^\circ = 60^\circ - 45^\circ$

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ with $A = 60^\circ$ and $B = 45^\circ$:

$\sin(60^\circ - 45^\circ) = \sin 60^\circ \cos 45^\circ - \cos 60^\circ \sin 45^\circ$

Substitute the known values:

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 60^\circ = \frac{1}{2}$

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

So,

$\sin 15^\circ = \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{2}\right) \times \left(\frac{1}{\sqrt{2}}\right)$

$\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$

$\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

Rationalize the denominator:

$\sin 15^\circ = \frac{(\sqrt{3} - 1) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Both methods give the same result.

To Find:

The value of $\tan \frac{13\pi}{12}$.

Solution:

We can express $\frac{13\pi}{12}$ as the sum of two standard angles. One way to do this is to write:

$\frac{13\pi}{12} = \frac{9\pi + 4\pi}{12} = \frac{9\pi}{12} + \frac{4\pi}{12} = \frac{3\pi}{4} + \frac{\pi}{3}$

Alternatively, we can write:

$\frac{13\pi}{12} = \frac{10\pi + 3\pi}{12} = \frac{10\pi}{12} + \frac{3\pi}{12} = \frac{5\pi}{6} + \frac{\pi}{4}$

Let's use the latter approach: $\frac{13\pi}{12} = \frac{5\pi}{6} + \frac{\pi}{4}$.

We use the trigonometric identity for the tangent of the sum of two angles:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

Let $A = \frac{5\pi}{6}$ and $B = \frac{\pi}{4}$. We need the tangent values for these angles:

$\tan A = \tan \frac{5\pi}{6}$

Since $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$, and tangent is negative in the second quadrant:

$\tan \frac{5\pi}{6} = \tan(\pi - \frac{\pi}{6}) = -\tan \frac{\pi}{6} = -\frac{1}{\sqrt{3}}$

$\tan B = \tan \frac{\pi}{4}$

$\tan \frac{\pi}{4} = 1$

Substitute these values into the tangent addition formula:

$\tan\left(\frac{5\pi}{6} + \frac{\pi}{4}\right) = \frac{\tan \frac{5\pi}{6} + \tan \frac{\pi}{4}}{1 - \tan \frac{5\pi}{6} \tan \frac{\pi}{4}}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{-\frac{1}{\sqrt{3}} + 1}{1 - \left(-\frac{1}{\sqrt{3}}\right) \times (1)}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$

To simplify the complex fraction, multiply the numerator and the denominator by $\sqrt{3}$:

$\tan\left(\frac{13\pi}{12}\right) = \frac{\left(1 - \frac{1}{\sqrt{3}}\right) \times \sqrt{3}}{\left(1 + \frac{1}{\sqrt{3}}\right) \times \sqrt{3}}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{1 \times \sqrt{3} - \frac{1}{\sqrt{3}} \times \sqrt{3}}{1 \times \sqrt{3} + \frac{1}{\sqrt{3}} \times \sqrt{3}}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Now, rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$:

$\tan\left(\frac{13\pi}{12}\right) = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} \times \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2}$

Expand the numerator using $(a-b)^2 = a^2 - 2ab + b^2$ and simplify the denominator:

$\tan\left(\frac{13\pi}{12}\right) = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{3 - 2\sqrt{3} + 1}{2}$

$\tan\left(\frac{13\pi}{12}\right) = \frac{4 - 2\sqrt{3}}{2}$

Factor out 2 from the numerator:

$\tan\left(\frac{13\pi}{12}\right) = \frac{2(2 - \sqrt{3})}{2}$

Cancel the common factor of 2:

$\tan\left(\frac{13\pi}{12}\right) = 2 - \sqrt{3}$

To Prove:

$\frac{\sin\;(x \;+\; y)}{\sin\;(x \;-\; y)} = \frac{\tan\;x \;+\; \tan\;y}{\tan\;x \;-\; \tan\;y}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \frac{\sin\;(x \;+\; y)}{\sin\;(x \;-\; y)}$

Using the sum and difference identities for sine, we have:

$\sin(x + y) = \sin x \cos y + \cos x \sin y$

$\sin(x - y) = \sin x \cos y - \cos x \sin y$

Substitute these expansions into the LHS expression:

LHS $= \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}$

To express the RHS in terms of tangent, we need to divide the numerator and the denominator by $\cos x \cos y$. Assuming $\cos x \ne 0$ and $\cos y \ne 0$:

LHS $= \frac{\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y}}{\frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y}}$

Separate the terms in the numerator and the denominator:

LHS $= \frac{\frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}}{\frac{\sin x \cos y}{\cos x \cos y} - \frac{\cos x \sin y}{\cos x \cos y}}$

Cancel out the common terms:

LHS $= \frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x} - \frac{\sin y}{\cos y}}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we get:

LHS $= \frac{\tan x + \tan y}{\tan x - \tan y}$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Show:

$\tan 3x \tan 2x \tan x = \tan 3x – \tan 2x – \tan x$

Proof:

We know that $3x = 2x + x$.

Consider the tangent of $3x$:

$\tan (3x) = \tan (2x + x)$

Using the tangent addition formula, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, with $A = 2x$ and $B = x$:

$\tan (2x + x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$

So, we have:

$\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$

Multiply both sides of the equation by $(1 - \tan 2x \tan x)$:

$\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x$

Distribute $\tan 3x$ on the left side:

$\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x$

Rearrange the terms to isolate $\tan 3x \tan 2x \tan x$ on one side. Add $\tan 3x \tan 2x \tan x$ to both sides and subtract $\tan 2x + \tan x$ from both sides:

$\tan 3x - (\tan 2x + \tan x) = \tan 3x \tan 2x \tan x$

$\tan 3x - \tan 2x - \tan x = \tan 3x \tan 2x \tan x$

This is the required identity.

Therefore, $\tan 3x \tan 2x \tan x = \tan 3x – \tan 2x – \tan x$ is shown.

$\cos\left( \frac{\pi}{4}+x \right) + \cos\left( \frac{\pi}{4}-x \right)= \sqrt{2} \;\cos \;x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos\left( \frac{\pi}{4}+x \right) + \cos\left( \frac{\pi}{4}-x \right)$

Using the sum and difference identities for cosine, we have:

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\cos(A-B) = \cos A \cos B + \sin A \sin B$

Apply these identities to the terms in the LHS:

$\cos\left( \frac{\pi}{4}+x \right) = \cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x$

$\cos\left( \frac{\pi}{4}-x \right) = \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x$

Substitute these expansions back into the LHS expression:

LHS $= (\cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x) + (\cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x)$

Remove the parentheses and combine like terms:

LHS $= \cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x$

The terms $-\sin \frac{\pi}{4} \sin x$ and $+\sin \frac{\pi}{4} \sin x$ cancel each other out:

LHS $= \cos \frac{\pi}{4} \cos x + \cos \frac{\pi}{4} \cos x$

LHS $= 2 \cos \frac{\pi}{4} \cos x$

We know the value of $\cos \frac{\pi}{4}$:

$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Substitute this value into the LHS expression:

LHS $= 2 \times \left(\frac{1}{\sqrt{2}}\right) \times \cos x$

LHS $= \frac{2}{\sqrt{2}} \cos x$

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

LHS $= \frac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \cos x$

LHS $= \frac{2\sqrt{2}}{2} \cos x$

Cancel the common factor of 2:

LHS $= \sqrt{2} \cos x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

Alternate Solution:

We can use the sum-to-product formula for cosine:

$\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$

Let $C = \frac{\pi}{4}+x$ and $D = \frac{\pi}{4}-x$.

First, find $C+D$ and $C-D$:

$C+D = \left(\frac{\pi}{4}+x\right) + \left(\frac{\pi}{4}-x\right) = \frac{\pi}{4} + x + \frac{\pi}{4} - x = \frac{\pi}{4} + \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$

$C-D = \left(\frac{\pi}{4}+x\right) - \left(\frac{\pi}{4}-x\right) = \frac{\pi}{4} + x - \frac{\pi}{4} + x = x + x = 2x$

Now, calculate $\frac{C+D}{2}$ and $\frac{C-D}{2}$:

$\frac{C+D}{2} = \frac{\pi/2}{2} = \frac{\pi}{4}$

$\frac{C-D}{2} = \frac{2x}{2} = x$

Substitute these into the sum-to-product formula:

LHS $= \cos\left( \frac{\pi}{4}+x \right) + \cos\left( \frac{\pi}{4}-x \right) = 2 \cos \left(\frac{\pi}{4}\right) \cos \left(x\right)$

We know that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

LHS $= 2 \times \frac{1}{\sqrt{2}} \times \cos x$

LHS $= \frac{2}{\sqrt{2}} \cos x = \sqrt{2} \cos x$

This is the RHS.

Thus, the identity is proven using the sum-to-product formula.

To Prove:

$\frac{\cos\;7x \;+\; \cos \;5x}{\sin \;7x \;-\; \sin \;5x} = \cot x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\cos\;7x \;+\; \cos \;5x}{\sin \;7x \;-\; \sin \;5x}$

We use the sum-to-product formulas:

$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

$\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

Applying the sum-to-product formula to the numerator with $C = 7x$ and $D = 5x$:

Numerator $= \cos\;7x \;+\; \cos \;5x = 2 \cos\left(\frac{7x+5x}{2}\right) \cos\left(\frac{7x-5x}{2}\right)$

Numerator $= 2 \cos\left(\frac{12x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \cos(6x) \cos(x)$

Applying the sum-to-product formula to the denominator with $C = 7x$ and $D = 5x$:

Denominator $= \sin\;7x \;-\; \sin \;5x = 2 \cos\left(\frac{7x+5x}{2}\right) \sin\left(\frac{7x-5x}{2}\right)$

Denominator $= 2 \cos\left(\frac{12x}{2}\right) \sin\left(\frac{2x}{2}\right) = 2 \cos(6x) \sin(x)$

Substitute these expressions back into the LHS:

$LHS = \frac{2 \cos(6x) \cos(x)}{2 \cos(6x) \sin(x)}$

Assuming $\cos(6x) \neq 0$, we can cancel out the common factors $2$ and $\cos(6x)$:

$LHS = \frac{\cos(x)}{\sin(x)}$

Using the identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$:

$LHS = \cot x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\frac{\sin\;5x \;-\; 2 \ \sin \;3x \;+\; \sin \;x}{\cos \;5x \;-\; \cos \;x} = \tan x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\sin\;5x \;-\; 2 \ \sin \;3x \;+\; \sin \;x}{\cos \;5x \;-\; \cos \;x}$

Rearrange the terms in the numerator:

$LHS = \frac{(\sin\;5x \;+\; \sin \;x) \;-\; 2 \ \sin \;3x}{\cos \;5x \;-\; \cos \;x}$

We use the sum-to-product and difference-to-product formulas:

$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

$\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

Apply the sum-to-product formula to the terms $(\sin\;5x \;+\; \sin \;x)$ in the numerator with $C = 5x$ and $D = x$:

$\sin\;5x \;+\; \sin \;x = 2 \sin\left(\frac{5x+x}{2}\right) \cos\left(\frac{5x-x}{2}\right)$

$\sin\;5x \;+\; \sin \;x = 2 \sin\left(\frac{6x}{2}\right) \cos\left(\frac{4x}{2}\right) = 2 \sin(3x) \cos(2x)$

Substitute this back into the numerator:

Numerator $= 2 \sin(3x) \cos(2x) \;-\; 2 \sin(3x)$

Factor out $2 \sin(3x)$ from the numerator:

Numerator $= 2 \sin(3x) (\cos(2x) \;-\; 1)$

Apply the difference-to-product formula to the denominator with $C = 5x$ and $D = x$:

Denominator $= \cos \;5x \;-\; \cos \;x = -2 \sin\left(\frac{5x+x}{2}\right) \sin\left(\frac{5x-x}{2}\right)$

Denominator $= -2 \sin\left(\frac{6x}{2}\right) \sin\left(\frac{4x}{2}\right) = -2 \sin(3x) \sin(2x)$

Substitute the simplified numerator and denominator back into the LHS expression:

$LHS = \frac{2 \sin(3x) (\cos(2x) \;-\; 1)}{-2 \sin(3x) \sin(2x)}$

Assuming $\sin(3x) \neq 0$ and $\sin(2x) \neq 0$, we can cancel out the common factors $2$ and $\sin(3x)$:

$LHS = \frac{\cos(2x) \;-\; 1}{-\sin(2x)}$

Rewrite the numerator: $\cos(2x) - 1 = - (1 - \cos(2x))$

$LHS = \frac{-(1 \;-\; \cos(2x))}{-\sin(2x)}$

The negative signs cancel out:

$LHS = \frac{1 \;-\; \cos(2x)}{\sin(2x)}$

We use the double angle identities:

$1 - \cos(2x) = 2 \sin^2(x)$

$\sin(2x) = 2 \sin(x) \cos(x)$

Substitute these identities into the LHS expression:

$LHS = \frac{2 \sin^2(x)}{2 \sin(x) \cos(x)}$

Assuming $\sin(x) \neq 0$ and $\cos(x) \neq 0$, we can cancel out the common factors $2$ and $\sin(x)$:

$LHS = \frac{\cancel{2} \sin^{\cancel{2}}(x)}{\cancel{2} \cancel{\sin(x)} \cos(x)}$

$LHS = \frac{\sin(x)}{\cos(x)}$

Using the identity $\tan x = \frac{\sin x}{\cos x}$:

$LHS = \tan x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4} = -\frac{1}{2}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}$

We know the values of the trigonometric ratios for the given angles:

$\sin\frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$

$\cos\frac{\pi}{3} = \cos 60^\circ = \frac{1}{2}$

$\tan\frac{\pi}{4} = \tan 45^\circ = 1$

Substitute these values into the LHS expression:

LHS $= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2$

Calculate the squares:

LHS $= \frac{1}{4} + \frac{1}{4} - 1$

Add the fractions:

LHS $= \frac{1+1}{4} - 1$

LHS $= \frac{2}{4} - 1$

Simplify the fraction:

LHS $= \frac{1}{2} - 1$

Combine the terms:

LHS $= \frac{1 - 2}{2}$

LHS $= -\frac{1}{2}$

The Right Hand Side (RHS) of the given equation is $-\frac{1}{2}$.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$2\sin^2\frac{\pi}{6}+\text{cosec}^2 \;\frac{7\pi}{6}\;\cos^2\frac{\pi}{3} = \frac{3}{2}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= 2\sin^2\frac{\pi}{6}+\text{cosec}^2 \;\frac{7\pi}{6}\;\cos^2\frac{\pi}{3}$

We need the values of $\sin\frac{\pi}{6}$, $\text{cosec}\frac{7\pi}{6}$, and $\cos\frac{\pi}{3}$.

$\frac{\pi}{6}$ corresponds to $30^\circ$.

$\sin\frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$

$\cos\frac{\pi}{3}$ corresponds to $60^\circ$.

$\cos\frac{\pi}{3} = \cos 60^\circ = \frac{1}{2}$

For $\text{cosec}\frac{7\pi}{6}$, we note that $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$.

The angle $\frac{7\pi}{6}$ lies in the third quadrant, where sine and cosecant are negative.

$\text{cosec}\left(\pi + \frac{\pi}{6}\right) = -\text{cosec}\frac{\pi}{6}$

Since $\text{cosec}\frac{\pi}{6} = \frac{1}{\sin(\pi/6)} = \frac{1}{1/2} = 2$, we have:

$\text{cosec}\frac{7\pi}{6} = -2$

Now substitute these values into the LHS expression:

LHS $= 2 \times \left(\sin\frac{\pi}{6}\right)^2 + \left(\text{cosec}\frac{7\pi}{6}\right)^2 \times \left(\cos\frac{\pi}{3}\right)^2$

LHS $= 2 \times \left(\frac{1}{2}\right)^2 + \left(-2\right)^2 \times \left(\frac{1}{2}\right)^2$

Calculate the squares:

LHS $= 2 \times \left(\frac{1}{4}\right) + \left(4\right) \times \left(\frac{1}{4}\right)$

Perform the multiplications:

LHS $= \frac{2}{4} + \frac{4}{4}$

Simplify the fractions:

LHS $= \frac{1}{2} + 1$

Add the terms:

LHS $= \frac{1}{2} + \frac{2}{2} = \frac{1+2}{2} = \frac{3}{2}$

The Right Hand Side (RHS) of the given equation is $\frac{3}{2}$.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cot^2\frac{\pi}{6}+\text{cosec} \;\frac{5\pi}{6}+3\tan^2\frac{\pi}{6} = 6$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cot^2\frac{\pi}{6}+\text{cosec} \;\frac{5\pi}{6}+3\tan^2\frac{\pi}{6}$

We need the values of $\cot\frac{\pi}{6}$, $\text{cosec}\frac{5\pi}{6}$, and $\tan\frac{\pi}{6}$.

$\frac{\pi}{6}$ corresponds to $30^\circ$.

$\cot\frac{\pi}{6} = \cot 30^\circ = \sqrt{3}$

For $\text{cosec}\frac{5\pi}{6}$, we note that $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$.

This angle is in the second quadrant ($150^\circ$), where cosecant is positive.

$\text{cosec}\frac{5\pi}{6} = \text{cosec}\left(\pi - \frac{\pi}{6}\right) = \text{cosec}\frac{\pi}{6}$

Since $\sin\frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$, $\text{cosec}\frac{\pi}{6} = \frac{1}{\sin(\pi/6)} = \frac{1}{1/2} = 2$.

So, $\text{cosec}\frac{5\pi}{6} = 2$

$\tan\frac{\pi}{6} = \tan 30^\circ = \frac{1}{\sqrt{3}}$

Substitute these values into the LHS expression:

LHS $= \left(\cot\frac{\pi}{6}\right)^2 + \text{cosec}\frac{5\pi}{6} + 3\left(\tan\frac{\pi}{6}\right)^2$

LHS $= \left(\sqrt{3}\right)^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2$

Calculate the squares:

LHS $= 3 + 2 + 3\left(\frac{1}{3}\right)$

Perform the multiplication:

LHS $= 3 + 2 + \frac{3}{3}$

LHS $= 3 + 2 + 1$

Add the terms:

LHS $= 6$

The Right Hand Side (RHS) of the given equation is $6$.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3} = 10$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= 2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3}$

We need the values of $\sin\frac{3\pi}{4}$, $\cos\frac{\pi}{4}$, and $\sec\frac{\pi}{3}$.

$\frac{\pi}{4}$ corresponds to $45^\circ$.

$\cos\frac{\pi}{4} = \cos 45^\circ = \frac{1}{\sqrt{2}}$

$\cos^2\frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$

$\frac{3\pi}{4} = \pi - \frac{\pi}{4}$ corresponds to $180^\circ - 45^\circ = 135^\circ$. The angle is in the second quadrant, where sine is positive.

$\sin\frac{3\pi}{4} = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\sin^2\frac{3\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$

$\frac{\pi}{3}$ corresponds to $60^\circ$.

$\sec\frac{\pi}{3} = \sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2$

$\sec^2\frac{\pi}{3} = (2)^2 = 4$

Substitute these values into the LHS expression:

LHS $= 2\left(\sin^2\frac{3\pi}{4}\right) + 2\left(\cos^2\frac{\pi}{4}\right) + 2\left(\sec^2\frac{\pi}{3}\right)$

LHS $= 2\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) + 2(4)$

Perform the multiplications:

LHS $= \frac{2}{2} + \frac{2}{2} + 8$

LHS $= 1 + 1 + 8$

Add the terms:

LHS $= 10$

The Right Hand Side (RHS) of the given equation is $10$.

Since LHS $=$ RHS, the given identity is proven.

To Find:

(i) The value of $\sin 75^\circ$.

Solution (i):

We can express $75^\circ$ as the sum of two standard angles, such as $45^\circ + 30^\circ$.

$\sin 75^\circ = \sin(45^\circ + 30^\circ)$

Using the sum identity for sine, $\sin(A + B) = \sin A \cos B + \cos A \sin B$, with $A = 45^\circ$ and $B = 30^\circ$:

$\sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$

Substitute the known values:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

So,

$\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)$

$\sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

Combine the fractions:

$\sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\sin 75^\circ = \frac{(\sqrt{3} + 1)\sqrt{2}}{(2\sqrt{2})\sqrt{2}} = \frac{\sqrt{3}\sqrt{2} + 1\sqrt{2}}{2(\sqrt{2})^2} = \frac{\sqrt{6} + \sqrt{2}}{2 \times 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

Thus, the value of $\sin 75^\circ$ is $\frac{\sqrt{6} + \sqrt{2}}{4}$.

To Find:

(ii) The value of $\tan 15^\circ$.

Solution (ii):

We can express $15^\circ$ as the difference of two standard angles, such as $45^\circ - 30^\circ$.

$\tan 15^\circ = \tan(45^\circ - 30^\circ)$

Using the difference identity for tangent, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, with $A = 45^\circ$ and $B = 30^\circ$:

$\tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

Substitute the known values:

$\tan 45^\circ = 1$

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

So,

$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$

To simplify, multiply the numerator and the denominator by $\sqrt{3}$:

$\tan 15^\circ = \frac{\left(1 - \frac{1}{\sqrt{3}}\right)\sqrt{3}}{\left(1 + \frac{1}{\sqrt{3}}\right)\sqrt{3}} = \frac{1 \times \sqrt{3} - \frac{1}{\sqrt{3}} \times \sqrt{3}}{1 \times \sqrt{3} + \frac{1}{\sqrt{3}} \times \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$:

$\tan 15^\circ = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} \times \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2}$

Expand the numerator using $(a-b)^2 = a^2 - 2ab + b^2$ and simplify the denominator:

$\tan 15^\circ = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2}$

Factor out 2 from the numerator and cancel:

$\tan 15^\circ = \frac{2(2 - \sqrt{3})}{2} = 2 - \sqrt{3}$

Thus, the value of $\tan 15^\circ$ is $2 - \sqrt{3}$.

To Prove:

$\cos\left( \frac{\pi}{4}-x \right) \cos\left( \frac{\pi}{4}-y \right) - \sin\left( \frac{\pi}{4}-x \right) \sin\left( \frac{\pi}{4}-y \right) = \sin \;(x + y)$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos\left( \frac{\pi}{4}-x \right) \cos\left( \frac{\pi}{4}-y \right) - \sin\left( \frac{\pi}{4}-x \right) \sin\left( \frac{\pi}{4}-y \right)$

This expression is in the form $\cos A \cos B - \sin A \sin B$, which is the expansion of $\cos(A+B)$.

Let $A = \frac{\pi}{4}-x$ and $B = \frac{\pi}{4}-y$.

Then, $A+B = \left(\frac{\pi}{4}-x\right) + \left(\frac{\pi}{4}-y\right)$

$A+B = \frac{\pi}{4} + \frac{\pi}{4} - x - y$

$A+B = \frac{2\pi}{4} - (x+y)$

$A+B = \frac{\pi}{2} - (x+y)$

So, the LHS can be written as:

LHS $= \cos(A+B) = \cos\left(\frac{\pi}{2} - (x+y)\right)$

Using the complementary angle identity $\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$, with $\theta = x+y$:

LHS $= \sin(x+y)$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\frac{\tan \;\left( \frac{\pi}{4} \;+\; x \right)}{\tan \;\left( \frac{\pi}{4} \;-\; x \right)}=\left( \frac{1\;+\;\tan \;x}{1\;-\;\tan \;x} \right)^{2}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \frac{\tan \;\left( \frac{\pi}{4} \;+\; x \right)}{\tan \;\left( \frac{\pi}{4} \;-\; x \right)}$

We use the sum and difference identities for tangent:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Apply the sum identity for the numerator with $A = \frac{\pi}{4}$ and $B = x$. We know $\tan\frac{\pi}{4} = 1$:

$\tan \;\left( \frac{\pi}{4} \;+\; x \right) = \frac{\tan \;\frac{\pi}{4} \;+\; \tan \;x}{1 \;-\; \tan \;\frac{\pi}{4} \;\tan \;x} = \frac{1 \;+\; \tan \;x}{1 \;-\; 1 \;\cdot\; \tan \;x} = \frac{1 \;+\; \tan \;x}{1 \;-\; \tan \;x}$

Apply the difference identity for the denominator with $A = \frac{\pi}{4}$ and $B = x$. We know $\tan\frac{\pi}{4} = 1$:

$\tan \;\left( \frac{\pi}{4} \;-\; x \right) = \frac{\tan \;\frac{\pi}{4} \;-\; \tan \;x}{1 \;+\; \tan \;\frac{\pi}{4} \;\tan \;x} = \frac{1 \;-\; \tan \;x}{1 \;+\; 1 \;\cdot\; \tan \;x} = \frac{1 \;-\; \tan \;x}{1 \;+\; \tan \;x}$

Substitute these expressions back into the LHS:

LHS $= \frac{\frac{1 \;+\; \tan \;x}{1 \;-\; \tan \;x}}{\frac{1 \;-\; \tan \;x}{1 \;+\; \tan \;x}}$

To simplify the complex fraction, multiply the numerator by the reciprocal of the denominator:

LHS $= \frac{1 \;+\; \tan \;x}{1 \;-\; \tan \;x} \times \frac{1 \;+\; \tan \;x}{1 \;-\; \tan \;x}$

LHS $= \frac{(1 \;+\; \tan \;x) \times (1 \;+\; \tan \;x)}{(1 \;-\; \tan \;x) \times (1 \;-\; \tan \;x)}$

LHS $= \frac{(1 \;+\; \tan \;x)^2}{(1 \;-\; \tan \;x)^2}$

LHS $= \left( \frac{1 \;+\; \tan \;x}{1 \;-\; \tan \;x} \right)^{2}$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\frac{\cos\;(\pi\;+\;x) \; \cos\;(-\;x)}{\sin\;(\pi\;-\;x) \; \cos\;\left( \frac{\pi}{2}\;+\;x \right)}=\cot^2x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \frac{\cos\;(\pi\;+\;x) \; \cos\;(-\;x)}{\sin\;(\pi\;-\;x) \; \cos\;\left( \frac{\pi}{2}\;+\;x \right)}$

Using the trigonometric identities for related angles:

$\cos(\pi + x) = -\cos x$ (Cosine is negative in the third quadrant)

$\cos(-x) = \cos x$ (Cosine is an even function)

$\sin(\pi - x) = \sin x$ (Sine is positive in the second quadrant)

$\cos\left( \frac{\pi}{2}\;+\;x \right) = -\sin x$ (Cosine is negative in the second quadrant)

Substitute these simplified terms into the LHS expression:

LHS $= \frac{(-\cos x) (\cos x)}{(\sin x) (-\sin x)}$

LHS $= \frac{-\cos^2 x}{-\sin^2 x}$

Cancel the negative signs:

LHS $= \frac{\cos^2 x}{\sin^2 x}$

Using the identity $\cot x = \frac{\cos x}{\sin x}$:

LHS $= \left(\frac{\cos x}{\sin x}\right)^2 = \cot^2 x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cos \left( \frac{3\pi}{2}+x \right) \cos (2\pi+x)\left[ \cot\left( \frac{3\pi}{2}-x \right)+ \cot (2\pi+x) \right]=1$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos \left( \frac{3\pi}{2}+x \right) \cos (2\pi+x)\left[ \cot\left( \frac{3\pi}{2}-x \right)+ \cot (2\pi+x) \right]$

Using the trigonometric identities for related angles:

$\cos \left( \frac{3\pi}{2}+x \right) = \sin x$ (Since $\frac{3\pi}{2}+x$ is in the 4th quadrant and $\cos(\frac{3\pi}{2}+\theta) = \sin\theta$)

$\cos (2\pi+x) = \cos x$ (Since $2\pi$ is a full period for cosine)

$\cot\left( \frac{3\pi}{2}-x \right) = \tan x$ (Since $\frac{3\pi}{2}-x$ is in the 3rd quadrant and $\cot(\frac{3\pi}{2}-\theta) = \tan\theta$)

$\cot (2\pi+x) = \cot x$ (Since $2\pi$ is a full period for cotangent)

Substitute these simplified terms into the LHS expression:

LHS $= (\sin x) (\cos x) [\tan x + \cot x]$

Rewrite $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:

LHS $= \sin x \cos x \left[ \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right]$

Combine the terms inside the square brackets by finding a common denominator:

LHS $= \sin x \cos x \left[ \frac{\sin x \cdot \sin x + \cos x \cdot \cos x}{\cos x \sin x} \right]$

LHS $= \sin x \cos x \left[ \frac{\sin^2 x + \cos^2 x}{\cos x \sin x} \right]$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

LHS $= \sin x \cos x \left[ \frac{1}{\cos x \sin x} \right]$

Multiply the terms:

LHS $= \frac{\sin x \cos x}{\cos x \sin x}$

Assuming $\sin x \neq 0$ and $\cos x \neq 0$, we can cancel out the common factors:

LHS $= 1$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x$

Rearrange the terms in the LHS:

LHS $= \cos (n + 1)x \cos (n + 2)x + \sin (n + 1)x \sin (n + 2)x$

This expression is in the form $\cos A \cos B + \sin A \sin B$, which is the expansion of $\cos(A - B)$.

Let $A = (n+2)x$ and $B = (n+1)x$.

Then, $A - B = (n+2)x - (n+1)x$

$A - B = (nx + 2x) - (nx + x)$

$A - B = nx + 2x - nx - x$

$A - B = x$

So, the LHS can be written as:

LHS $= \cos(A - B) = \cos((n+2)x - (n+1)x)$

LHS $= \cos(x)$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cos \left( \frac{3\pi}{4} +x\right) - \cos \left( \frac{3\pi}{4} -x\right)=-\sqrt{2} \;\sin \;x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos \left( \frac{3\pi}{4} +x\right) - \cos \left( \frac{3\pi}{4} -x\right)$

We use the difference-to-product formula for cosine:

$\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

Let $C = \frac{3\pi}{4} + x$ and $D = \frac{3\pi}{4} - x$.

First, calculate the sum and difference of $C$ and $D$:

$C+D = \left(\frac{3\pi}{4} + x\right) + \left(\frac{3\pi}{4} - x\right) = \frac{3\pi}{4} + x + \frac{3\pi}{4} - x = \frac{6\pi}{4} = \frac{3\pi}{2}$

$C-D = \left(\frac{3\pi}{4} + x\right) - \left(\frac{3\pi}{4} - x\right) = \frac{3\pi}{4} + x - \frac{3\pi}{4} + x = 2x$

Now, calculate $\frac{C+D}{2}$ and $\frac{C-D}{2}$:

$\frac{C+D}{2} = \frac{3\pi/2}{2} = \frac{3\pi}{4}$

$\frac{C-D}{2} = \frac{2x}{2} = x$

Substitute these into the difference-to-product formula:

LHS $= -2 \sin\left(\frac{3\pi}{4}\right) \sin\left(x\right)$

We need the value of $\sin\left(\frac{3\pi}{4}\right)$. The angle $\frac{3\pi}{4}$ is in the second quadrant, and $\frac{3\pi}{4} = \pi - \frac{\pi}{4}$.

$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Substitute this value back into the LHS expression:

LHS $= -2 \times \left(\frac{1}{\sqrt{2}}\right) \times \sin x$

LHS $= -\frac{2}{\sqrt{2}} \sin x$

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

LHS $= -\frac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \sin x = -\frac{2\sqrt{2}}{2} \sin x = -\sqrt{2} \sin x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

Alternate Solution:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos \left( \frac{3\pi}{4} +x\right) - \cos \left( \frac{3\pi}{4} -x\right)$

Using the sum and difference identities for cosine:

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\cos(A-B) = \cos A \cos B + \sin A \sin B$

Apply these identities with $A = \frac{3\pi}{4}$ and $B = x$:

$\cos \left( \frac{3\pi}{4} +x\right) = \cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x$

$\cos \left( \frac{3\pi}{4} -x\right) = \cos \frac{3\pi}{4} \cos x + \sin \frac{3\pi}{4} \sin x$

Substitute these into the LHS expression:

LHS $= (\cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x) - (\cos \frac{3\pi}{4} \cos x + \sin \frac{3\pi}{4} \sin x)$

Remove the parentheses:

LHS $= \cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x - \cos \frac{3\pi}{4} \cos x - \sin \frac{3\pi}{4} \sin x$

Combine like terms:

LHS $= (\cos \frac{3\pi}{4} \cos x - \cos \frac{3\pi}{4} \cos x) + (-\sin \frac{3\pi}{4} \sin x - \sin \frac{3\pi}{4} \sin x)$

LHS $= 0 - 2 \sin \frac{3\pi}{4} \sin x$

LHS $= -2 \sin \frac{3\pi}{4} \sin x$

We know that $\sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

LHS $= -2 \times \left(\frac{1}{\sqrt{2}}\right) \times \sin x$

LHS $= -\frac{2}{\sqrt{2}} \sin x = -\sqrt{2} \sin x$

This is the RHS.

Thus, the identity is proven using sum and difference formulas.

To Prove:

$\sin^2 6x – \sin^2 4x = \sin 2x \sin 10x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \sin^2 6x – \sin^2 4x$

We use the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$.

Let $A = 6x$ and $B = 4x$.

Then, $A+B = 6x + 4x = 10x$

And, $A-B = 6x - 4x = 2x$

Substitute these into the identity:

LHS $= \sin(6x+4x)\sin(6x-4x)$

LHS $= \sin(10x)\sin(2x)$

LHS $= \sin 2x \sin 10x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

Alternate Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \sin^2 6x – \sin^2 4x$

We can use the double angle identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$.

LHS $= \frac{1 - \cos(2 \cdot 6x)}{2} - \frac{1 - \cos(2 \cdot 4x)}{2}$

LHS $= \frac{1 - \cos 12x}{2} - \frac{1 - \cos 8x}{2}$

Combine the terms over a common denominator:

LHS $= \frac{(1 - \cos 12x) - (1 - \cos 8x)}{2}$

LHS $= \frac{1 - \cos 12x - 1 + \cos 8x}{2}$

LHS $= \frac{\cos 8x - \cos 12x}{2}$

Rearrange the terms in the numerator to use the difference-to-product formula $\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ with $C = 8x$ and $D = 12x$.

Numerator $= \cos 8x - \cos 12x = -2 \sin\left(\frac{8x+12x}{2}\right) \sin\left(\frac{8x-12x}{2}\right)$

Numerator $= -2 \sin\left(\frac{20x}{2}\right) \sin\left(\frac{-4x}{2}\right)$

Numerator $= -2 \sin(10x) \sin(-2x)$

Using the property $\sin(-\theta) = -\sin\theta$:

Numerator $= -2 \sin(10x) (-\sin(2x))$

Numerator $= 2 \sin(10x) \sin(2x)$

Substitute the numerator back into the LHS expression:

LHS $= \frac{2 \sin(10x) \sin(2x)}{2}$

Cancel the common factor of 2:

LHS $= \sin(10x) \sin(2x)$

LHS $= \sin 2x \sin 10x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos^2 2x – \cos^2 6x$

Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$, we can rewrite the LHS:

LHS $= (1 - \sin^2 2x) - (1 - \sin^2 6x)$

LHS $= 1 - \sin^2 2x - 1 + \sin^2 6x$

LHS $= \sin^2 6x - \sin^2 2x$

Now, we use the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$.

Let $A = 6x$ and $B = 2x$.

Then, $A+B = 6x + 2x = 8x$

And, $A-B = 6x - 2x = 4x$

Substitute these into the identity:

LHS $= \sin(6x+4x)\sin(6x-4x)$

LHS $= \sin(10x)\sin(4x)$

LHS $= \sin 4x \sin 10x$

This does not match the RHS directly as stated in the question, which is $\sin 4x \sin 8x$. Let me recheck the provided identity. Ah, the identity is $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$, which is correct. Let me recheck the question itself or the calculation.

The question states $\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x$. My derivation led to $\sin 4x \sin 10x$. There seems to be a potential typo in the question's RHS, it should likely be $\sin 4x \sin 8x$. Let's assume the RHS in the question $\sin 4x \sin 8x$ is correct and see if we can reach it from LHS. Oh, wait, I used $A=6x, B=2x$. The identity is $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$. So my substitution is correct for $\sin^2 6x - \sin^2 2x$. Let's re-evaluate the initial step carefully.

LHS $= \cos^2 2x – \cos^2 6x$

Using $\cos^2 \theta = 1 - \sin^2 \theta$:

LHS $= (1 - \sin^2 2x) - (1 - \sin^2 6x)$

LHS $= 1 - \sin^2 2x - 1 + \sin^2 6x$

LHS $= \sin^2 6x - \sin^2 2x$

Using $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$ with $A=6x$ and $B=2x$:

LHS $= \sin(6x+2x)\sin(6x-2x)$

LHS $= \sin(8x)\sin(4x)$

LHS $= \sin 4x \sin 8x$

This matches the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

Alternate Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos^2 2x – \cos^2 6x$

We use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.

LHS $= \frac{1 + \cos(2 \cdot 2x)}{2} - \frac{1 + \cos(2 \cdot 6x)}{2}$

LHS $= \frac{1 + \cos 4x}{2} - \frac{1 + \cos 12x}{2}$

Combine the terms over a common denominator:

LHS $= \frac{(1 + \cos 4x) - (1 + \cos 12x)}{2}$

LHS $= \frac{1 + \cos 4x - 1 - \cos 12x}{2}$

LHS $= \frac{\cos 4x - \cos 12x}{2}$

Use the difference-to-product formula $\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$.

Let $C = 4x$ and $D = 12x$.

Then, $\frac{C+D}{2} = \frac{4x+12x}{2} = \frac{16x}{2} = 8x$

And, $\frac{C-D}{2} = \frac{4x-12x}{2} = \frac{-8x}{2} = -4x$

Substitute these into the formula:

Numerator $= -2 \sin\left(8x\right) \sin\left(-4x\right)$

Using the property $\sin(-\theta) = -\sin\theta$:

Numerator $= -2 \sin(8x) (-\sin(4x))$

Numerator $= 2 \sin(8x) \sin(4x)$

Substitute the numerator back into the LHS expression:

LHS $= \frac{2 \sin(8x) \sin(4x)}{2}$

Cancel the common factor of 2:

LHS $= \sin(8x) \sin(4x)$

LHS $= \sin 4x \sin 8x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \sin 2x + 2 \sin 4x + \sin 6x$

Group the terms $\sin 6x$ and $\sin 2x$ together:

LHS $= (\sin 6x + \sin 2x) + 2 \sin 4x$

Use the sum-to-product formula: $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

Let $C = 6x$ and $D = 2x$.

Then, $\frac{C+D}{2} = \frac{6x+2x}{2} = \frac{8x}{2} = 4x$

And, $\frac{C-D}{2} = \frac{6x-2x}{2} = \frac{4x}{2} = 2x$

Substitute these into the sum-to-product formula:

$\sin 6x + \sin 2x = 2 \sin(4x) \cos(2x)$

Substitute this expression back into the LHS:

LHS $= 2 \sin(4x) \cos(2x) + 2 \sin 4x$

Factor out the common term $2 \sin 4x$:

LHS $= 2 \sin 4x (\cos 2x + 1)$

Use the double angle identity: $\cos 2x = 2 \cos^2 x - 1$.

Rearranging this identity, we get: $1 + \cos 2x = 2 \cos^2 x$.

Substitute this into the LHS expression:

LHS $= 2 \sin 4x (2 \cos^2 x)$

Multiply the terms:

LHS $= 4 \cos^2 x \sin 4x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x – \sin 3x)$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cot 4x (\sin 5x + \sin 3x)$

Using the sum-to-product formula $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$ for $(\sin 5x + \sin 3x)$:

$\sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)$

$\sin 5x + \sin 3x = 2 \sin\left(\frac{8x}{2}\right) \cos\left(\frac{2x}{2}\right)$

$\sin 5x + \sin 3x = 2 \sin(4x) \cos(x)$

Substitute this back into the LHS expression:

LHS $= \cot 4x (2 \sin 4x \cos x)$

Rewrite $\cot 4x$ as $\frac{\cos 4x}{\sin 4x}$:

LHS $= \frac{\cos 4x}{\sin 4x} (2 \sin 4x \cos x)$

Assuming $\sin 4x \neq 0$, cancel $\sin 4x$:

LHS $= 2 \cos 4x \cos x$

Now consider the Right Hand Side (RHS) of the equation:

RHS $= \cot x (\sin 5x – \sin 3x)$

Using the difference-to-product formula $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ for $(\sin 5x - \sin 3x)$:

$\sin 5x - \sin 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \sin\left(\frac{5x-3x}{2}\right)$

$\sin 5x - \sin 3x = 2 \cos\left(\frac{8x}{2}\right) \sin\left(\frac{2x}{2}\right)$

$\sin 5x - \sin 3x = 2 \cos(4x) \sin(x)$

Substitute this back into the RHS expression:

RHS $= \cot x (2 \cos 4x \sin x)$

Rewrite $\cot x$ as $\frac{\cos x}{\sin x}$:

RHS $= \frac{\cos x}{\sin x} (2 \cos 4x \sin x)$

Assuming $\sin x \neq 0$, cancel $\sin x$:

RHS $= 2 \cos x \cos 4x$

RHS $= 2 \cos 4x \cos x$

Comparing the simplified LHS and RHS:

LHS $= 2 \cos 4x \cos x$

RHS $= 2 \cos 4x \cos x$

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\frac{\cos \;9x \;-\; \cos \;5x}{\sin \;17x \;-\; \sin \;3x}$ = $-\frac{\sin \;2x}{\cos \;10x}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\cos \;9x \;-\; \cos \;5x}{\sin \;17x \;-\; \sin \;3x}$

We use the difference-to-product formulas:

$\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

$\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

Applying the difference-to-product formula to the numerator with $C = 9x$ and $D = 5x$:

Numerator $= \cos\;9x \;-\; \cos \;5x = -2 \sin\left(\frac{9x+5x}{2}\right) \sin\left(\frac{9x-5x}{2}\right)$

Numerator $= -2 \sin\left(\frac{14x}{2}\right) \sin\left(\frac{4x}{2}\right) = -2 \sin(7x) \sin(2x)$

Applying the difference-to-product formula to the denominator with $C = 17x$ and $D = 3x$:

Denominator $= \sin \;17x \;-\; \sin \;3x = 2 \cos\left(\frac{17x+3x}{2}\right) \sin\left(\frac{17x-3x}{2}\right)$

Denominator $= 2 \cos\left(\frac{20x}{2}\right) \sin\left(\frac{14x}{2}\right) = 2 \cos(10x) \sin(7x)$

Substitute these expressions back into the LHS:

$LHS = \frac{-2 \sin(7x) \sin(2x)}{2 \cos(10x) \sin(7x)}$

Assuming $\sin(7x) \neq 0$ and $\cos(10x) \neq 0$, we can cancel out the common factors $2$ and $\sin(7x)$:

$LHS = \frac{-\cancel{2} \cancel{\sin(7x)} \sin(2x)}{\cancel{2} \cos(10x) \cancel{\sin(7x)}}$

$LHS = -\frac{\sin(2x)}{\cos(10x)}$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\frac{\sin \;5x \;+\; \sin \;3x}{\cos \;5x \;+\; \cos \;3x}$ = tan 4x

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\sin \;5x \;+\; \sin \;3x}{\cos \;5x \;+\; \cos \;3x}$

We use the sum-to-product formulas:

$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

Applying the sum-to-product formula to the numerator with $C = 5x$ and $D = 3x$:

Numerator $= \sin\;5x \;+\; \sin \;3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)$

Numerator $= 2 \sin\left(\frac{8x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \sin(4x) \cos(x)$

Applying the sum-to-product formula to the denominator with $C = 5x$ and $D = 3x$:

Denominator $= \cos\;5x \;+\; \cos \;3x = 2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)$

Denominator $= 2 \cos\left(\frac{8x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \cos(4x) \cos(x)$

Substitute these expressions back into the LHS:

$LHS = \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}$

Assuming $\cos(4x) \neq 0$ and $\cos(x) \neq 0$, we can cancel out the common factors $2$ and $\cos(x)$:

$LHS = \frac{\cancel{2} \sin(4x) \cancel{\cos(x)}}{\cancel{2} \cos(4x) \cancel{\cos(x)}}$

$LHS = \frac{\sin(4x)}{\cos(4x)}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$LHS = \tan(4x)$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\frac{\sin \;x \;-\; \sin \;y}{\cos \;x \;+\; \cos \;y} = \tan \frac{x \;-\; y}{2}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\sin \;x \;-\; \sin \;y}{\cos \;x \;+\; \cos \;y}$

We use the difference-to-product and sum-to-product formulas:

$\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

Applying the difference-to-product formula to the numerator with $C = x$ and $D = y$:

Numerator $= \sin\;x \;-\; \sin \;y = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)$

Applying the sum-to-product formula to the denominator with $C = x$ and $D = y$:

Denominator $= \cos\;x \;+\; \cos \;y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$

Substitute these expressions back into the LHS:

$LHS = \frac{2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)}{2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)}$

Assuming $\cos\left(\frac{x+y}{2}\right) \neq 0$ and $\cos\left(\frac{x-y}{2}\right) \neq 0$, we can cancel out the common factors $2$ and $\cos\left(\frac{x+y}{2}\right)$:

$LHS = \frac{\cancel{2} \cancel{\cos\left(\frac{x+y}{2}\right)} \sin\left(\frac{x-y}{2}\right)}{\cancel{2} \cancel{\cos\left(\frac{x+y}{2}\right)} \cos\left(\frac{x-y}{2}\right)}$

$LHS = \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$LHS = \tan\left(\frac{x-y}{2}\right)$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\frac{\sin \;x \;+\; \sin \;3x}{\cos \;x \;+\; \cos \;3x} = \tan 2x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\sin \;x \;+\; \sin \;3x}{\cos \;x \;+\; \cos \;3x}$

We use the sum-to-product formulas:

$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

Applying the sum-to-product formula to the numerator with $C = 3x$ and $D = x$:

Numerator $= \sin\;3x \;+\; \sin \;x = 2 \sin\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right)$

Numerator $= 2 \sin\left(\frac{4x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \sin(2x) \cos(x)$

Applying the sum-to-product formula to the denominator with $C = 3x$ and $D = x$:

Denominator $= \cos\;3x \;+\; \cos \;x = 2 \cos\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right)$

Denominator $= 2 \cos\left(\frac{4x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \cos(2x) \cos(x)$

Substitute these expressions back into the LHS:

$LHS = \frac{2 \sin(2x) \cos(x)}{2 \cos(2x) \cos(x)}$

Assuming $\cos(2x) \neq 0$ and $\cos(x) \neq 0$, we can cancel out the common factors $2$ and $\cos(x)$:

$LHS = \frac{\cancel{2} \sin(2x) \cancel{\cos(x)}}{\cancel{2} \cos(2x) \cancel{\cos(x)}}$

$LHS = \frac{\sin(2x)}{\cos(2x)}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$LHS = \tan(2x)$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\frac{\sin \;x\;-\;\sin\;3x}{\sin^{2}\;x\;-\;\cos^{2}\;x}=2\sin \;x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\sin \;x\;-\;\sin\;3x}{\sin^{2}\;x\;-\;\cos^{2}\;x}$

We use the difference-to-product formula for the numerator:

$\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$

Let $C = x$ and $D = 3x$.

Numerator $= \sin\;x \;-\; \sin\;3x = 2 \cos\left(\frac{x+3x}{2}\right) \sin\left(\frac{x-3x}{2}\right)$

Numerator $= 2 \cos\left(\frac{4x}{2}\right) \sin\left(\frac{-2x}{2}\right)$

Numerator $= 2 \cos(2x) \sin(-x)$

Using the identity $\sin(-\theta) = -\sin\theta$:

Numerator $= 2 \cos(2x) (-\sin x) = -2 \sin x \cos 2x$

Now consider the denominator. We use the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$.

Denominator $= \sin^2\;x\;-\;\cos^2\;x = -(\cos^2\;x\;-\;\sin^2\;x)$

Denominator $= -\cos(2x)$

Substitute the simplified numerator and denominator back into the LHS expression:

$LHS = \frac{-2 \sin x \cos 2x}{-\cos 2x}$

Assuming $\cos(2x) \neq 0$, we can cancel out the common factor $\cos(2x)$ and the negative signs:

$LHS = \frac{-\cancel{2} \sin x \cancel{\cos 2x}}{-\cancel{\cos 2x}}$

$LHS = 2 \sin x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\frac{\cos \;4x \;+\; \cos \;3x \;+\; \cos \;2x}{\sin \;4x \;+\; \sin \;3x \;+\; \sin \;2x} = \cot \;3x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

$LHS = \frac{\cos \;4x \;+\; \cos \;3x \;+\; \cos \;2x}{\sin \;4x \;+\; \sin \;3x \;+\; \sin \;2x}$

Group the terms in the numerator and denominator:

$LHS = \frac{(\cos \;4x \;+\; \cos \;2x) \;+\; \cos \;3x}{(\sin \;4x \;+\; \sin \;2x) \;+\; \sin \;3x}$

Use the sum-to-product formulas:

$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

Apply these formulas to the grouped terms in the numerator and denominator with $C = 4x$ and $D = 2x$:

$\cos \;4x \;+\; \cos \;2x = 2 \cos\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) = 2 \cos(3x) \cos(x)$

$\sin \;4x \;+\; \sin \;2x = 2 \sin\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) = 2 \sin(3x) \cos(x)$

Substitute these results back into the LHS expression:

$LHS = \frac{2 \cos(3x) \cos(x) \;+\; \cos \;3x}{2 \sin(3x) \cos(x) \;+\; \sin \;3x}$

Factor out the common term $\cos 3x$ from the numerator and $\sin 3x$ from the denominator:

$LHS = \frac{\cos \;3x \;(2 \cos \;x \;+\; 1)}{\sin \;3x \;(2 \cos \;x \;+\; 1)}$

Assuming $2 \cos x + 1 \neq 0$, cancel the common factor $(2 \cos x + 1)$:

$LHS = \frac{\cos \;3x}{\sin \;3x}$

Using the identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$:

$LHS = \cot \;3x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the identity is proven.

To Prove:

$\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$

Proof:

We know that $3x = 2x + x$.

Consider the tangent of $3x$ written as the sum of $2x$ and $x$:

$\tan(3x) = \tan(2x + x)$

Using the tangent addition formula, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, with $A = 2x$ and $B = x$:

$\tan(3x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$

Now, convert the tangent terms to cotangent using $\tan \theta = \frac{1}{\cot \theta}$:

$\frac{1}{\cot 3x} = \frac{\frac{1}{\cot 2x} + \frac{1}{\cot x}}{1 - \frac{1}{\cot 2x} \cdot \frac{1}{\cot x}}$

Simplify the right-hand side by finding a common denominator in the numerator and the denominator:

$\frac{1}{\cot 3x} = \frac{\frac{\cot x + \cot 2x}{\cot 2x \cot x}}{1 - \frac{1}{\cot 2x \cot x}}$

$\frac{1}{\cot 3x} = \frac{\frac{\cot x + \cot 2x}{\cot 2x \cot x}}{\frac{\cot 2x \cot x - 1}{\cot 2x \cot x}}$

Multiply the numerator by the reciprocal of the denominator:

$\frac{1}{\cot 3x} = \frac{\cot x + \cot 2x}{\cot 2x \cot x} \times \frac{\cot 2x \cot x}{\cot 2x \cot x - 1}$

Assuming $\cot 2x \cot x \neq 0$, cancel the common term $\cot 2x \cot x$:

$\frac{1}{\cot 3x} = \frac{\cot x + \cot 2x}{\cot 2x \cot x - 1}$

Cross-multiply:

$1 \cdot (\cot 2x \cot x - 1) = \cot 3x \cdot (\cot x + \cot 2x)$

$\cot 2x \cot x - 1 = \cot 3x \cot x + \cot 3x \cot 2x$

Rearrange the terms to match the required identity. Move the terms involving $\cot 3x$ to the left side and $-1$ to the right side:

$\cot 2x \cot x - \cot 3x \cot x - \cot 3x \cot 2x = 1$

Rewrite the terms in the order shown in the question:

$\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\tan \;4x = \frac{4 \;\tan \;x (1\;-\;\tan^{2}\;x)}{1\;-\;6\;\tan^{2}\;x\;+\;\tan^{4}\;x}$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \tan \;4x$

We can write $4x$ as $2 \times 2x$. Using the double angle formula for tangent, $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$, with $\theta = 2x$:

LHS $= \tan (2 \times 2x) = \frac{2 \tan 2x}{1 - \tan^2 2x}$

Now, we need to express $\tan 2x$ in terms of $\tan x$. Using the same double angle formula with $\theta = x$:

$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$

Substitute this expression for $\tan 2x$ into the LHS:

LHS $= \frac{2 \left(\frac{2 \tan x}{1 - \tan^2 x}\right)}{1 - \left(\frac{2 \tan x}{1 - \tan^2 x}\right)^2}$

Simplify the numerator and the term being squared in the denominator:

Numerator $= 2 \times \frac{2 \tan x}{1 - \tan^2 x} = \frac{4 \tan x}{1 - \tan^2 x}$

$\left(\frac{2 \tan x}{1 - \tan^2 x}\right)^2 = \frac{(2 \tan x)^2}{(1 - \tan^2 x)^2} = \frac{4 \tan^2 x}{(1 - \tan^2 x)^2}$

Substitute these back into the LHS expression:

LHS $= \frac{\frac{4 \tan x}{1 - \tan^2 x}}{1 - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2}}$

Combine the terms in the denominator by finding a common denominator, which is $(1 - \tan^2 x)^2$:

Denominator $= \frac{(1 - \tan^2 x)^2}{(1 - \tan^2 x)^2} - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2} = \frac{(1 - \tan^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2}

Expand the term $(1 - \tan^2 x)^2 = 1 - 2 \tan^2 x + \tan^4 x$:

Denominator $= \frac{(1 - 2 \tan^2 x + \tan^4 x) - 4 \tan^2 x}{(1 - \tan^2 x)^2} = \frac{1 - 6 \tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}

Now, substitute the simplified numerator and denominator back into the LHS expression:

LHS $= \frac{\frac{4 \tan x}{1 - \tan^2 x}}{\frac{1 - 6 \tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}}$

To divide the fractions, multiply the numerator by the reciprocal of the denominator:

LHS $= \frac{4 \tan x}{1 - \tan^2 x} \times \frac{(1 - \tan^2 x)^2}{1 - 6 \tan^2 x + \tan^4 x}$

Cancel out one factor of $(1 - \tan^2 x)$ from the numerator and the denominator:

LHS $= \frac{4 \tan x}{\cancel{1 - \tan^2 x}} \times \frac{(1 - \tan^2 x)^{\cancel{2}}}{1 - 6 \tan^2 x + \tan^4 x}

LHS $= \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x}$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cos 4x = 1 – 8\sin^2 x \cos^2 x$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos 4x$

We can write $4x$ as $2 \times 2x$. Using the double angle formula for cosine, $\cos 2\theta = 1 - 2\sin^2 \theta$, with $\theta = 2x$:

LHS $= \cos (2 \times 2x)$

LHS $= 1 - 2\sin^2(2x)$

Now, we use the double angle formula for sine, $\sin 2x = 2\sin x \cos x$:

Substitute this expression for $\sin 2x$ into the LHS:

LHS $= 1 - 2(2\sin x \cos x)^2$

Square the term inside the parenthesis:

LHS $= 1 - 2(4\sin^2 x \cos^2 x)$

Multiply by 2:

LHS $= 1 - 8\sin^2 x \cos^2 x$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

To Prove:

$\cos 6x = 32 \cos^6 x – 48\cos^4 x + 18 \cos^2 x – 1$

Proof:

Consider the Left Hand Side (LHS) of the equation:

LHS $= \cos 6x$

We can write $\cos 6x$ as $\cos(3 \times 2x)$. Using the triple angle formula for cosine, $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$, with $\theta = 2x$:

LHS $= \cos(3 \cdot 2x) = 4\cos^3(2x) - 3\cos(2x)$

Now, we need to express $\cos 2x$ in terms of $\cos x$. Using the double angle formula for cosine, $\cos 2x = 2\cos^2 x - 1$:

Substitute this expression for $\cos 2x$ into the LHS:

LHS $= 4(2\cos^2 x - 1)^3 - 3(2\cos^2 x - 1)$

Expand $(2\cos^2 x - 1)^3$ using the binomial expansion $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, with $a = 2\cos^2 x$ and $b = 1$:

$(2\cos^2 x - 1)^3 = (2\cos^2 x)^3 - 3(2\cos^2 x)^2(1) + 3(2\cos^2 x)(1)^2 - 1^3$

$ = 8\cos^6 x - 3(4\cos^4 x) + 6\cos^2 x - 1$

$ = 8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1$

Substitute this expanded form back into the LHS expression:

LHS $= 4(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) - 3(2\cos^2 x - 1)$

Distribute the 4 and the -3:

LHS $= 32\cos^6 x - 48\cos^4 x + 24\cos^2 x - 4 - 6\cos^2 x + 3$

Combine the like terms ($24\cos^2 x - 6\cos^2 x$ and $-4 + 3$):

LHS $= 32\cos^6 x - 48\cos^4 x + (24 - 6)\cos^2 x + (-4 + 3)$

LHS $= 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$

This is the Right Hand Side (RHS) of the given equation.

Since LHS $=$ RHS, the given identity is proven.

The equation is $\sin x = \frac{\sqrt{3}}{2}$.

We know that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.

Since $\frac{\pi}{3}$ lies in the interval $[0, 2\pi)$, it is one principal solution.

Also, we know that $\sin (\pi - \theta) = \sin \theta$.

Therefore, $\sin (\pi - \frac{\pi}{3}) = \sin \frac{2\pi}{3} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.

Since $\frac{2\pi}{3}$ also lies in the interval $[0, 2\pi)$, it is another principal solution.

The principal solutions of $\sin x = \frac{\sqrt{3}}{2}$ are the values of $x \in [0, 2\pi)$ that satisfy the equation.

These are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.

Thus, the principal solutions are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.

The given equation is $\tan x = -\frac{1}{\sqrt{3}}$.

We know that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

Since $\tan x$ is negative, the principal solutions lie in the second and fourth quadrants.

In the second quadrant, the angle is $\pi - \theta$. Using the reference angle $\frac{\pi}{6}$, we get:

$x = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

We check that $\tan \frac{5\pi}{6} = \tan (\pi - \frac{\pi}{6}) = -\tan \frac{\pi}{6} = -\frac{1}{\sqrt{3}}$.

Since $0 \leq \frac{5\pi}{6} < 2\pi$, $\frac{5\pi}{6}$ is a principal solution.

In the fourth quadrant, the angle is $2\pi - \theta$. Using the reference angle $\frac{\pi}{6}$, we get:

$x = 2\pi - \frac{\pi}{6} = \frac{12\pi - \pi}{6} = \frac{11\pi}{6}$.

We check that $\tan \frac{11\pi}{6} = \tan (2\pi - \frac{\pi}{6}) = -\tan \frac{\pi}{6} = -\frac{1}{\sqrt{3}}$.

Since $0 \leq \frac{11\pi}{6} < 2\pi$, $\frac{11\pi}{6}$ is a principal solution.

The principal solutions of $\tan x = -\frac{1}{\sqrt{3}}$ are the values of $x \in [0, 2\pi)$ that satisfy the equation.

These are $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$.

Thus, the principal solutions are $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$.

The given equation is $\sin x = -\frac{\sqrt{3}}{2}$.

We know that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.

Since $\sin x$ is negative, the values of $x$ lie in the third and fourth quadrants.

We can write $-\frac{\sqrt{3}}{2}$ as $\sin (-\frac{\pi}{3})$.

So, the equation becomes $\sin x = \sin (-\frac{\pi}{3})$.

The general solution for $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$.

Here, $y = -\frac{\pi}{3}$. Substituting this value into the general solution formula, we get:

$x = n\pi + (-1)^n (-\frac{\pi}{3})$

$x = n\pi - (-1)^n \frac{\pi}{3}$

$x = n\pi + (-1)^{n+1} \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = n\pi + (-1)^{n+1} \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

The given equation is $\cos x = \frac{1}{2}$.

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

So, the equation can be written as $\cos x = \cos \frac{\pi}{3}$.

The general solution for $\cos x = \cos y$ is given by $x = 2n\pi \pm y$, where $n \in \mathbb{Z}$.

Here, $y = \frac{\pi}{3}$. Substituting this value into the general solution formula, we get:

$x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

The given equation is $\tan 2x = -\cot \left( x+\frac{\pi}{3} \right)$.

We know that $\cot \theta = \tan \left( \frac{\pi}{2} - \theta \right)$.

So, $\cot \left( x+\frac{\pi}{3} \right) = \tan \left( \frac{\pi}{2} - \left( x+\frac{\pi}{3} \right) \right)$

$= \tan \left( \frac{\pi}{2} - x - \frac{\pi}{3} \right)$

$= \tan \left( \frac{3\pi - 2\pi}{6} - x \right)$

$= \tan \left( \frac{\pi}{6} - x \right)$.

The equation becomes $\tan 2x = -\tan \left( \frac{\pi}{6} - x \right)$.

We know that $-\tan \theta = \tan (-\theta)$.

So, $-\tan \left( \frac{\pi}{6} - x \right) = \tan \left( -\left( \frac{\pi}{6} - x \right) \right) = \tan \left( x - \frac{\pi}{6} \right)$.

Alternatively, we can use $-\tan \theta = \tan(\pi - \theta)$.

$-\tan \left( \frac{\pi}{6} - x \right) = \tan \left( \pi - \left( \frac{\pi}{6} - x \right) \right)$

$= \tan \left( \pi - \frac{\pi}{6} + x \right)$

$= \tan \left( \frac{5\pi}{6} + x \right)$.

Let's use the second form: $\tan 2x = \tan \left( x + \frac{5\pi}{6} \right)$.

The general solution for $\tan A = \tan B$ is $A = n\pi + B$, where $n \in \mathbb{Z}$.

Here, $A = 2x$ and $B = x + \frac{5\pi}{6}$.

So, $2x = n\pi + x + \frac{5\pi}{6}$.

Now, we solve for $x$:

$2x - x = n\pi + \frac{5\pi}{6}$

$x = n\pi + \frac{5\pi}{6}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = n\pi + \frac{5\pi}{6}$, where $n \in \mathbb{Z}$.

The given equation is $\sin 2x – \sin 4x + \sin 6x = 0$.

Rearrange the terms:

$\sin 6x + \sin 2x - \sin 4x = 0$

Use the sum-to-product formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ on $(\sin 6x + \sin 2x)$.

$C = 6x$, $D = 2x$.

$\frac{C+D}{2} = \frac{6x+2x}{2} = \frac{8x}{2} = 4x$

$\frac{C-D}{2} = \frac{6x-2x}{2} = \frac{4x}{2} = 2x$

So, $\sin 6x + \sin 2x = 2 \sin 4x \cos 2x$.

Substitute this back into the equation:

$2 \sin 4x \cos 2x - \sin 4x = 0$

Factor out $\sin 4x$:

$\sin 4x (2 \cos 2x - 1) = 0$

This equation holds if either $\sin 4x = 0$ or $2 \cos 2x - 1 = 0$.

Case 1: $\sin 4x = 0$

The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n \in \mathbb{Z}$.

Here, $\theta = 4x$.

$4x = n\pi$

$x = \frac{n\pi}{4}$, where $n \in \mathbb{Z}$.

Case 2: $2 \cos 2x - 1 = 0$

$2 \cos 2x = 1$

$\cos 2x = \frac{1}{2}$

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

So, $\cos 2x = \cos \frac{\pi}{3}$.

The general solution for $\cos \theta = \cos y$ is $\theta = 2m\pi \pm y$, where $m \in \mathbb{Z}$.

Here, $\theta = 2x$ and $y = \frac{\pi}{3}$.

$2x = 2m\pi \pm \frac{\pi}{3}$

$x = m\pi \pm \frac{\pi}{6}$, where $m \in \mathbb{Z}$.

Combining the solutions from both cases, the general solution is:

$x = \frac{n\pi}{4}$ or $x = m\pi \pm \frac{\pi}{6}$, where $n, m \in \mathbb{Z}$.

The given equation is $2 \cos^2 x + 3 \sin x = 0$.

We know the identity $\cos^2 x + \sin^2 x = 1$, which means $\cos^2 x = 1 - \sin^2 x$.

Substitute this into the given equation:

$2(1 - \sin^2 x) + 3 \sin x = 0$

$2 - 2 \sin^2 x + 3 \sin x = 0$

Multiply by -1 and rearrange the terms:

$2 \sin^2 x - 3 \sin x - 2 = 0$

This is a quadratic equation in $\sin x$. Let $y = \sin x$. The equation becomes $2y^2 - 3y - 2 = 0$.

We can factor this quadratic equation:

$2y^2 - 4y + y - 2 = 0$

$2y(y - 2) + 1(y - 2) = 0$

$(y - 2)(2y + 1) = 0$

Substitute back $y = \sin x$:

$(\sin x - 2)(2 \sin x + 1) = 0$

This gives two possible cases:

Case 1: $\sin x - 2 = 0 \implies \sin x = 2$.

Since the range of $\sin x$ is $[-1, 1]$, there is no real value of $x$ for which $\sin x = 2$. Thus, this case yields no solution.

Case 2: $2 \sin x + 1 = 0 \implies 2 \sin x = -1 \implies \sin x = -\frac{1}{2}$.

We need to find the general solution for $\sin x = -\frac{1}{2}$.

We know that $\sin \frac{\pi}{6} = \frac{1}{2}$.

Since $\sin x$ is negative, the principal value lies in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, which is $-\frac{\pi}{6}$, because $\sin (-\frac{\pi}{6}) = -\sin \frac{\pi}{6} = -\frac{1}{2}$.

The general solution for $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$.

Here, $y = -\frac{\pi}{6}$. Substituting this value into the general solution formula, we get:

$x = n\pi + (-1)^n (-\frac{\pi}{6})$

$x = n\pi - (-1)^n \frac{\pi}{6}$

$x = n\pi + (-1)^{n+1} \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = n\pi + (-1)^{n+1} \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

The given equation is $\tan x = \sqrt{3}$.

Principal Solutions:

We know that $\tan \frac{\pi}{3} = \sqrt{3}$.

The tangent function is positive in the first and third quadrants.

The principal value in the first quadrant is $x = \frac{\pi}{3}$.

We check that $0 \leq \frac{\pi}{3} < 2\pi$, so $\frac{\pi}{3}$ is a principal solution.

In the third quadrant, the angle is $\pi + \theta$. Using the reference angle $\frac{\pi}{3}$, we get:

$x = \pi + \frac{\pi}{3} = \frac{3\pi + \pi}{3} = \frac{4\pi}{3}$.

We check that $0 \leq \frac{4\pi}{3} < 2\pi$, so $\frac{4\pi}{3}$ is a principal solution.

The principal solutions are the values of $x \in [0, 2\pi)$ that satisfy the equation. These are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$.

General Solution:

The equation is $\tan x = \sqrt{3}$.

We can write this as $\tan x = \tan \frac{\pi}{3}$.

The general solution for $\tan x = \tan y$ is given by $x = n\pi + y$, where $n \in \mathbb{Z}$.

Here, $y = \frac{\pi}{3}$. Substituting this value into the general solution formula, we get:

$x = n\pi + \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

Thus, the principal solutions are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$, and the general solution is $x = n\pi + \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

The given equation is $\sec x = 2$.

We can rewrite the equation in terms of $\cos x$ since $\sec x = \frac{1}{\cos x}$.

So, $\frac{1}{\cos x} = 2$, which implies $\cos x = \frac{1}{2}$.

Principal Solutions:

We need to find the values of $x \in [0, 2\pi)$ such that $\cos x = \frac{1}{2}$.

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

Since $\cos x$ is positive, the principal solutions lie in the first and fourth quadrants.

In the first quadrant, the angle is $x = \frac{\pi}{3}$.

We check that $0 \leq \frac{\pi}{3} < 2\pi$, so $\frac{\pi}{3}$ is a principal solution.

In the fourth quadrant, the angle is $2\pi - \theta$. Using the reference angle $\frac{\pi}{3}$, we get:

$x = 2\pi - \frac{\pi}{3} = \frac{6\pi - \pi}{3} = \frac{5\pi}{3}$.

We check that $0 \leq \frac{5\pi}{3} < 2\pi$, so $\frac{5\pi}{3}$ is a principal solution.

The principal solutions are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

General Solution:

The equation is $\cos x = \frac{1}{2}$.

We can write this as $\cos x = \cos \frac{\pi}{3}$.

The general solution for $\cos x = \cos y$ is given by $x = 2n\pi \pm y$, where $n \in \mathbb{Z}$.

Here, $y = \frac{\pi}{3}$. Substituting this value into the general solution formula, we get:

$x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

The given equation is $\cot x = -\sqrt{3}$.

Principal Solutions:

We need to find the values of $x \in [0, 2\pi)$ such that $\cot x = -\sqrt{3}$.

We know that $\cot \frac{\pi}{6} = \sqrt{3}$.

Since $\cot x$ is negative, the principal solutions lie in the second and fourth quadrants.

In the second quadrant, the angle is $\pi - \theta$. Using the reference angle $\frac{\pi}{6}$, we get:

$x = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

We check that $\cot \frac{5\pi}{6} = \cot (\pi - \frac{\pi}{6}) = -\cot \frac{\pi}{6} = -\sqrt{3}$.

Since $0 \leq \frac{5\pi}{6} < 2\pi$, $\frac{5\pi}{6}$ is a principal solution.

In the fourth quadrant, the angle is $2\pi - \theta$. Using the reference angle $\frac{\pi}{6}$, we get:

$x = 2\pi - \frac{\pi}{6} = \frac{12\pi - \pi}{6} = \frac{11\pi}{6}$.

We check that $\cot \frac{11\pi}{6} = \cot (2\pi - \frac{\pi}{6}) = -\cot \frac{\pi}{6} = -\sqrt{3}$.

Since $0 \leq \frac{11\pi}{6} < 2\pi$, $\frac{11\pi}{6}$ is a principal solution.

The principal solutions are $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$.

General Solution:

The equation is $\cot x = -\sqrt{3}$.

We can write this as $\cot x = \cot \frac{5\pi}{6}$ (using the principal value from $[0, \pi)$).

The general solution for $\cot x = \cot y$ is given by $x = n\pi + y$, where $n \in \mathbb{Z}$.

Here, $y = \frac{5\pi}{6}$. Substituting this value into the general solution formula, we get:

$x = n\pi + \frac{5\pi}{6}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = n\pi + \frac{5\pi}{6}$, where $n \in \mathbb{Z}$.

The given equation is $\text{cosec } x = -2$.

We can rewrite the equation in terms of $\sin x$ since $\text{cosec } x = \frac{1}{\sin x}$.

So, $\frac{1}{\sin x} = -2$, which implies $\sin x = -\frac{1}{2}$.

Principal Solutions:

We need to find the values of $x \in [0, 2\pi)$ such that $\sin x = -\frac{1}{2}$.

We know that $\sin \frac{\pi}{6} = \frac{1}{2}$.

Since $\sin x$ is negative, the principal solutions lie in the third and fourth quadrants.

In the third quadrant, the angle is $\pi + \theta$. Using the reference angle $\frac{\pi}{6}$, we get:

$x = \pi + \frac{\pi}{6} = \frac{6\pi + \pi}{6} = \frac{7\pi}{6}$.

We check that $\sin \frac{7\pi}{6} = \sin (\pi + \frac{\pi}{6}) = -\sin \frac{\pi}{6} = -\frac{1}{2}$.

Since $0 \leq \frac{7\pi}{6} < 2\pi$, $\frac{7\pi}{6}$ is a principal solution.

In the fourth quadrant, the angle is $2\pi - \theta$. Using the reference angle $\frac{\pi}{6}$, we get:

$x = 2\pi - \frac{\pi}{6} = \frac{12\pi - \pi}{6} = \frac{11\pi}{6}$.

We check that $\sin \frac{11\pi}{6} = \sin (2\pi - \frac{\pi}{6}) = -\sin \frac{\pi}{6} = -\frac{1}{2}$.

Since $0 \leq \frac{11\pi}{6} < 2\pi$, $\frac{11\pi}{6}$ is a principal solution.

The principal solutions are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.

General Solution:

The equation is $\sin x = -\frac{1}{2}$.

We know that $\sin (-\frac{\pi}{6}) = -\frac{1}{2}$.

So, we can write the equation as $\sin x = \sin (-\frac{\pi}{6})$.

The general solution for $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$.

Here, $y = -\frac{\pi}{6}$. Substituting this value into the general solution formula, we get:

$x = n\pi + (-1)^n (-\frac{\pi}{6})$

$x = n\pi - (-1)^n \frac{\pi}{6}$

$x = n\pi + (-1)^{n+1} \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

Thus, the general solution is $x = n\pi + (-1)^{n+1} \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

The given equation is $\cos 4x = \cos 2x$.

The general solution for $\cos A = \cos B$ is given by $A = 2n\pi \pm B$, where $n \in \mathbb{Z}$.

Here, $A = 4x$ and $B = 2x$. Substituting these values, we get:

$4x = 2n\pi \pm 2x$, where $n \in \mathbb{Z}$.

We need to consider the two cases arising from the $\pm$ sign.

Case 1: $4x = 2n\pi + 2x$

$4x - 2x = 2n\pi$

$2x = 2n\pi$

Divide both sides by 2:

$x = n\pi$, where $n \in \mathbb{Z}$.

Case 2: $4x = 2n\pi - 2x$

$4x + 2x = 2n\pi$

$6x = 2n\pi$

Divide both sides by 6:

$x = \frac{2n\pi}{6}$

$x = \frac{n\pi}{3}$, where $n \in \mathbb{Z}$.

The solutions are $x = n\pi$ and $x = \frac{n\pi}{3}$.

Notice that the solutions from Case 1 ($x=n\pi$) are included in the solutions from Case 2 ($x=\frac{n\pi}{3}$) when $n$ is a multiple of 3. For example, if $n=3k$ in the second case, $x = \frac{3k\pi}{3} = k\pi$.

So the general solution is given by the second case only.

Thus, the general solution is $x = \frac{n\pi}{3}$, where $n \in \mathbb{Z}$.

The given equation is $\cos 3x + \cos x – \cos 2x = 0$.

Group the first two terms and apply the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$.

Here, $A = 3x$ and $B = x$.

$\frac{A+B}{2} = \frac{3x+x}{2} = \frac{4x}{2} = 2x$

$\frac{A-B}{2} = \frac{3x-x}{2} = \frac{2x}{2} = x$

So, $\cos 3x + \cos x = 2 \cos 2x \cos x$.

Substitute this into the equation:

$2 \cos 2x \cos x - \cos 2x = 0$

Factor out $\cos 2x$:

$\cos 2x (2 \cos x - 1) = 0$

This equation holds if either $\cos 2x = 0$ or $2 \cos x - 1 = 0$.

Case 1: $\cos 2x = 0$

The general solution for $\cos \theta = 0$ is $\theta = (2n+1)\frac{\pi}{2}$, where $n \in \mathbb{Z}$.

Here, $\theta = 2x$.

$2x = (2n+1)\frac{\pi}{2}$

Divide by 2:

$x = (2n+1)\frac{\pi}{4}$, where $n \in \mathbb{Z}$.

Case 2: $2 \cos x - 1 = 0$

$2 \cos x = 1$

$\cos x = \frac{1}{2}$

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

So, the equation is $\cos x = \cos \frac{\pi}{3}$.

The general solution for $\cos x = \cos y$ is $x = 2m\pi \pm y$, where $m \in \mathbb{Z}$.

Here, $y = \frac{\pi}{3}$.

$x = 2m\pi \pm \frac{\pi}{3}$, where $m \in \mathbb{Z}$.

Combining the solutions from both cases, the general solution is:

$x = (2n+1)\frac{\pi}{4}$ or $x = 2m\pi \pm \frac{\pi}{3}$, where $n, m \in \mathbb{Z}$.

The given equation is $\sin 2x + \cos x = 0$.

Use the double angle identity for $\sin 2x$: $\sin 2x = 2 \sin x \cos x$.

Substitute this into the equation:

$2 \sin x \cos x + \cos x = 0$

Factor out the common term $\cos x$:

$\cos x (2 \sin x + 1) = 0$

This equation holds if either $\cos x = 0$ or $2 \sin x + 1 = 0$.

Case 1: $\cos x = 0$

The general solution for $\cos x = 0$ is $x = (2n+1)\frac{\pi}{2}$, where $n \in \mathbb{Z}$.

Case 2: $2 \sin x + 1 = 0$

$2 \sin x = -1$

$\sin x = -\frac{1}{2}$

We know that $\sin \frac{\pi}{6} = \frac{1}{2}$. Since $\sin x$ is negative, the principal value can be taken as $-\frac{\pi}{6}$.

So, the equation is $\sin x = \sin (-\frac{\pi}{6})$.

The general solution for $\sin x = \sin y$ is given by $x = m\pi + (-1)^m y$, where $m \in \mathbb{Z}$.

Here, $y = -\frac{\pi}{6}$. Substituting this value, we get:

$x = m\pi + (-1)^m (-\frac{\pi}{6})$

$x = m\pi - (-1)^m \frac{\pi}{6}$

$x = m\pi + (-1)^{m+1} \frac{\pi}{6}$, where $m \in \mathbb{Z}$.

Combining the solutions from both cases, the general solution is:

$x = (2n+1)\frac{\pi}{2}$ or $x = m\pi + (-1)^{m+1} \frac{\pi}{6}$, where $n, m \in \mathbb{Z}$.

The given equation is $\sec^2 2x = 1 – \tan 2x$.

We use the identity $\sec^2 \theta = 1 + \tan^2 \theta$. Here, $\theta = 2x$.

So, $\sec^2 2x = 1 + \tan^2 2x$.

Substitute this into the given equation:

$1 + \tan^2 2x = 1 - \tan 2x$

Rearrange the terms to form a quadratic equation in $\tan 2x$:

$\tan^2 2x + \tan 2x + 1 - 1 = 0$

$\tan^2 2x + \tan 2x = 0$

Factor out $\tan 2x$:

$\tan 2x (\tan 2x + 1) = 0$

This equation holds if either $\tan 2x = 0$ or $\tan 2x + 1 = 0$.

Case 1: $\tan 2x = 0$

The general solution for $\tan \theta = 0$ is $\theta = n\pi$, where $n \in \mathbb{Z}$.

Here, $\theta = 2x$.

$2x = n\pi$

Divide both sides by 2:

$x = \frac{n\pi}{2}$, where $n \in \mathbb{Z}$.

Case 2: $\tan 2x + 1 = 0$

$\tan 2x = -1$

We know that $\tan \frac{\pi}{4} = 1$. Since $\tan 2x$ is negative, we can use the angle in the second quadrant: $\tan (\pi - \frac{\pi}{4}) = \tan \frac{3\pi}{4} = -1$.

So, the equation is $\tan 2x = \tan \frac{3\pi}{4}$.

The general solution for $\tan \theta = \tan y$ is $\theta = m\pi + y$, where $m \in \mathbb{Z}$.

Here, $\theta = 2x$ and $y = \frac{3\pi}{4}$.

$2x = m\pi + \frac{3\pi}{4}$

Divide both sides by 2:

$x = \frac{m\pi}{2} + \frac{3\pi}{8}$, where $m \in \mathbb{Z}$.

Combining the solutions from both cases, the general solution is:

$x = \frac{n\pi}{2}$ or $x = \frac{m\pi}{2} + \frac{3\pi}{8}$, where $n, m \in \mathbb{Z}$.

The given equation is $\sin x + \sin 3x + \sin 5x = 0$.

Rearrange the terms to group $\sin 5x$ and $\sin x$:

$\sin 5x + \sin x + \sin 3x = 0$

Apply the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ to the first two terms ($\sin 5x + \sin x$).

Let $A = 5x$ and $B = x$.

$\frac{A+B}{2} = \frac{5x+x}{2} = \frac{6x}{2} = 3x$

$\frac{A-B}{2} = \frac{5x-x}{2} = \frac{4x}{2} = 2x$

So, $\sin 5x + \sin x = 2 \sin 3x \cos 2x$.

Substitute this back into the equation:

$2 \sin 3x \cos 2x + \sin 3x = 0$

Factor out the common term $\sin 3x$:

$\sin 3x (2 \cos 2x + 1) = 0$

This equation is satisfied if either $\sin 3x = 0$ or $2 \cos 2x + 1 = 0$.

Case 1: $\sin 3x = 0$

The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n \in \mathbb{Z}$.

Here, $\theta = 3x$.

$3x = n\pi$

Divide both sides by 3:

$x = \frac{n\pi}{3}$, where $n \in \mathbb{Z}$.

Case 2: $2 \cos 2x + 1 = 0$

$2 \cos 2x = -1$

$\cos 2x = -\frac{1}{2}$

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$. Since $\cos 2x$ is negative, the angles are in the second and third quadrants. We can write $-\frac{1}{2}$ as $\cos \frac{2\pi}{3}$, because $\cos \frac{2\pi}{3} = \cos (\pi - \frac{\pi}{3}) = -\cos \frac{\pi}{3} = -\frac{1}{2}$.

So, the equation is $\cos 2x = \cos \frac{2\pi}{3}$.

The general solution for $\cos \theta = \cos y$ is $\theta = 2m\pi \pm y$, where $m \in \mathbb{Z}$.

Here, $\theta = 2x$ and $y = \frac{2\pi}{3}$.

$2x = 2m\pi \pm \frac{2\pi}{3}$

Divide both sides by 2:

$x = \frac{2m\pi}{2} \pm \frac{2\pi}{6}$

$x = m\pi \pm \frac{\pi}{3}$, where $m \in \mathbb{Z}$.

Combining the solutions from both cases, the general solution is:

$x = \frac{n\pi}{3}$ or $x = m\pi \pm \frac{\pi}{3}$, where $n, m \in \mathbb{Z}$.

Given that $\sin x = \frac{3}{5}$ and $\cos y = -\frac{12}{13}$.

Both $x$ and $y$ lie in the second quadrant.

In the second quadrant, $\sin x$ is positive and $\cos x$ is negative. Also, $\sin y$ is positive and $\cos y$ is negative.

We need to find $\cos x$ and $\sin y$.

For angle $x$ in the second quadrant, we use the identity $\sin^2 x + \cos^2 x = 1$.

$\cos^2 x = 1 - \sin^2 x$

$\cos^2 x = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$

$\cos x = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$.

Since $x$ is in the second quadrant, $\cos x$ is negative.

So, $\cos x = -\frac{4}{5}$.

For angle $y$ in the second quadrant, we use the identity $\sin^2 y + \cos^2 y = 1$.

$\sin^2 y = 1 - \cos^2 y$

$\sin^2 y = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$

$\sin y = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$.

Since $y$ is in the second quadrant, $\sin y$ is positive.

So, $\sin y = \frac{5}{13}$.

Now we want to find $\sin(x+y)$. The formula for $\sin(x+y)$ is:

$\sin(x+y) = \sin x \cos y + \cos x \sin y$

Substitute the values we found:

$\sin(x+y) = \left(\frac{3}{5}\right) \left(-\frac{12}{13}\right) + \left(-\frac{4}{5}\right) \left(\frac{5}{13}\right)$

$\sin(x+y) = -\frac{3 \times 12}{5 \times 13} + \left(-\frac{4 \times 5}{5 \times 13}\right)$

$\sin(x+y) = -\frac{36}{65} - \frac{20}{65}$

$\sin(x+y) = \frac{-36 - 20}{65}$

$\sin(x+y) = \frac{-56}{65}$

$\sin(x+y) = -\frac{56}{65}$.

Thus, the value of $\sin(x+y)$ is $-\frac{56}{65}$.

We need to prove the identity $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$.

Consider the Left Hand Side (LHS):

LHS $= \cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2}$

We use the product-to-sum formula: $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.

Applying this to the first term $\cos 2x \cos \frac{x}{2}$:

$A = 2x, B = \frac{x}{2}$

$A+B = 2x + \frac{x}{2} = \frac{5x}{2}$

$A-B = 2x - \frac{x}{2} = \frac{3x}{2}$

So, $\cos 2x \cos \frac{x}{2} = \frac{1}{2} \left( \cos \frac{5x}{2} + \cos \frac{3x}{2} \right)$.

Applying the same formula to the second term $\cos 3x \cos \frac{9x}{2}$:

$A = 3x, B = \frac{9x}{2}$

$A+B = 3x + \frac{9x}{2} = \frac{15x}{2}$

$A-B = 3x - \frac{9x}{2} = -\frac{3x}{2}$

So, $\cos 3x \cos \frac{9x}{2} = \frac{1}{2} \left( \cos \frac{15x}{2} + \cos \left(-\frac{3x}{2}\right) \right)$.

Since $\cos(-\theta) = \cos \theta$, we have $\cos(-\frac{3x}{2}) = \cos \frac{3x}{2}$.

Thus, $\cos 3x \cos \frac{9x}{2} = \frac{1}{2} \left( \cos \frac{15x}{2} + \cos \frac{3x}{2} \right)$.

Substitute these results back into the LHS:

LHS $= \frac{1}{2} \left( \cos \frac{5x}{2} + \cos \frac{3x}{2} \right) - \frac{1}{2} \left( \cos \frac{15x}{2} + \cos \frac{3x}{2} \right)$

LHS $= \frac{1}{2} \left( \cos \frac{5x}{2} + \cos \frac{3x}{2} - \cos \frac{15x}{2} - \cos \frac{3x}{2} \right)$

LHS $= \frac{1}{2} \left( \cos \frac{5x}{2} - \cos \frac{15x}{2} \right)$

Now, we use the difference-to-product formula: $\cos C - \cos D = 2 \sin \frac{D+C}{2} \sin \frac{D-C}{2}$.

Here, $C = \frac{5x}{2}$ and $D = \frac{15x}{2}$.

$\frac{D+C}{2} = \frac{\frac{15x}{2} + \frac{5x}{2}}{2} = \frac{\frac{20x}{2}}{2} = \frac{10x}{2} = 5x$

$\frac{D-C}{2} = \frac{\frac{15x}{2} - \frac{5x}{2}}{2} = \frac{\frac{10x}{2}}{2} = \frac{5x}{2}$

Substitute these into the expression for LHS:

LHS $= \frac{1}{2} \left( 2 \sin \left( 5x \right) \sin \left( \frac{5x}{2} \right) \right)$

LHS $= \sin \left( 5x \right) \sin \left( \frac{5x}{2} \right)$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

We want to find the value of $\tan \frac{\pi}{8}$.

We notice that $\frac{\pi}{8}$ is half of $\frac{\pi}{4}$. We can use the half-angle formula for tangent.

The half-angle formula is given by $\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}$.

Let $\theta = \frac{\pi}{4}$. Then $\frac{\theta}{2} = \frac{\pi}{8}$.

We know the values of $\sin \frac{\pi}{4}$ and $\cos \frac{\pi}{4}$:

$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Substitute $\theta = \frac{\pi}{4}$ into the half-angle formula:

$\tan \frac{\pi}{8} = \tan \frac{\frac{\pi}{4}}{2} = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}$

Substitute the values of $\cos \frac{\pi}{4}$ and $\sin \frac{\pi}{4}$:

$\tan \frac{\pi}{8} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$

Simplify the expression. Multiply the numerator and denominator by $\sqrt{2}$:

$\tan \frac{\pi}{8} = \frac{\sqrt{2} \left( 1 - \frac{1}{\sqrt{2}} \right)}{\sqrt{2} \left( \frac{1}{\sqrt{2}} \right)}$

$\tan \frac{\pi}{8} = \frac{\sqrt{2} \times 1 - \sqrt{2} \times \frac{1}{\sqrt{2}}}{1}$

$\tan \frac{\pi}{8} = \frac{\sqrt{2} - 1}{1}$

$\tan \frac{\pi}{8} = \sqrt{2} - 1$.

Since $\frac{\pi}{8}$ is in the first quadrant ($0 < \frac{\pi}{8} < \frac{\pi}{2}$), $\tan \frac{\pi}{8}$ must be positive. The value $\sqrt{2} - 1$ is positive as $\sqrt{2} \approx 1.414 > 1$.

Thus, the value of $\tan \frac{\pi}{8}$ is $\sqrt{2} - 1$.

Given that $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$.

The condition $\pi < x < \frac{3\pi}{2}$ means that the angle $x$ lies in the third quadrant.

In the third quadrant, $\tan x$ is positive, $\sin x$ is negative, and $\cos x$ is negative.

We know that $1 + \tan^2 x = \sec^2 x$.

$\sec^2 x = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{16 + 9}{16} = \frac{25}{16}$.

$\sec x = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}$.

Since $x$ is in the third quadrant, $\cos x$ is negative, and thus $\sec x = \frac{1}{\cos x}$ is also negative.

So, $\sec x = -\frac{5}{4}$.

From $\sec x = -\frac{5}{4}$, we find $\cos x = \frac{1}{\sec x} = -\frac{4}{5}$.

Now we find $\sin x$. We know that $\tan x = \frac{\sin x}{\cos x}$.

$\sin x = \tan x \times \cos x = \left(\frac{3}{4}\right) \times \left(-\frac{4}{5}\right) = -\frac{12}{20} = -\frac{3}{5}$.

Now we need to find the values of $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$.

First, determine the range of $\frac{x}{2}$. Given $\pi < x < \frac{3\pi}{2}$, divide the inequality by 2:

$\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.

This means $\frac{x}{2}$ lies in the second quadrant.

In the second quadrant, $\sin \frac{x}{2}$ is positive, $\cos \frac{x}{2}$ is negative, and $\tan \frac{x}{2}$ is negative.

We use the half-angle formulas:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

$\sin^2 \frac{x}{2} = \frac{1 - (-\frac{4}{5})}{2} = \frac{1 + \frac{4}{5}}{2} = \frac{\frac{5+4}{5}}{2} = \frac{\frac{9}{5}}{2} = \frac{9}{10}$.

$\sin \frac{x}{2} = \sqrt{\frac{9}{10}}$ (since $\sin \frac{x}{2}$ is positive in the second quadrant)

$\sin \frac{x}{2} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$.

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

$\cos^2 \frac{x}{2} = \frac{1 + (-\frac{4}{5})}{2} = \frac{1 - \frac{4}{5}}{2} = \frac{\frac{5-4}{5}}{2} = \frac{\frac{1}{5}}{2} = \frac{1}{10}$.

$\cos \frac{x}{2} = -\sqrt{\frac{1}{10}}$ (since $\cos \frac{x}{2}$ is negative in the second quadrant)

$\cos \frac{x}{2} = -\frac{1}{\sqrt{10}} = -\frac{\sqrt{10}}{10}$.

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$

$\tan \frac{x}{2} = \frac{\frac{3}{\sqrt{10}}}{-\frac{1}{\sqrt{10}}} = -3$.

Alternatively, using the formula $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$:

$\tan \frac{x}{2} = \frac{1 - (-\frac{4}{5})}{-\frac{3}{5}} = \frac{1 + \frac{4}{5}}{-\frac{3}{5}} = \frac{\frac{9}{5}}{-\frac{3}{5}} = \frac{9}{5} \times (-\frac{5}{3}) = -\frac{9}{3} = -3$.

The sign is negative, consistent with $\frac{x}{2}$ being in the second quadrant.

The values are:

$\sin \frac{x}{2} = \frac{3\sqrt{10}}{10}$

$\cos \frac{x}{2} = -\frac{\sqrt{10}}{10}$

$\tan \frac{x}{2} = -3$

We need to prove the identity $\cos^2 x + \cos^2 \left( x+\frac{\pi}{3} \right) + \cos^2 \left( x-\frac{\pi}{3} \right) = \frac{3}{2}$.

Consider the Left Hand Side (LHS):

LHS $= \cos^2 x + \cos^2 \left( x+\frac{\pi}{3} \right) + \cos^2 \left( x-\frac{\pi}{3} \right)$

We use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$. Applying this to each term:

$\cos^2 x = \frac{1 + \cos 2x}{2}$

$\cos^2 \left( x+\frac{\pi}{3} \right) = \frac{1 + \cos 2\left( x+\frac{\pi}{3} \right)}{2} = \frac{1 + \cos \left( 2x+\frac{2\pi}{3} \right)}{2}$

$\cos^2 \left( x-\frac{\pi}{3} \right) = \frac{1 + \cos 2\left( x-\frac{\pi}{3} \right)}{2} = \frac{1 + \cos \left( 2x-\frac{2\pi}{3} \right)}{2}$

Substitute these into the LHS:

LHS $= \frac{1 + \cos 2x}{2} + \frac{1 + \cos \left( 2x+\frac{2\pi}{3} \right)}{2} + \frac{1 + \cos \left( 2x-\frac{2\pi}{3} \right)}{2}$

LHS $= \frac{1}{2} \left[ (1 + \cos 2x) + (1 + \cos \left( 2x+\frac{2\pi}{3} \right)) + (1 + \cos \left( 2x-\frac{2\pi}{3} \right)) \right]$

LHS $= \frac{1}{2} \left[ 3 + \cos 2x + \cos \left( 2x+\frac{2\pi}{3} \right) + \cos \left( 2x-\frac{2\pi}{3} \right) \right]$

We use the sum-to-product formula: $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$.

Let $A = 2x$ and $B = \frac{2\pi}{3}$. Then $\cos \left( 2x+\frac{2\pi}{3} \right) + \cos \left( 2x-\frac{2\pi}{3} \right) = 2 \cos (2x) \cos \left(\frac{2\pi}{3}\right)$.

We know that $\cos \frac{2\pi}{3} = \cos (\pi - \frac{\pi}{3}) = -\cos \frac{\pi}{3} = -\frac{1}{2}$.

So, $2 \cos (2x) \cos \left(\frac{2\pi}{3}\right) = 2 \cos (2x) \left(-\frac{1}{2}\right) = -\cos 2x$.

Substitute this back into the expression for LHS:

LHS $= \frac{1}{2} \left[ 3 + \cos 2x + (-\cos 2x) \right]$

LHS $= \frac{1}{2} \left[ 3 + \cos 2x - \cos 2x \right]$

LHS $= \frac{1}{2} [3]$

LHS $= \frac{3}{2}$.

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$.

To Prove: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13}$

Use the product-to-sum formula: $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.

Apply this to the first term $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}$. Let $A = \frac{\pi}{13}$ and $B = \frac{9 \pi}{13}$.

$2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} = \cos \left(\frac{\pi}{13} + \frac{9\pi}{13}\right) + \cos \left(\frac{\pi}{13} - \frac{9\pi}{13}\right)$

$= \cos \frac{10\pi}{13} + \cos \left(-\frac{8\pi}{13}\right)$

Since $\cos(-\theta) = \cos \theta$, we have $\cos \left(-\frac{8\pi}{13}\right) = \cos \frac{8\pi}{13}$.

So, $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} = \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13}$.

Substitute this back into the LHS:

LHS $= \left(\cos \frac{10\pi}{13} + \cos \frac{8\pi}{13}\right) + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13}$

Rearrange the terms:

LHS $= \cos \frac{10\pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{5 \pi}{13}$

Now, use the sum-to-product formula: $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$.

Apply this to the first pair $\cos \frac{10\pi}{13} + \cos \frac{3 \pi}{13}$:

Let $C = \frac{10\pi}{13}$ and $D = \frac{3 \pi}{13}$.

$\frac{C+D}{2} = \frac{\frac{10\pi}{13} + \frac{3\pi}{13}}{2} = \frac{\frac{13\pi}{13}}{2} = \frac{\pi}{2}$

$\frac{C-D}{2} = \frac{\frac{10\pi}{13} - \frac{3\pi}{13}}{2} = \frac{\frac{7\pi}{13}}{2} = \frac{7\pi}{26}$

So, $\cos \frac{10\pi}{13} + \cos \frac{3 \pi}{13} = 2 \cos \frac{\pi}{2} \cos \frac{7\pi}{26}$.

Apply the sum-to-product formula to the second pair $\cos \frac{8\pi}{13} + \cos \frac{5 \pi}{13}$:

Let $C = \frac{8\pi}{13}$ and $D = \frac{5 \pi}{13}$.

$\frac{C+D}{2} = \frac{\frac{8\pi}{13} + \frac{5\pi}{13}}{2} = \frac{\frac{13\pi}{13}}{2} = \frac{\pi}{2}$

$\frac{C-D}{2} = \frac{\frac{8\pi}{13} - \frac{5\pi}{13}}{2} = \frac{\frac{3\pi}{13}}{2} = \frac{3\pi}{26}$

So, $\cos \frac{8\pi}{13} + \cos \frac{5 \pi}{13} = 2 \cos \frac{\pi}{2} \cos \frac{3\pi}{26}$.

Substitute these results back into the expression for LHS:

LHS $= \left(2 \cos \frac{\pi}{2} \cos \frac{7\pi}{26}\right) + \left(2 \cos \frac{\pi}{2} \cos \frac{3\pi}{26}\right)$

We know that $\cos \frac{\pi}{2} = 0$.

LHS $= 2(0) \cos \frac{7\pi}{26} + 2(0) \cos \frac{3\pi}{26}$

LHS $= 0 + 0$

LHS $= 0$

This is equal to the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $(\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x = 0$.

To Prove: $(\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x = 0$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= (\sin 3x + \sin x) \sin x + (\cos 3x – \cos x) \cos x$

Distribute $\sin x$ into the first term and $\cos x$ into the second term:

LHS $= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$

Rearrange the terms:

LHS $= \cos 3x \cos x + \sin 3x \sin x - (\cos^2 x - \sin^2 x)$

We use the identities:

• $\cos A \cos B + \sin A \sin B = \cos(A-B)$
• $\cos^2 x - \sin^2 x = \cos 2x$

Apply the first identity to the terms $\cos 3x \cos x + \sin 3x \sin x$. Let $A = 3x$ and $B = x$.

$\cos 3x \cos x + \sin 3x \sin x = \cos (3x - x) = \cos 2x$.

Substitute this and the second identity back into the LHS:

LHS $= \cos 2x - (\cos 2x)$

LHS $= \cos 2x - \cos 2x$

LHS $= 0$

This is equal to the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4 \cos^2 \frac{x \;+\; y}{2}$.

To Prove: $(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4 \cos^2 \frac{x \;+\; y}{2}$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= (\cos x + \cos y)^2 + (\sin x – \sin y)^2$

Expand the squares:

$(\cos x + \cos y)^2 = \cos^2 x + 2 \cos x \cos y + \cos^2 y$

$(\sin x – \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$

Substitute these expansions back into the LHS expression:

LHS $= (\cos^2 x + 2 \cos x \cos y + \cos^2 y) + (\sin^2 x - 2 \sin x \sin y + \sin^2 y)$

Rearrange and group terms using the identity $\sin^2 \theta + \cos^2 \theta = 1$:

LHS $= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2 \cos x \cos y - 2 \sin x \sin y$

LHS $= 1 + 1 + 2 (\cos x \cos y - \sin x \sin y)$

LHS $= 2 + 2 (\cos x \cos y - \sin x \sin y)$

Use the angle subtraction formula for cosine: $\cos(A+B) = \cos A \cos B - \sin A \sin B$.

Let $A=x$ and $B=y$. Then $\cos x \cos y - \sin x \sin y = \cos(x+y)$.

Substitute this into the LHS expression:

LHS $= 2 + 2 \cos(x+y)$

Factor out 2:

LHS $= 2 (1 + \cos(x+y))$

Use the double angle identity $1 + \cos 2\theta = 2 \cos^2 \theta$. Let $2\theta = x+y$, so $\theta = \frac{x+y}{2}$.

Then $1 + \cos(x+y) = 2 \cos^2 \left(\frac{x+y}{2}\right)$.

Substitute this into the LHS expression:

LHS $= 2 \left( 2 \cos^2 \left(\frac{x+y}{2}\right) \right)$

LHS $= 4 \cos^2 \left(\frac{x+y}{2}\right)$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $(\cos x – \cos y)^2 + (\sin x – \sin y)^2 = 4 \sin^2 \frac{x \;-\; y}{2}$.

To Prove: $(\cos x – \cos y)^2 + (\sin x – \sin y)^2 = 4 \sin^2 \frac{x \;-\; y}{2}$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= (\cos x – \cos y)^2 + (\sin x – \sin y)^2$

Expand the squares:

$(\cos x – \cos y)^2 = \cos^2 x - 2 \cos x \cos y + \cos^2 y$

$(\sin x – \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$

Substitute these expansions back into the LHS expression:

LHS $= (\cos^2 x - 2 \cos x \cos y + \cos^2 y) + (\sin^2 x - 2 \sin x \sin y + \sin^2 y)$

Rearrange and group terms using the identity $\sin^2 \theta + \cos^2 \theta = 1$:

LHS $= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2 \cos x \cos y - 2 \sin x \sin y$

LHS $= 1 + 1 - 2 (\cos x \cos y + \sin x \sin y)$

LHS $= 2 - 2 (\cos x \cos y + \sin x \sin y)$

Use the angle subtraction formula for cosine: $\cos(A-B) = \cos A \cos B + \sin A \sin B$.

Let $A=x$ and $B=y$. Then $\cos x \cos y + \sin x \sin y = \cos(x-y)$.

Substitute this into the LHS expression:

LHS $= 2 - 2 \cos(x-y)$

Factor out 2:

LHS $= 2 (1 - \cos(x-y))$

Use the half-angle identity $1 - \cos 2\theta = 2 \sin^2 \theta$. Let $2\theta = x-y$, so $\theta = \frac{x-y}{2}$.

Then $1 - \cos(x-y) = 2 \sin^2 \left(\frac{x-y}{2}\right)$.

Substitute this into the LHS expression:

LHS $= 2 \left( 2 \sin^2 \left(\frac{x-y}{2}\right) \right)$

LHS $= 4 \sin^2 \left(\frac{x-y}{2}\right)$

This is the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$.

To Prove: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= \sin x + \sin 3x + \sin 5x + \sin 7x$

Group the terms: $(\sin 7x + \sin x) + (\sin 5x + \sin 3x)$.

Use the sum-to-product formula: $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$.

For the first group $(\sin 7x + \sin x)$, let $A = 7x$ and $B = x$:

$\sin 7x + \sin x = 2 \sin \frac{7x+x}{2} \cos \frac{7x-x}{2} = 2 \sin \frac{8x}{2} \cos \frac{6x}{2} = 2 \sin 4x \cos 3x$.

For the second group $(\sin 5x + \sin 3x)$, let $A = 5x$ and $B = 3x$:

$\sin 5x + \sin 3x = 2 \sin \frac{5x+3x}{2} \cos \frac{5x-3x}{2} = 2 \sin \frac{8x}{2} \cos \frac{2x}{2} = 2 \sin 4x \cos x$.

Substitute these back into the LHS:

LHS $= 2 \sin 4x \cos 3x + 2 \sin 4x \cos x$

Factor out the common term $2 \sin 4x$:

LHS $= 2 \sin 4x (\cos 3x + \cos x)$

Now, use the sum-to-product formula for cosine: $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$.

Apply this to $(\cos 3x + \cos x)$, let $C = 3x$ and $D = x$:

$\cos 3x + \cos x = 2 \cos \frac{3x+x}{2} \cos \frac{3x-x}{2} = 2 \cos \frac{4x}{2} \cos \frac{2x}{2} = 2 \cos 2x \cos x$.

Substitute this back into the LHS expression:

LHS $= 2 \sin 4x (2 \cos 2x \cos x)$

LHS $= 4 \sin 4x \cos 2x \cos x$

Rearrange the terms to match the RHS:

LHS $= 4 \cos x \cos 2x \sin 4x$

This is equal to the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $\frac{(\sin \;7x \;+\; \sin \;5x) \;+\; (\sin\; 9x \;+\; \sin \;3x)}{(\cos \;7x \;+\; \cos \;5x) \;+\; (\cos \;9x \;+\; \cos \;3x)} = \tan\;6x$.

To Prove: $\frac{(\sin \;7x \;+\; \sin \;5x) \;+\; (\sin\; 9x \;+\; \sin \;3x)}{(\cos \;7x \;+\; \cos \;5x) \;+\; (\cos \;9x \;+\; \cos \;3x)} = \tan\;6x$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= \frac{(\sin \;7x \;+\; \sin \;5x) \;+\; (\sin\; 9x \;+\; \sin \;3x)}{(\cos \;7x \;+\; \cos \;5x) \;+\; (\cos \;9x \;+\; \cos \;3x)}$

Use the sum-to-product formulas:

$\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$

$\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

Apply the sum-to-product formulas to the terms in the numerator:

$\sin 7x + \sin 5x = 2 \sin \frac{7x+5x}{2} \cos \frac{7x-5x}{2} = 2 \sin 6x \cos x$

$\sin 9x + \sin 3x = 2 \sin \frac{9x+3x}{2} \cos \frac{9x-3x}{2} = 2 \sin 6x \cos 3x$

Substitute these into the numerator:

Numerator $= (2 \sin 6x \cos x) + (2 \sin 6x \cos 3x)$

Factor out $2 \sin 6x$ from the numerator:

Numerator $= 2 \sin 6x (\cos x + \cos 3x)$

Apply the sum-to-product formulas to the terms in the denominator:

$\cos 7x + \cos 5x = 2 \cos \frac{7x+5x}{2} \cos \frac{7x-5x}{2} = 2 \cos 6x \cos x$

$\cos 9x + \cos 3x = 2 \cos \frac{9x+3x}{2} \cos \frac{9x-3x}{2} = 2 \cos 6x \cos 3x$

Substitute these into the denominator:

Denominator $= (2 \cos 6x \cos x) + (2 \cos 6x \cos 3x)$

Factor out $2 \cos 6x$ from the denominator:

Denominator $= 2 \cos 6x (\cos x + \cos 3x)$

Substitute the simplified numerator and denominator back into the LHS expression:

LHS $= \frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)}$

Assuming that $\cos x + \cos 3x \neq 0$, we can cancel the common term $(\cos x + \cos 3x)$:

LHS $= \frac{2 \sin 6x}{2 \cos 6x}$

LHS $= \frac{\sin 6x}{\cos 6x}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we have:

LHS $= \tan 6x$

This is equal to the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given: The identity $\sin 3x + \sin 2x – \sin x = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}$.

To Prove: $\sin 3x + \sin 2x – \sin x = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}$.

Proof:

Consider the Left Hand Side (LHS):

LHS $= \sin 3x + \sin 2x – \sin x$

Rearrange the terms:

LHS $= (\sin 3x - \sin x) + \sin 2x$

Use the difference-to-product formula: $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$.

Apply this to $(\sin 3x - \sin x)$ with $A = 3x$ and $B = x$:

$\sin 3x - \sin x = 2 \cos \frac{3x+x}{2} \sin \frac{3x-x}{2} = 2 \cos \frac{4x}{2} \sin \frac{2x}{2} = 2 \cos 2x \sin x$.

Substitute this back into the LHS:

LHS $= 2 \cos 2x \sin x + \sin 2x$

Use the double angle identity: $\sin 2x = 2 \sin x \cos x$.

LHS $= 2 \cos 2x \sin x + 2 \sin x \cos x$

Factor out the common term $2 \sin x$:

LHS $= 2 \sin x (\cos 2x + \cos x)$

Use the sum-to-product formula for cosine: $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$.

Apply this to $(\cos 2x + \cos x)$ with $C = 2x$ and $D = x$:

$\cos 2x + \cos x = 2 \cos \frac{2x+x}{2} \cos \frac{2x-x}{2} = 2 \cos \frac{3x}{2} \cos \frac{x}{2}$.

Substitute this back into the LHS expression:

LHS $= 2 \sin x \left(2 \cos \frac{3x}{2} \cos \frac{x}{2}\right)$

LHS $= 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

This is equal to the Right Hand Side (RHS).

Since LHS = RHS, the identity is proved.

Hence Proved.

Given that $\tan x = -\frac{4}{3}$ and $x$ is in quadrant II.

Since $x$ is in quadrant II, we have $\frac{\pi}{2} < x < \pi$.

Dividing the inequality by 2, we get $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.

This means that $\frac{x}{2}$ is in quadrant I.

In quadrant I, $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive.

We are given $\tan x = -\frac{4}{3}$. We use the identity $1 + \tan^2 x = \sec^2 x$ to find $\sec x$.

$\sec^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{9+16}{9} = \frac{25}{9}$.

$\sec x = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3}$.

Since $x$ is in quadrant II, $\cos x$ is negative, so $\sec x$ is negative.

Thus, $\sec x = -\frac{5}{3}$.

Now, we find $\cos x$: $\cos x = \frac{1}{\sec x} = \frac{1}{-\frac{5}{3}} = -\frac{3}{5}$.

We can find $\sin x$ using $\tan x = \frac{\sin x}{\cos x}$.

$\sin x = \tan x \times \cos x = \left(-\frac{4}{3}\right) \times \left(-\frac{3}{5}\right) = \frac{12}{15} = \frac{4}{5}$.

Since $x$ is in quadrant II, $\sin x$ is positive, which is consistent with our result.

Now we use the half-angle formulas:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

$\sin^2 \frac{x}{2} = \frac{1 - (-\frac{3}{5})}{2} = \frac{1 + \frac{3}{5}}{2} = \frac{\frac{5+3}{5}}{2} = \frac{\frac{8}{5}}{2} = \frac{8}{10} = \frac{4}{5}$.

Since $\frac{x}{2}$ is in quadrant I, $\sin \frac{x}{2} > 0$.

$\sin \frac{x}{2} = \sqrt{\frac{4}{5}} = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.

Rationalizing the denominator: $\sin \frac{x}{2} = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$.

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

$\cos^2 \frac{x}{2} = \frac{1 + (-\frac{3}{5})}{2} = \frac{1 - \frac{3}{5}}{2} = \frac{\frac{5-3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{2}{10} = \frac{1}{5}$.

Since $\frac{x}{2}$ is in quadrant I, $\cos \frac{x}{2} > 0$.

$\cos \frac{x}{2} = \sqrt{\frac{1}{5}} = \frac{\sqrt{1}}{\sqrt{5}} = \frac{1}{\sqrt{5}}$.

Rationalizing the denominator: $\cos \frac{x}{2} = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$

$\tan \frac{x}{2} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{1} = 2$.

Alternatively, using the formula $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$:

$\tan \frac{x}{2} = \frac{1 - (-\frac{3}{5})}{\frac{4}{5}} = \frac{1 + \frac{3}{5}}{\frac{4}{5}} = \frac{\frac{8}{5}}{\frac{4}{5}} = \frac{8}{5} \times \frac{5}{4} = \frac{\cancel{8}^{2} \times \cancel{5}^{1}}{\cancel{5}^{1} \times \cancel{4}^{1}} = 2$.

The values are:

$\sin \frac{x}{2} = \frac{2\sqrt{5}}{5}$

$\cos \frac{x}{2} = \frac{\sqrt{5}}{5}$

$\tan \frac{x}{2} = 2$

Given that $\cos x = -\frac{1}{3}$ and $x$ is in quadrant III.

Since $x$ is in quadrant III, we have $\pi < x < \frac{3\pi}{2}$.

Dividing the inequality by 2, we get $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$.

This means that $\frac{x}{2}$ lies in the second quadrant.

In the second quadrant, $\sin \frac{x}{2}$ is positive, $\cos \frac{x}{2}$ is negative, and $\tan \frac{x}{2}$ is negative.

We use the half-angle formulas for $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

Substitute the value of $\cos x = -\frac{1}{3}$:

$\sin^2 \frac{x}{2} = \frac{1 - (-\frac{1}{3})}{2} = \frac{1 + \frac{1}{3}}{2} = \frac{\frac{3+1}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3}$.

Since $\frac{x}{2}$ is in quadrant II, $\sin \frac{x}{2} > 0$.

$\sin \frac{x}{2} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.

Rationalize the denominator:

$\sin \frac{x}{2} = \frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$.

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

Substitute the value of $\cos x = -\frac{1}{3}$:

$\cos^2 \frac{x}{2} = \frac{1 + (-\frac{1}{3})}{2} = \frac{1 - \frac{1}{3}}{2} = \frac{\frac{3-1}{3}}{2} = \frac{\frac{2}{3}}{2} = \frac{2}{6} = \frac{1}{3}$.

Since $\frac{x}{2}$ is in quadrant II, $\cos \frac{x}{2} < 0$.

$\cos \frac{x}{2} = -\sqrt{\frac{1}{3}} = -\frac{1}{\sqrt{3}}$.

Rationalize the denominator:

$\cos \frac{x}{2} = -\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$.

Now we find $\tan \frac{x}{2}$ using the values of $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$:

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{\sqrt{6}}{3}}{-\frac{\sqrt{3}}{3}} = \frac{\sqrt{6}}{3} \times \left(-\frac{3}{\sqrt{3}}\right) = -\frac{\sqrt{6}}{\sqrt{3}} = -\sqrt{\frac{6}{3}} = -\sqrt{2}$.

Alternatively, we can use the half-angle formula $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$.

First, we need to find $\sin x$. Since $x$ is in quadrant III, $\sin x$ is negative.

$\sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{9-1}{9} = \frac{8}{9}$.

$\sin x = -\sqrt{\frac{8}{9}} = -\frac{\sqrt{8}}{\sqrt{9}} = -\frac{2\sqrt{2}}{3}$.

Now substitute into the formula for $\tan \frac{x}{2}$:

$\tan \frac{x}{2} = \frac{1 - (-\frac{1}{3})}{-\frac{2\sqrt{2}}{3}} = \frac{1 + \frac{1}{3}}{-\frac{2\sqrt{2}}{3}} = \frac{\frac{4}{3}}{-\frac{2\sqrt{2}}{3}} = \frac{4}{3} \times \left(-\frac{3}{2\sqrt{2}}\right) = -\frac{4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\frac{2\sqrt{2}}{2} = -\sqrt{2}$.

The values are:

$\sin \frac{x}{2} = \frac{\sqrt{6}}{3}$

$\cos \frac{x}{2} = -\frac{\sqrt{3}}{3}$

$\tan \frac{x}{2} = -\sqrt{2}$

Given that $\sin x = \frac{1}{4}$ and $x$ is in quadrant II.

Since $x$ is in quadrant II, we have $\frac{\pi}{2} < x < \pi$.

Dividing the inequality by 2, we get $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.

This means that $\frac{x}{2}$ lies in the first quadrant.

In the first quadrant, $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive.

We are given $\sin x = \frac{1}{4}$. We use the identity $\sin^2 x + \cos^2 x = 1$ to find $\cos x$.

$\cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{16-1}{16} = \frac{15}{16}$.

Since $x$ is in quadrant II, $\cos x$ is negative.

Thus, $\cos x = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$.

Now we use the half-angle formulas:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

Substitute the value of $\cos x = -\frac{\sqrt{15}}{4}$:

$\sin^2 \frac{x}{2} = \frac{1 - (-\frac{\sqrt{15}}{4})}{2} = \frac{1 + \frac{\sqrt{15}}{4}}{2} = \frac{\frac{4+\sqrt{15}}{4}}{2} = \frac{4+\sqrt{15}}{8}$.

Since $\frac{x}{2}$ is in quadrant I, $\sin \frac{x}{2} > 0$.

$\sin \frac{x}{2} = \sqrt{\frac{4+\sqrt{15}}{8}} = \frac{\sqrt{4+\sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}$.

Rationalizing the denominator:

$\sin \frac{x}{2} = \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2(4+\sqrt{15})}}{4} = \frac{\sqrt{8+2\sqrt{15}}}{4}$.

To simplify $\sqrt{8+2\sqrt{15}}$, we look for two numbers whose sum is 8 and product is 15 (which are 5 and 3).

So, $\sqrt{8+2\sqrt{15}} = \sqrt{(\sqrt{5})^2 + (\sqrt{3})^2 + 2\sqrt{5}\sqrt{3}} = \sqrt{(\sqrt{5}+\sqrt{3})^2} = \sqrt{5}+\sqrt{3}$.

Therefore, $\sin \frac{x}{2} = \frac{\sqrt{5} + \sqrt{3}}{4}$.

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

Substitute the value of $\cos x = -\frac{\sqrt{15}}{4}$:

$\cos^2 \frac{x}{2} = \frac{1 + (-\frac{\sqrt{15}}{4})}{2} = \frac{1 - \frac{\sqrt{15}}{4}}{2} = \frac{\frac{4-\sqrt{15}}{4}}{2} = \frac{4-\sqrt{15}}{8}$.

Since $\frac{x}{2}$ is in quadrant I, $\cos \frac{x}{2} > 0$.

$\cos \frac{x}{2} = \sqrt{\frac{4-\sqrt{15}}{8}} = \frac{\sqrt{4-\sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}}$.

Rationalizing the denominator:

$\cos \frac{x}{2} = \frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2(4-\sqrt{15})}}{4} = \frac{\sqrt{8-2\sqrt{15}}}{4}$.

To simplify $\sqrt{8-2\sqrt{15}}$, we look for two numbers whose sum is 8 and product is 15 (which are 5 and 3).

So, $\sqrt{8-2\sqrt{15}} = \sqrt{(\sqrt{5})^2 + (\sqrt{3})^2 - 2\sqrt{5}\sqrt{3}} = \sqrt{(\sqrt{5}-\sqrt{3})^2} = |\sqrt{5}-\sqrt{3}|$.

Since $\sqrt{5} > \sqrt{3}$, $|\sqrt{5}-\sqrt{3}| = \sqrt{5}-\sqrt{3}$.

Therefore, $\cos \frac{x}{2} = \frac{\sqrt{5} - \sqrt{3}}{4}$.

Now we find $\tan \frac{x}{2}$:

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{\sqrt{5} + \sqrt{3}}{4}}{\frac{\sqrt{5} - \sqrt{3}}{4}} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$.

Rationalize the denominator:

$\tan \frac{x}{2} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{5 + 2\sqrt{15} + 3}{5 - 3} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$.

The values are:

$\sin \frac{x}{2} = \frac{\sqrt{5} + \sqrt{3}}{4}$

$\cos \frac{x}{2} = \frac{\sqrt{5} - \sqrt{3}}{4}$

$\tan \frac{x}{2} = 4 + \sqrt{15}$