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1. Sets 2. Relations and Functions 3. Trigonometric Functions
4. Principle of Mathematical Induction 5. Complex Numbers and Quadratic Equations 6. Linear Inequalities
7. Permutations and Combinations 8. Binomial Theorem 9. Sequences and Series
10. Straight Lines 11. Conic Sections 12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives 14. Mathematical Reasoning 15. Statistics
16. Probability

Content On This Page
Angle and Its Measurement Length of an Arc and Area of a Sector Trigonometric Functions of a Real Number
Graph of Trigonometric Functions Trigonometric Functions of Sum and Difference Trigonometric Equations


Chapter 3 Trigonometric Functions (Concepts)

Welcome to this significantly expanded exploration of Trigonometry, which dramatically broadens the scope beyond the confines of right-angled triangles studied in Class 10. While our previous work focused on trigonometric ratios defined by side lengths, this chapter generalizes these concepts, allowing us to analyze angles of any magnitude and explore the deeper, periodic nature of trigonometric relationships. This forms the essential basis for advanced applications in calculus, physics, engineering, and beyond.

We begin by redefining an angle not just as a static vertex between two lines, but as a dynamic measure of rotation of a ray around a fixed point (the vertex). This allows for angles greater than $90^\circ$ or even negative angles (representing rotation in the opposite direction). Alongside this, we introduce a crucial alternative unit for measuring angles: the radian. While degrees are familiar, radians provide a more natural mathematical measure, especially in calculus. We establish the fundamental conversion factor $\mathbf{\pi \text{ radians} = 180^\circ}$ and practice converting between these two systems ($\text{radians} = \text{degrees} \times \frac{\pi}{180}$ and $\text{degrees} = \text{radians} \times \frac{180}{\pi}$).

The core conceptual shift involves generalizing the trigonometric functions (sine, cosine, tangent, cosecant, secant, cotangent) beyond the SOH CAH TOA definitions. We utilize the unit circle approach. Consider a circle centered at the origin with radius 1 ($x^2 + y^2 = 1$). For any angle $\theta$ measured in standard position (initial side along the positive x-axis, rotation counter-clockwise), the terminal side of the angle intersects the unit circle at a unique point with coordinates $(x, y)$. We then define the primary trigonometric functions for any angle $\theta$ as:

The remaining functions are defined based on these: $\tan \theta = \frac{y}{x}$ (for $x \neq 0$), $\sec \theta = \frac{1}{x}$ (for $x \neq 0$), $\csc \theta = \frac{1}{y}$ (for $y \neq 0$), and $\cot \theta = \frac{x}{y}$ (for $y \neq 0$). This powerful definition seamlessly extends trigonometry to all angles and naturally reveals the signs of the trigonometric functions in the four different quadrants (often remembered by the mnemonic ASTC: All positive in Quad I, Sine positive in Quad II, Tangent positive in Quad III, Cosine positive in Quad IV). We also formally determine the domain (permissible input angles) and range (possible output values) for each of the six functions.

This generalized definition unveils the inherent periodic nature of trigonometric functions. Since adding a full rotation ($360^\circ$ or $2\pi$ radians) brings the terminal side back to the same position, the function values repeat. For example, $\sin(\theta + 2\pi) = \sin \theta$ and $\cos(\theta + 2\pi) = \cos \theta$ (period $2\pi$), while $\tan(\theta + \pi) = \tan \theta$ (period $\pi$). We study the characteristic graphs of these functions (like the sine wave), noting key features such as amplitude and period, which visually represent this repetition.

A major component of this chapter is the derivation and application of a vast array of Trigonometric Identities. Building upon the revisited reciprocal, quotient, and Pythagorean identities ($\sin^2\theta + \cos^2\theta = 1$, etc.), we develop formulas for:

These identities are indispensable tools for simplifying complex trigonometric expressions and proving further relationships.

Finally, the chapter focuses on solving Trigonometric Equations – equations involving trigonometric functions of unknown angles. We learn methods to find:

We derive the general solution formulas for equations of the form $\sin x = \sin y$, $\cos x = \cos y$, and $\tan x = \tan y$, providing a complete framework for solving basic trigonometric equations.



Angle and Its Measurement

The foundation of trigonometry lies in the concept of an angle. In geometry, an angle is often seen as a static figure formed by two rays meeting at a common endpoint. However, in trigonometry, an angle is defined in a more dynamic sense: it is the measure of rotation of a ray about its initial point.

Imagine a ray, which we can call OA. If this ray rotates around its fixed starting point O, it sweeps out an angle. The key components of this angle are:

A diagram showing the formation of an angle. Ray OA is the initial side, ray OB is the terminal side, and O is the vertex. An arc with an arrow shows the direction of rotation.

Positive and Negative Angles

The amount of rotation has a magnitude, but in trigonometry, its direction is equally important. This direction determines whether the angle is considered positive or negative, providing a crucial distinction for analyzing circular motion and periodic functions.

1. Positive Angle

An angle is considered positive if the direction of rotation from the initial side to the terminal side is anti-clockwise (or counter-clockwise). This is the direction opposite to the movement of the hands of a clock.

2. Negative Angle

An angle is considered negative if the direction of rotation from the initial side to the terminal side is clockwise. This is the same direction in which the hands of a clock move.

Two diagrams side-by-side. The left shows a positive angle with an anti-clockwise arrow. The right shows a negative angle with a clockwise arrow. Both start from the same initial side.

Angle in Standard Position

To have a consistent and universal way of representing angles, we place them in a standard position on the Cartesian coordinate plane (the x-y plane).

An angle is said to be in standard position if it satisfies two conditions:

  1. Its vertex is at the origin (0, 0) of the coordinate system.
  2. Its initial side lies along the positive x-axis.

When an angle is in standard position, the terminal side can lie in any of the four quadrants or on one of the axes. This standardized placement allows us to easily compare and analyze different angles.

The Cartesian plane with an angle in standard position. The vertex is at the origin, the initial side is on the positive x-axis, and the terminal side is in Quadrant II.

Quadrants and Quadrantal Angles

The x and y axes divide the Cartesian plane into four regions called quadrants. When an angle is in standard position, its terminal side determines which quadrant the angle lies in.

If the terminal side of an angle in standard position lies on one of the axes (e.g., at 0°, 90°, 180°, 270°, 360°), it is called a quadrantal angle.

The Cartesian plane showing the four quadrants labeled I, II, III, and IV, with the corresponding angle ranges for each.

Systems of Measuring Angles:

There are two primary systems used for measuring angles: the Degree Measure and the Radian Measure.

1. Degree Measure (Sexagesimal System):

In the Degree system, a complete rotation of the ray is divided into 360 equal parts. Each part is called a degree. The symbol for degree is $^\circ$.

So, a complete revolution corresponds to $360^\circ$. A straight angle (half a revolution) is $180^\circ$, and a right angle (a quarter of a revolution) is $90^\circ$.

A circle in the Cartesian plane showing the measurement of angles in degrees. A full circle is 360 degrees, a straight line is 180 degrees, and a right angle is 90 degrees. The four quadrantal angles are marked.

$1 \text{ revolution} = 360^\circ$

For finer measurements, a degree is subdivided into smaller units:

For example, an angle of 30 degrees, 15 minutes, and 45 seconds is written as $30^\circ 15' 45''$.

2. Radian Measure (Circular System):

The Radian system is the standard unit of angular measurement used in many areas of mathematics and physics, particularly in calculus. One radian is defined as the measure of the angle subtended at the center of a circle by an arc whose length is equal to the radius of the circle.

Consider a circle with center O and radius $r$. Let AB be an arc on the circle such that its length $l$ is equal to the radius $r$. The angle $\angle \text{AOB}$ subtended by this arc at the center is defined as 1 radian (1 rad).

Diagram defining 1 radian as the angle subtended at the center of a circle by an arc equal in length to the radius

For a circle of radius $r$, an arc of length $l$ subtends an angle $\theta$ at the center. The measure of this angle in radians is given by the ratio of the arc length to the radius:

$$\theta = \frac{\text{Arc Length}}{\text{Radius}} = \frac{l}{r}$$

From this, we get the relationship $l = r\theta$. This formula holds when $\theta$ is measured in radians. Note that since $\theta$ is the ratio of two lengths, it is technically a dimensionless quantity, but we use the unit "radian" for clarity.


Relation Between Degree and Radian Measures

In trigonometry and geometry, angles can be measured in two primary units: degrees and radians. To work effectively with angles, it's crucial to understand the relationship between these two systems and how to convert from one to the other. The relationship is established by considering a complete revolution or a full circle.


Deriving the Fundamental Relationship

The core of the conversion lies in equating the measure of a full circle in both systems.

  1. Full Revolution in Degrees: By definition, a complete revolution around a central point is divided into 360 degrees.

    $1 \text{ complete revolution} = 360^\circ$

  2. Full Revolution in Radians: The radian measure of an angle is defined by the formula $\theta = \frac{l}{r}$, where $l$ is the arc length and $r$ is the radius of the circle.

    For a complete revolution, the arc length is the entire circumference of the circle, which is given by the formula $l = 2\pi r$.

    Substituting this arc length into the radian formula, we get:

    $\theta = \frac{2\pi r}{r} = 2\pi$

    Thus, a complete revolution is equal to $2\pi$ radians.

  3. Equating the Measures: Since both $360^\circ$ and $2\pi$ radians represent the same physical quantity (a full circle), we can set them equal to each other:

    $360^\circ = 2\pi \text{ radians}$

    Dividing both sides by 2 gives us the most common and fundamental relationship for conversion:

    $180^\circ = \pi \text{ radians}$


Conversion Formulas

From the fundamental relationship $180^\circ = \pi \text{ radians}$, we can derive the specific conversion factors.

1. To Convert from Degrees to Radians:

To find the value of one degree, we can divide the fundamental relationship by 180:

$1^\circ = \frac{\pi}{180} \text{ radians}$

Therefore, to convert any angle from degrees to radians, you must multiply the degree measure by the conversion factor $\frac{\pi}{180}$.

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

2. To Convert from Radians to Degrees:

To find the value of one radian, we can divide the fundamental relationship by $\pi$:

$1 \text{ radian} = \frac{180^\circ}{\pi}$

Using the approximation $\pi \approx 3.14159$, one radian is approximately $57.2958^\circ$, or more accurately $57^\circ 16'$.

Therefore, to convert any angle from radians to degrees, you must multiply the radian measure by the conversion factor $\frac{180}{\pi}$.

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


Summary Table of Key Relations

To Convert Multiply by Example: $60^\circ$ to Radians
Degrees to Radians $\frac{\pi}{180}$ $60 \times \frac{\pi}{180} = \frac{\pi}{3}$ radians
Radians to Degrees $\frac{180}{\pi}$ $\frac{\pi}{3} \times \frac{180}{\pi} = 60^\circ$

Common Conversions:

Here are some common angle measures expressed in both degrees and radians:

Degree Radian
$0^\circ$ $0$
$30^\circ$ $30 \times \frac{\pi}{180} = \frac{\pi}{6}$
$45^\circ$ $45 \times \frac{\pi}{180} = \frac{\pi}{4}$
$60^\circ$ $60 \times \frac{\pi}{180} = \frac{\pi}{3}$
$90^\circ$ $90 \times \frac{\pi}{180} = \frac{\pi}{2}$
$180^\circ$ $180 \times \frac{\pi}{180} = \pi$
$270^\circ$ $270 \times \frac{\pi}{180} = \frac{3\pi}{2}$
$360^\circ$ $360 \times \frac{\pi}{180} = 2\pi$

Example 1. Convert $45^\circ$ to radian measure.

Answer:

Given:

Angle in degree measure $= 45^\circ$.

To Find:

Equivalent angle in radian measure.

Solution:

We use the conversion formula from degrees to radians (Formula 5):

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

Substitute the given degree measure:

Radian measure = $45 \times \frac{\pi}{180}$

Simplify the expression:

Radian measure = $\frac{45\pi}{180}$

We can simplify the fraction $\frac{45}{180}$ by dividing both numerator and denominator by their greatest common divisor, which is 45.

$\frac{\cancel{45}^{1} \times \pi}{\cancel{180}_{4}} = \frac{\pi}{4}$

So, $45^\circ$ is equal to $\frac{\pi}{4}$ radians.

The final answer is $\mathbf{\frac{\pi}{4} radians}$.


Example 2. Convert $\frac{2\pi}{3}$ radians to degree measure.

Answer:

Given:

Angle in radian measure $= \frac{2\pi}{3}$.

To Find:

Equivalent angle in degree measure.

Solution:

We use the conversion formula from radians to degrees (Formula 6):

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

Substitute the given radian measure:

Degree measure = $\frac{2\pi}{3} \times \frac{180}{\pi}$

Simplify the expression by cancelling $\pi$ from numerator and denominator, and simplifying the numbers:

$\frac{2\cancel{\pi}}{\cancel{3}_{1}} \times \frac{\cancel{180}^{60}}{\cancel{\pi}} = 2 \times 60$

= $120$

So, $\frac{2\pi}{3}$ radians is equal to $120^\circ$.

The final answer is $\textbf{120}^\circ$.


Example 3. Convert $6^\circ$ into radians.

Answer:

Given:

Angle in degree measure $= 6^\circ$.

To Find:

Equivalent angle in radian measure.

Solution:

Using the conversion formula from degrees to radians (Formula 5):

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

Substitute the given degree measure:

Radian measure = $6 \times \frac{\pi}{180}$

Simplify the fraction:

$\frac{\cancel{6}^{1} \times \pi}{\cancel{180}_{30}} = \frac{\pi}{30}$

So, $6^\circ$ is equal to $\frac{\pi}{30}$ radians.

The final answer is $\mathbf{\frac{\pi}{30} radians}$.



Length of an Arc and Area of a Sector

The radian measure of an angle provides a very natural and convenient relationship between the angle, the radius of a circle, and the length of the arc subtended by the angle at the center. This relationship is fundamental for solving problems involving sectors and arcs of circles.

A diagram of a circular sector. The circle has its center at O. Two radii, OA and OB, form the sector. The radius is labeled 'r'. The angle AOB at the center is labeled 'θ'. The curved line from A to B is the arc, and its length is labeled 'l'.

Length of an Arc

Consider a circle with radius $r$. An arc of this circle subtends a central angle $\theta$ at the center. The length of this arc, denoted by $l$, is directly proportional to both the radius $r$ and the central angle $\theta$.

The formula relating arc length ($l$), radius ($r$), and the central angle ($\theta$) is:

$l = r\theta$

This formula is valid only when the angle $\theta$ is measured in radians. If the angle is given in degrees, it must first be converted to radians by multiplying by the conversion factor $\frac{\pi}{180}$.

Derivation of the Arc Length Formula ($l = r\theta$)

The derivation is based on the principle of proportionality. In any circle, the ratio of the length of an arc to the total circumference is equal to the ratio of the angle subtended by that arc to the angle of a full circle ($360^\circ$ or $2\pi$ radians).

We can set up a proportion using radian measures:

$\frac{\text{Length of Arc}}{\text{Circumference}} = \frac{\text{Central Angle of Arc}}{\text{Angle of Full Circle}}$

Substituting the corresponding values:

$\frac{l}{2\pi r} = \frac{\theta}{2\pi}$

To solve for $l$, we multiply both sides by $2\pi r$:

$l = \left(\frac{\theta}{2\pi}\right) \times 2\pi r$

By canceling the $2\pi$ terms, we arrive at the formula:

$l = \theta \times r \implies \boldsymbol{l = r\theta}$


Area of a Sector

A sector of a circle is the pie-shaped region enclosed by two radii and the arc between them. Similar to arc length, the area of a sector is directly proportional to its central angle $\theta$.

For a circle of radius $r$, the area ($A$) of a sector with a central angle $\theta$ is given by the formula:

$A = \frac{1}{2}r^2\theta$

As with the arc length formula, the angle $\theta$ in this formula must be in radians.

Derivation of the Area of a Sector Formula ($A = \frac{1}{2}r^2\theta$)

The derivation uses the same proportionality principle. The ratio of the area of a sector to the area of the entire circle is equal to the ratio of the sector's central angle to the angle of a full circle.

$\frac{\text{Area of Sector}}{\text{Area of Full Circle}} = \frac{\text{Central Angle of Sector}}{\text{Angle of Full Circle}}$

The area of a full circle is $\pi r^2$. Substituting the values into the proportion:

$\frac{A}{\pi r^2} = \frac{\theta}{2\pi}$

To solve for $A$, we multiply both sides by $\pi r^2$:

$A = \left(\frac{\theta}{2\pi}\right) \times \pi r^2$

We can cancel the $\pi$ term from the numerator and denominator:

$A = \frac{\theta}{2\cancel{\pi}} \times \cancel{\pi} r^2 = \frac{\theta r^2}{2}$

Rearranging this gives the standard form of the formula:

$\boldsymbol{A = \frac{1}{2}r^2\theta}$


Example 1. A circle has a radius of 5 cm. Find the length of the arc subtended by a central angle of $60^\circ$.

Answer:

Given:

Radius of the circle, $r = 5$ cm.

Central angle $= 60^\circ$.

To Find:

Length of the arc ($l$).

Solution:

The formula for arc length $l = r\theta$ requires the angle $\theta$ to be in radians. The given angle is in degrees, so we first convert $60^\circ$ to radians using the conversion formula (Formula 5 from Section I1):

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

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

Simplify the expression:

$\theta = \frac{60\pi}{180}$ radians

$\theta = \frac{\cancel{60}^{1} \times \pi}{\cancel{180}_{3}}$ radians

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

Now that the angle is in radians, we can use the arc length formula $l = r\theta$ (Formula 1):

$l = r\theta$

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

$l = 5 \times \frac{\pi}{3}$

$l = \frac{5\pi}{3}$

The length of the arc is $\frac{5\pi}{3}$ cm.

The final answer is $\mathbf{\frac{5\pi}{3} cm}$.


Example 2. A sector of a circle with radius 8 cm has a central angle of $\frac{3\pi}{4}$ radians. Find the area of the sector.

Answer:

Given:

Radius of the circle, $r = 8$ cm.

Central angle, $\theta = \frac{3\pi}{4}$ radians.

To Find:

Area of the sector ($A$).

Solution:

The central angle is given in radians ($\frac{3\pi}{4}$ radians), so we can directly use the formula for the area of a sector $A = \frac{1}{2}r^2\theta$ (Formula 2):

$A = \frac{1}{2}r^2\theta$

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

$A = \frac{1}{2} \times (8)^2 \times \frac{3\pi}{4}$

Calculate $(8)^2$:

$A = \frac{1}{2} \times 64 \times \frac{3\pi}{4}$

Simplify the expression. We can multiply $\frac{1}{2}$ and 64, and then simplify with the denominator 4:

$A = 32 \times \frac{3\pi}{4}$

$A = \cancel{32}^{8} \times \frac{3\pi}{\cancel{4}_{1}}$

$A = 8 \times 3\pi$

$A = 24\pi$

The area of the sector is $24\pi$ square centimeters.

The final answer is $\mathbf{24\pi cm}^\mathbf{2}$.



Trigonometric Functions of a Real Number

In earlier classes, you learned about trigonometric ratios of acute angles in right-angled triangles (angles between $0^\circ$ and $90^\circ$). These ratios ($\sin \theta, \cos \theta, \tan \theta$, etc.) relate the angles of a right triangle to the lengths of its sides. However, to analyze periodic phenomena and use trigonometry in calculus and other advanced topics, we need to extend the definitions of these ratios to angles of any magnitude (positive or negative) and define them as functions of a real number.

This extension is done using the concept of the unit circle and defining trigonometric functions based on the coordinates of a point on this circle.


The Unit Circle: A Broader Definition of Trigonometric Functions

While mnemonics like Pandit Badri Prasad or SOH-CAH-TOA work perfectly for defining trigonometric ratios for acute angles within a right-angled triangle, this approach is inherently limited. The unit circle provides a more powerful and comprehensive way to define trigonometric functions for any angle, regardless of its size or sign.


What is the Unit Circle?

The unit circle is a circle drawn on the Cartesian coordinate plane with two specific properties:

  1. Its center is located at the origin, point $(0, 0)$.
  2. Its radius ($r$) is exactly 1 unit.

Because its radius is 1, any point $(x, y)$ on the circumference of the unit circle must satisfy its equation: $x^2 + y^2 = 1^2$, or simply $\boldsymbol{x^2 + y^2 = 1}$.


Defining Sine and Cosine using the Unit Circle

Imagine an angle $\theta$ placed in a "standard position" on the Cartesian plane. This means its vertex is at the origin $(0, 0)$ and its starting side (initial side) lies along the positive x-axis. The other side (terminal side) rotates counter-clockwise by an amount $\theta$.

This rotating terminal side will eventually intersect the circumference of the unit circle at a unique point, which we can label $P(x, y)$.

Unit circle with point P(x,y) on its circumference, showing the angle theta from the positive x-axis to the radius OP, and the coordinates (x,y)

Connecting to Right-Triangle Trigonometry

To understand why the coordinates of this point are so special, let's create a right-angled triangle by dropping a perpendicular line from the point $P(x, y)$ down to the x-axis. This gives us a triangle with:

Now, let's apply our standard trigonometric definitions ("Pandit Badri Prasad" or SOH-CAH-TOA) to this triangle:

This reveals the crucial connection: the x and y coordinates of the point on the unit circle are precisely the cosine and sine of the angle $\theta$. The definition works so cleanly because the hypotenuse is 1, simplifying the ratios.


The Formal Definition and Its Implication

Because of this relationship, we can formally define sine and cosine for any angle $\theta$ based on the coordinates of the intersection point $P(x, y)$ on the unit circle:

This is a powerful concept. It means that any point on the unit circle can be represented not just by its Cartesian coordinates $(x, y)$, but also by its trigonometric coordinates $(\cos \theta, \sin \theta)$, where $\theta$ is the angle that leads to that point.


The Fundamental Pythagorean Identity

This unit circle definition leads directly to the most fundamental identity in trigonometry. We know that every point $(x, y)$ on the unit circle must satisfy its equation:

$x^2 + y^2 = 1$

By substituting our new definitions for $x$ and $y$ into this equation, we get:

$(\cos \theta)^2 + (\sin \theta)^2 = 1$

This is universally written as the Pythagorean Identity:

$\boldsymbol{\cos^2 \theta + \sin^2 \theta = 1}$

This identity holds true for any real number $\theta$, making it a cornerstone of trigonometry.


Trigonometric Ratios in All Four Quadrants using the Unit Circle

The unit circle provides a powerful framework for understanding the values and, critically, the signs (positive or negative) of trigonometric functions for any angle. The sign of each function is determined by the quadrant in which the angle's terminal side lies.


The Four Quadrants and Coordinate Signs

The Cartesian plane is divided into four quadrants by the x and y axes. An angle $\theta$ is in a specific quadrant if its terminal side lies in that quadrant.

The four quadrants of the Cartesian plane. Quadrant I (top right), Quadrant II (top left), Quadrant III (bottom left), Quadrant IV (bottom right).

The key is to remember the sign of the x and y coordinates in each quadrant. Since on the unit circle, $\boldsymbol{\cos \theta = x}$ and $\boldsymbol{\sin \theta = y}$, the signs of sine and cosine directly follow the signs of the coordinates.


Signs of Trigonometric Ratios by Quadrant

We can now determine the sign of all six trigonometric ratios in each quadrant. Remember that $\tan \theta = \frac{\sin \theta}{\cos \theta}$, and the reciprocal functions ($\text{cosec}, \sec, \cot$) will have the same sign as their parent functions ($\sin, \cos, \tan$).


Quadrant I $(+, +)$

A unit circle showing an angle theta in Quadrant I. The point P(x,y) on the circle has a positive x-coordinate and a positive y-coordinate.

Quadrant II $(-, +)$

A unit circle showing an angle theta in Quadrant II. The point P(x,y) on the circle has a negative x-coordinate and a positive y-coordinate.

Quadrant III $(-, -)$

A unit circle showing an angle theta in Quadrant III. The point P(x,y) on the circle has a negative x-coordinate and a negative y-coordinate.

Quadrant IV $(+, -)$

A unit circle showing an angle theta in Quadrant IV. The point P(x,y) on the circle has a positive x-coordinate and a negative y-coordinate.

The ASTC Rule: A Mnemonic for Remembering the Signs

A simple way to remember which functions are positive in each quadrant is the ASTC rule, often remembered with the phrase "All Students Take Calculus" or "Add Sugar To Coffee". Starting from Quadrant I and moving counter-clockwise:

Summary Table

Quadrant Angle Range (Degrees) Angle Range (Radians) Signs of (x, y) Positive Ratios
I $0^\circ - 90^\circ$ $0 - \frac{\pi}{2}$ $(+, +)$ All ($\sin, \cos, \tan, \text{cosec}, \sec, \cot$)
II $90^\circ - 180^\circ$ $\frac{\pi}{2} - \pi$ $(-, +)$ Sine, Cosecant
III $180^\circ - 270^\circ$ $\pi - \frac{3\pi}{2}$ $(-, -)$ Tangent, Cotangent
IV $270^\circ - 360^\circ$ $\frac{3\pi}{2} - 2\pi$ $(+, -)$ Cosine, Secant

Why This Definition is Powerful

The unit circle definition extends the concept of sine and cosine beyond acute angles ($0^\circ$ to $90^\circ$). It allows us to find the values of trigonometric functions for any angle, including:

It also naturally explains the signs (positive or negative) of sine and cosine in each of the four quadrants based on the signs of the x and y coordinates in those quadrants.


Definition of Trigonometric Functions:

Using the unit circle and the coordinates $(x, y)$ of the point $P$ corresponding to a real number $\theta$ (the angle in radians), we define the six trigonometric functions for any real number $\theta$ as follows:

Let $\theta$ be any real number, and let $P(x, y)$ be the point on the unit circle such that the angle from the positive x-axis to OP is $\theta$ radians.

1. Sine Function: The sine of $\theta$ is the y-coordinate of the point $P$.

$\sin \theta = y$

2. Cosine Function: The cosine of $\theta$ is the x-coordinate of the point $P$.

$\cos \theta = x$

3. Tangent Function: The tangent of $\theta$ is the ratio of the y-coordinate to the x-coordinate.

$\tan \theta = \frac{y}{x}$, provided $x \neq 0$. Since $x = \cos \theta$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$, provided $\cos \theta \neq 0$.

4. Cosecant Function: The cosecant of $\theta$ is the reciprocal of the y-coordinate.

$\text{cosec} \theta = \frac{1}{y}$, provided $y \neq 0$. Since $y = \sin \theta$, $\text{cosec} \theta = \frac{1}{\sin \theta}$, provided $\sin \theta \neq 0$.

5. Secant Function: The secant of $\theta$ is the reciprocal of the x-coordinate.

$\sec \theta = \frac{1}{x}$, provided $x \neq 0$. Since $x = \cos \theta$, $\sec \theta = \frac{1}{\cos \theta}$, provided $\cos \theta \neq 0$.

6. Cotangent Function: The cotangent of $\theta$ is the ratio of the x-coordinate to the y-coordinate.

$\cot \theta = \frac{x}{y}$, provided $y \neq 0$. Since $y = \sin \theta$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$, provided $\sin \theta \neq 0$. This also means $\cot \theta = \frac{1}{\tan \theta}$.

Basic Trigonometric Identities (Reciprocal and Ratio Identities):

From the definitions above, we can see the reciprocal relationships and ratio relationships:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\cot \theta = \frac{\cos \theta}{\sin \theta}$

$\cot \theta = \frac{1}{\tan \theta}$ (provided $\tan \theta \neq 0$)

$\text{cosec} \theta = \frac{1}{\sin \theta}$ (provided $\sin \theta \neq 0$)

$\sec \theta = \frac{1}{\cos \theta}$ (provided $\cos \theta \neq 0$)

Pythagorean Identities Derived from $\cos^2 \theta + \sin^2 \theta = 1$:

From the fundamental identity $\cos^2 \theta + \sin^2 \theta = 1$ (Equation 3), we can derive two more important identities:

1. Divide Equation (3) by $\cos^2 \theta$, assuming $\cos \theta \neq 0$:

$\frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$

$1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \left(\frac{1}{\cos \theta}\right)^2$

Using the definitions $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$, we get:

$1 + \tan^2 \theta = \sec^2 \theta$

... (4)

This identity is valid for all real numbers $\theta$ where $\cos \theta \neq 0$, i.e., $\theta \neq (2n+1)\frac{\pi}{2}$ for integer $n$.

2. Divide Equation (3) by $\sin^2 \theta$, assuming $\sin \theta \neq 0$:

$\frac{\cos^2 \theta}{\sin^2 \theta} + \frac{\sin^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}$

$\left(\frac{\cos \theta}{\sin \theta}\right)^2 + 1 = \left(\frac{1}{\sin \theta}\right)^2$

Using the definitions $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\text{cosec} \theta = \frac{1}{\sin \theta}$, we get:

$\cot^2 \theta + 1 = \text{cosec}^2 \theta$

... (5)

This identity is valid for all real numbers $\theta$ where $\sin \theta \neq 0$, i.e., $\theta \neq n\pi$ for integer $n$.


Signs of Trigonometric Functions in Different Quadrants:

The sign of the trigonometric functions depends on the quadrant in which the terminal side of the angle $\theta$ lies when placed in standard position (initial side on the positive x-axis, vertex at the origin). This is determined by the signs of the coordinates $(x, y)$ of the point $P$ on the unit circle, since $\cos \theta = x$ and $\sin \theta = y$.

Quadrant Angle $\theta$ (Radians) x-coord ($= \cos \theta$) y-coord ($= \sin \theta$) $\sin \theta$ $\cos \theta$ $\tan \theta = \frac{y}{x}$ $\text{cosec} \theta = \frac{1}{y}$ $\sec \theta = \frac{1}{x}$ $\cot \theta = \frac{x}{y}$
I $0 < \theta < \frac{\pi}{2}$ + + + + + + + +
II $\frac{\pi}{2} < \theta < \pi$ - + + - - + - -
III $\pi < \theta < \frac{3\pi}{2}$ - - - - + - - +
IV $\frac{3\pi}{2} < \theta < 2\pi$ + - - + - - + -

To easily remember which functions are positive in each quadrant, you can use the mnemonic "All Students Take Coffee" or "Add Sugar To Coffee", starting from Quadrant I and moving anti-clockwise:


Values of Trigonometric Functions at Standard Angles:

The values of trigonometric functions for certain standard angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$, and their multiples) are frequently used and should be memorized or be easily derivable. These values can be found using right triangles (for acute angles) or directly from the coordinates on the unit circle for angles that terminate on the axes.

Angle $\theta$ (Degrees) Angle $\theta$ (Radians) $\sin \theta$ $\cos \theta$ $\tan \theta$ $\text{cosec} \theta$ $\sec \theta$ $\cot \theta$
$0^\circ$ $0$ $0$ $1$ $0$ Undefined $1$ Undefined
$30^\circ$ $\frac{\pi}{6}$ $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{3}}$ $2$ $\frac{2}{\sqrt{3}}$ $\sqrt{3}$
$45^\circ$ $\frac{\pi}{4}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{2}}$ $1$ $\sqrt{2}$ $\sqrt{2}$ $1$
$60^\circ$ $\frac{\pi}{3}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\sqrt{3}$ $\frac{2}{\sqrt{3}}$ $2$ $\frac{1}{\sqrt{3}}$
$90^\circ$ $\frac{\pi}{2}$ $1$ $0$ Undefined $1$ Undefined $0$
$180^\circ$ $\pi$ $0$ $-1$ $0$ Undefined $-1$ Undefined
$270^\circ$ $\frac{3\pi}{2}$ $-1$ $0$ Undefined $-1$ Undefined $0$
$360^\circ$ $2\pi$ $0$ $1$ $0$ Undefined $1$ Undefined

Notes on Undefined Values:

Certain trigonometric functions are undefined at angles where their denominators (based on the unit circle definitions) are zero.


Periodicity of Trigonometric Functions

A function $f(x)$ is called periodic if its values repeat at regular intervals. The smallest positive interval over which the function repeats is called the period. This property is easily visualized on the unit circle.

When we find the value of a trigonometric function for an angle $\theta$, we are looking at the coordinates of a point $P(x, y)$ on the unit circle. If we add a full rotation ($360^\circ$ or $2\pi$ radians) to our angle, we land on the exact same point $P(x, y)$. This means the values of sine and cosine repeat every $2\pi$ radians.

A unit circle showing that an angle theta and an angle theta + 2*pi both land on the same point P(x,y), illustrating the 2*pi periodicity of sine and cosine.

1. Sine, Cosine, Cosecant, and Secant (Period $2\pi$)

Since a full rotation of $2\pi$ brings the terminal side back to its original position, the sine and cosine values repeat. The same applies to their reciprocals, cosecant and secant.

For any integer $n$ (representing $n$ full rotations):

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

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

$\text{cosec}(x + 2n\pi) = \text{cosec } x$

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

2. Tangent and Cotangent (Period $\pi$)

The tangent and cotangent functions have a shorter period. If an angle $\theta$ corresponds to the point $P(x, y)$, then rotating by half a circle ($\pi$ radians or $180^\circ$) leads to the diametrically opposite point $P'(-x, -y)$.

Let's check the value of tangent for this new angle:

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

Since the value repeats after just $\pi$ radians, the period of tangent (and its reciprocal, cotangent) is $\pi$.

A unit circle showing an angle theta pointing to P(x,y) and an angle theta + pi pointing to the diametrically opposite point P'(-x,-y), illustrating the pi periodicity of tangent.

For any integer $n$ (representing $n$ half rotations):

$\tan(x + n\pi) = \tan x$

$\cot(x + n\pi) = \cot x$


Trigonometric Functions for Negative Angles (Even and Odd Properties)

The relationship between the trigonometric values of a positive angle $\theta$ and a negative angle $-\theta$ can be understood by looking at their positions on the unit circle. An angle $-\theta$ is a clockwise rotation of the same magnitude as the counter-clockwise rotation for $\theta$.

This results in two points that are reflections of each other across the x-axis. If $\theta$ corresponds to point $P(x, y)$, then $-\theta$ corresponds to point $P'(x, -y)$.

A unit circle showing that angle theta corresponds to point P(x,y) while angle negative-theta corresponds to point P'(x,-y), a reflection across the x-axis.

From this symmetry, we can establish the following identities:

This leads to the classification of trigonometric functions as either even ($f(-x) = f(x)$) or odd ($f(-x) = -f(x)$).

Even Functions (Symmetric about the y-axis)

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

$\sec(-x) = \frac{1}{\cos(-x)} = \frac{1}{\cos x} = \sec x$

Odd Functions (Symmetric about the origin)

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

$\tan(-x) = \frac{\sin(-x)}{\cos(-x)} = \frac{-\sin x}{\cos x} = -\tan x$

$\text{cosec}(-x) = \frac{1}{\sin(-x)} = \frac{1}{-\sin x} = -\text{cosec } x$

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



Graph of Trigonometric Functions

Understanding the graphs of the six trigonometric functions ($\sin x, \cos x, \tan x, \text{cosec } x, \sec x, \cot x$) is crucial as it provides a visual insight into their properties, such as domain, range, periodicity, amplitude, and asymptotes. The graph of a trigonometric function is a representation of the set of ordered pairs $(x, f(x))$, where $x$ is a real number (the angle in radians) in the function's domain, and $f(x)$ is the corresponding output value.

Since trigonometric functions are periodic, their graphs repeat over intervals corresponding to their periods. This makes it sufficient to study the graph over one period to understand its entire behavior.


1. Graph of $y = \sin x$:

The sine function $f(x) = \sin x$ maps a real number $x$ (radian angle) to the y-coordinate of the point on the unit circle at angle $x$.

Domain: The sine function is defined for all real numbers. Domain($\sin x$) $= \mathbb{R}$.

Range: From the unit circle definition, the y-coordinate of any point on the unit circle varies between -1 and 1. Thus, the range of $\sin x$ is the closed interval $[-1, 1]$.

Range($\sin x$) $= [-1, 1]$

Period: The values of $\sin x$ repeat every $2\pi$. The period is $2\pi$.

The graph of $y = \sin x$ is a smooth, continuous wave that oscillates between $y=-1$ and $y=1$. It passes through the origin $(0, 0)$.

Graph of y = sin x showing a sinusoidal wave oscillating between y=-1 and y=1, crossing the x-axis at multiples of pi, with a peak at pi/2 and a trough at 3pi/2 in the interval [0, 2pi]

Key points on the graph of the sine function over one period, which is the interval $[0, 2\pi]$, are:


2. Graph of $y = \cos x$:

The cosine function $f(x) = \cos x$ maps a real number $x$ to the x-coordinate of the point on the unit circle at angle $x$.

Domain: The cosine function is defined for all real numbers. Domain($\cos x$) $= \mathbb{R}$.

Range: From the unit circle definition, the x-coordinate of any point on the unit circle varies between -1 and 1. Thus, the range of $\cos x$ is the closed interval $[-1, 1]$.

Range($\cos x$) $= [-1, 1]$

Period: The values of $\cos x$ repeat every $2\pi$. The period is $2\pi$.

The graph of $y = \cos x$ is also a smooth, continuous wave that oscillates between $y=-1$ and $y=1$. It passes through the point $(0, 1)$. The graph of $\cos x$ is identical to the graph of $\sin x$ shifted $\pi/2$ units to the left ($\cos x = \sin(x + \pi/2)$).

Graph of y = cos x showing a sinusoidal wave oscillating between y=-1 and y=1, starting at y=1 for x=0, crossing the x-axis at odd multiples of pi/2, with a trough at pi and a peak at 2pi

Key points on the graph of the cosine function over one period, which is the interval $[0, 2\pi]$, are:


3. Graph of $y = \tan x$:

The tangent function $f(x) = \tan x = \frac{\sin x}{\cos x}$.

Domain: The tangent function is defined for all real numbers $x$ except where $\cos x = 0$. This occurs at $x = (2n+1)\frac{\pi}{2}$ for any integer $n$.

Domain($\tan x$) $= \{ x \in \mathbb{R} : x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} \}$

Range: The value of $\tan x$ can be any real number. Range($\tan x$) $= \mathbb{R}$.

Period: The values of $\tan x$ repeat every $\pi$. The period is $\pi$.

The graph of $y = \tan x$ has vertical asymptotes at the values of $x$ where it is undefined (odd multiples of $\pi/2$). The graph consists of repeating S-shaped curves between these asymptotes, passing through the x-axis at integer multiples of $\pi$.

Graph of y = tan x showing vertical asymptotes at x=pi/2, 3pi/2, etc. and x=-pi/2, -3pi/2 etc. and repeating curves increasing within each vertical strip, crossing the x-axis at multiples of pi

The graph illustrates that $\tan x$ approaches $\infty$ as $x$ approaches $(2n+1)\frac{\pi}{2}$ from the left, and approaches $-\infty$ as $x$ approaches $(2n+1)\frac{\pi}{2}$ from the right.


4. Graph of $y = \text{cosec } x$:

The cosecant function $f(x) = \text{cosec } x = \frac{1}{\sin x}$.

Domain: The cosecant function is defined for all real numbers $x$ except where $\sin x = 0$. This occurs at $x = n\pi$ for any integer $n$.

Domain($\text{cosec } x$) $= \{ x \in \mathbb{R} : x \neq n\pi, n \in \mathbb{Z} \}$

Range: Since $\sin x$ varies between -1 and 1 (excluding 0 where cosec is undefined), $1/\sin x$ will have values greater than or equal to 1 (when $\sin x$ is positive) or less than or equal to -1 (when $\sin x$ is negative). The range is $(-\infty, -1] \cup [1, \infty)$.

Range($\text{cosec } x$) $= (-\infty, -1] \cup [1, \infty)$

Period: The period of $\text{cosec } x$ is $2\pi$.

The graph of $y = \text{cosec } x$ has vertical asymptotes at the values of $x$ where $\sin x = 0$ (integer multiples of $\pi$). The graph consists of repeating U-shaped curves, opening upwards above $y=1$ and downwards below $y=-1$. The graph is related to the $\sin x$ graph; where $\sin x$ has a maximum (1), $\text{cosec } x$ has a minimum (1), and where $\sin x$ has a minimum (-1), $\text{cosec } x$ has a maximum (-1).

Graph of y = cosec x showing vertical asymptotes at multiples of pi and U-shaped curves between them, tangent to y=1 and y=-1 at points where sin x is 1 and -1

5. Graph of $y = \sec x$:

The secant function $f(x) = \sec x = \frac{1}{\cos x}$.

Domain: The secant function is defined for all real numbers $x$ except where $\cos x = 0$. This occurs at $x = (2n+1)\frac{\pi}{2}$ for any integer $n$.

Domain($\sec x$) $= \{ x \in \mathbb{R} : x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} \}$

Range: Since $\cos x$ varies between -1 and 1 (excluding 0 where sec is undefined), $1/\cos x$ will have values greater than or equal to 1 or less than or equal to -1. The range is $(-\infty, -1] \cup [1, \infty)$.

Range($\sec x$) $= (-\infty, -1] \cup [1, \infty)$

Period: The period of $\sec x$ is $2\pi$.

The graph of $y = \sec x$ has vertical asymptotes at the values of $x$ where $\cos x = 0$ (odd multiples of $\pi/2$). The graph consists of repeating U-shaped curves, opening upwards above $y=1$ and downwards below $y=-1$. The graph is related to the $\cos x$ graph; where $\cos x$ has a maximum (1), $\sec x$ has a minimum (1), and where $\cos x$ has a minimum (-1), $\sec x$ has a maximum (-1).

Graph of y = sec x showing vertical asymptotes at odd multiples of pi/2 and U-shaped curves between them, tangent to y=1 and y=-1 at points where cos x is 1 and -1, symmetric about the y-axis

6. Graph of $y = \cot x$:

The cotangent function $f(x) = \cot x = \frac{\cos x}{\sin x}$.

Domain: The cotangent function is defined for all real numbers $x$ except where $\sin x = 0$. This occurs at $x = n\pi$ for any integer $n$.

Domain($\cot x$) $= \{ x \in \mathbb{R} : x \neq n\pi, n \in \mathbb{Z} \}$

Range: The value of $\cot x$ can be any real number. Range($\cot x$) $= \mathbb{R}$.

Period: The period of $\cot x$ is $\pi$.

The graph of $y = \cot x$ has vertical asymptotes at the values of $x$ where it is undefined (integer multiples of $\pi$). The graph consists of repeating curves that decrease within each interval between asymptotes, passing through the x-axis at odd multiples of $\pi/2$.

Graph of y = cot x showing vertical asymptotes at integer multiples of pi and repeating curves decreasing within each vertical strip, crossing the x-axis at odd multiples of pi/2

The graph illustrates that $\cot x$ approaches $\infty$ as $x$ approaches $n\pi$ from the right, and approaches $-\infty$ as $x$ approaches $n\pi$ from the left.



Trigonometric Functions of Sum and Difference

The ability to express trigonometric functions of angles that are the sum or difference of two other angles is essential for simplifying expressions, solving trigonometric equations, and deriving further identities. These formulas are often referred to as compound angle formulas or addition and subtraction formulas.


Fundamental Formulas:

The most fundamental of these formulas are those for the cosine and sine of the sum and difference of two angles.

1. Cosine of Sum and Difference of Two Angles:

For any real numbers $x$ and $y$ (representing angles in radians):

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

... (1)

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

... (2)

These are central identities from which many others can be derived.

Derivation of the Cosine Difference Formula: $\cos(x - y)$

We can derive this fundamental formula using the unit circle and the distance formula. This geometric proof connects algebra and trigonometry.

Step 1: Set up the points on the Unit Circle

Consider a unit circle centered at the origin $O(0,0)$. Let's place two points, P and Q, on the circle corresponding to angles $x$ and $y$ respectively.

The angle between the segments OP and OQ is $\angle \text{POQ} = x - y$.

A unit circle used to derive the formula for cos(x-y). Points P and Q correspond to angles x and y. The system is then rotated so the angle x-y is in standard position.

Step 2: Calculate the distance PQ using the Distance Formula

The square of the distance between points P and Q is given by:

$PQ^2 = (\cos x - \cos y)^2 + (\sin x - \sin y)^2$

Now, expand the squared terms:

$PQ^2 = (\cos^2 x - 2\cos x \cos y + \cos^2 y) + (\sin^2 x - 2\sin x \sin y + \sin^2 y)$

Group the terms using the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$:

$PQ^2 = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2\cos x \cos y - 2\sin x \sin y$

$PQ^2 = 1 + 1 - 2(\cos x \cos y + \sin x \sin y)$

$PQ^2 = 2 - 2(\cos x \cos y + \sin x \sin y)$

... (A)

Step 3: Rotate the system and calculate the distance again

Now, let's rotate the entire setup by an angle of $-y$ so that point Q moves to the position $Q'(1, 0)$ on the x-axis. Since rotation is a rigid transformation, the distance between the points remains the same. The point P will move to a new position P', which now corresponds to the angle $x-y$.

The square of the distance between these new points P' and Q' is:

$(P'Q')^2 = (\cos(x-y) - 1)^2 + (\sin(x-y) - 0)^2$

Expand the terms:

$(P'Q')^2 = (\cos^2(x-y) - 2\cos(x-y) + 1) + \sin^2(x-y)$

Again, group the terms using the Pythagorean identity:

$(P'Q')^2 = (\cos^2(x-y) + \sin^2(x-y)) + 1 - 2\cos(x-y)$

$(P'Q')^2 = 1 + 1 - 2\cos(x-y)$

$(P'Q')^2 = 2 - 2\cos(x-y)$

... (B)

Step 4: Equate the two expressions for the distance

Since the rotation did not change the distance between the points, $PQ^2 = (P'Q')^2$. We can now equate expressions (A) and (B):

$2 - 2(\cos x \cos y + \sin x \sin y) = 2 - 2\cos(x-y)$

Subtract 2 from both sides:

$-2(\cos x \cos y + \sin x \sin y) = -2\cos(x-y)$

Finally, divide both sides by -2 to arrive at the formula:

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

Derivation of the Cosine Sum Formula: $\cos(x + y)$

To derive the formula for $\cos(x + y)$, we can cleverly use the formula for $\cos(x - y)$, which we have already proven. The key is to rewrite the sum $x+y$ as a difference: $x - (-y)$.

A unit circle showing the angle x+y, which is formed by first rotating by an angle x from the positive x-axis and then by a further angle y.

We express the sum as follows:

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

Now, we can apply the cosine difference formula, $\cos(A - B) = \cos A \cos B + \sin A \sin B$, where $A=x$ and $B=-y$:

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

Next, we use the properties of negative angles (also known as even/odd function properties):

Substituting these identities into our expression:

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

This simplifies directly to the final formula for the cosine of a sum:

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


2. Sine of Sum and Difference of Two Angles:

We can derive the formulas for sine using the co-function identity $\sin \theta = \cos(\pi/2 - \theta)$ and the cosine sum/difference formulas.

Derivation of $\sin(x+y)$:

$\sin(x + y) = \cos(\frac{\pi}{2} - (x + y))$

$= \cos((\frac{\pi}{2} - x) - y)$

Apply the $\cos(A - B)$ formula (Formula 2) with $A = (\pi/2 - x)$ and $B = y$:

$= \cos(\frac{\pi}{2} - x) \cos y + \sin(\frac{\pi}{2} - x) \sin y$

Using the co-function identities $\cos(\pi/2 - x) = \sin x$ and $\sin(\pi/2 - x) = \cos x$:

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

... (3)

Derivation of $\sin(x-y)$:

To derive the formula for $\sin(x - y)$, substitute $-y$ for $y$ in the formula for $\sin(x + y)$ (Formula 3):

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

Apply Formula (3) with the second angle being $-y$:

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

Using the negative angle identities $\cos(-y) = \cos y$ and $\sin(-y) = -\sin y$:

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

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

... (4)


3. Tangent of Sum and Difference of Two Angles:

We can derive the formulas for the tangent of sum and difference using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and the formulas for sine and cosine sums/differences.

Derivation of $\tan(x+y)$:

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

Substitute the formulas for $\sin(x+y)$ (Formula 3) and $\cos(x+y)$ (Formula 1):

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

To express this in terms of $\tan x$ and $\tan y$, divide the numerator and the denominator by $\cos x \cos y$, assuming $\cos x \neq 0$ and $\cos y \neq 0$:

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

Simplify the terms (e.g., $\frac{\sin x \cos y}{\cos x \cos y} = \frac{\sin x}{\cos x} = \tan x$):

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

... (5)

This formula is valid when $\cos x \neq 0$, $\cos y \neq 0$, and $\cos(x+y) \neq 0$.

Derivation of $\tan(x-y)$:

Substitute $-y$ for $y$ in the formula for $\tan(x + y)$ (Formula 5):

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

Using the negative angle identity $\tan(-y) = -\tan y$:

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

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

... (6)

This formula is valid when $\cos x \neq 0$, $\cos y \neq 0$, and $\cos(x-y) \neq 0$.


4. Cotangent of Sum and Difference of Two Angles:

Similar derivations using $\cot \theta = \frac{\cos \theta}{\sin \theta}$ or $\cot \theta = \frac{1}{\tan \theta}$ lead to the formulas for cotangent.

Derivation of $\cot(x+y)$ (using $\tan(x+y)$):

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

To express this in terms of $\cot x$ and $\cot y$, divide the numerator and denominator by $\tan x \tan y$ (assuming $\tan x \neq 0, \tan y \neq 0$):

$= \frac{\frac{1}{\tan x \tan y} - 1}{\frac{\tan x}{\tan x \tan y} + \frac{\tan y}{\tan x \tan y}} = \frac{\cot x \cot y - 1}{\cot y + \cot x}$

... (7)

This formula is valid when $\sin x \neq 0$, $\sin y \neq 0$, and $\sin(x+y) \neq 0$.

Derivation of $\cot(x-y)$ (using $\tan(x-y)$):

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

Divide the numerator and denominator by $\tan x \tan y$ (assuming $\tan x \neq 0, \tan y \neq 0$):

$= \frac{\frac{1}{\tan x \tan y} + 1}{\frac{\tan x}{\tan x \tan y} - \frac{\tan y}{\tan x \tan y}} = \frac{\cot x \cot y + 1}{\cot y - \cot x}$

... (8)

This formula is valid when $\sin x \neq 0$, $\sin y \neq 0$, and $\sin(x-y) \neq 0$.


Identities derived from Sum/Difference Formulas:

The sum and difference formulas are fundamental and can be used to derive many other important trigonometric identities.

Double Angle Formulas:

By setting $y = x$ in the sum formulas (Formulas 1, 3, and 5), we obtain the double angle formulas:

$\sin(x + x) = \sin x \cos x + \cos x \sin x = 2 \sin x \cos x$

$\sin(2x) = 2 \sin x \cos x$

... (9)

$\cos(x + x) = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$

Using the identity $\sin^2 x + \cos^2 x = 1$, we can write $\cos^2 x = 1 - \sin^2 x$ or $\sin^2 x = 1 - \cos^2 x$ to get alternate forms:

$\cos(2x) = \cos^2 x - \sin^2 x$

... (10a)

$\cos(2x) = (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x$

... (10b)

$\cos(2x) = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1$

... (10c)

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

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

... (11)

These double angle formulas are valid for values of $x$ for which the terms are defined.

Product-to-Sum Formulas:

By adding or subtracting the sum and difference formulas, we can express products of sines and cosines as sums or differences of sines and cosines.

Add (3) and (4):

$$\begin{array}{r c l} & \sin(x+y) & = & \sin x \cos y + \cos x \sin y \\ + & \sin(x-y) & = & \sin x \cos y - \cos x \sin y \\ \hline & \sin(x+y) + \sin(x-y) & = & 2 \sin x \cos y \\ \hline \end{array}$$

$2 \sin x \cos y = \sin(x+y) + \sin(x-y)$

... (12)

Subtract (4) from (3):

$$\begin{array}{r c l} & \sin(x+y) & = & \sin x \cos y + \cos x \sin y \\ - & (\sin(x-y) & = & \sin x \cos y - \cos x \sin y) \\ \hline + & -\sin(x-y) & = & -\sin x \cos y + \cos x \sin y \\ \hline & \sin(x+y) - \sin(x-y) & = & 2 \cos x \sin y \\ \hline \end{array}$$

$2 \cos x \sin y = \sin(x+y) - \sin(x-y)$

... (13)

Add (1) and (2):

$$\begin{array}{r c l} & \cos(x+y) & = & \cos x \cos y - \sin x \sin y \\ + & \cos(x-y) & = & \cos x \cos y + \sin x \sin y \\ \hline & \cos(x+y) + \cos(x-y) & = & 2 \cos x \cos y \\ \hline \end{array}$$

$2 \cos x \cos y = \cos(x+y) + \cos(x-y)$

... (14)

Subtract (1) from (2):

$$\begin{array}{r c l} & \cos(x-y) & = & \cos x \cos y + \sin x \sin y \\ - & (\cos(x+y) & = & \cos x \cos y - \sin x \sin y) \\ \hline + & -\cos(x+y) & = & -\cos x \cos y + \sin x \sin y \\ \hline & \cos(x-y) - \cos(x+y) & = & 2 \sin x \sin y \\ \hline \end{array}$$

$2 \sin x \sin y = \cos(x-y) - \cos(x+y)$

... (15)

These four formulas are the product-to-sum identities.

Sum-to-Product Formulas:

By substituting $C = x+y$ and $D = x-y$ into the product-to-sum formulas, we can derive formulas that express sums or differences of sines and cosines as products.


Derivation of Substitution Variables

Let's define two new variables, C and D, as follows:

$C = x+y$

...(a)

$D = x-y$

...(b)

Adding equations (a) and (b) gives:

$C+D = (x+y) + (x-y) = 2x$

$\implies x = \frac{C+D}{2}$

Subtracting equation (b) from (a) gives:

$C-D = (x+y) - (x-y) = 2y$

$\implies y = \frac{C-D}{2}$

We now substitute these expressions for $x$ and $y$ back into the product-to-sum formulas (assuming they are numbered 12 to 15).


The Sum-to-Product Identities

From the formula $2 \sin x \cos y = \sin(x+y) + \sin(x-y)$:

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

... (16)

From the formula $2 \cos x \sin y = \sin(x+y) - \sin(x-y)$:

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

... (17)

From the formula $2 \cos x \cos y = \cos(x+y) + \cos(x-y)$:

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

... (18)

From the formula $2 \sin x \sin y = \cos(x-y) - \cos(x+y)$:

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

... (19)

Note that $\cos D - \cos C = -(\cos C - \cos D)$. So, formula (19) is often written with a negative sign:

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

... (19')

These four formulas are known as the sum-to-product identities.


Mastering these sum, difference, double angle, product-to-sum, and sum-to-product formulas is fundamental for solving various problems in trigonometry.


Trigonometric Functions of Allied Angles

Allied angles are angles that are related to a given angle, $x$, by adding or subtracting multiples of $\pi/2$ (or $90^\circ$). The trigonometric functions of these allied angles can be expressed in terms of the trigonometric functions of $x$ itself. These identities are extremely useful for simplifying expressions and are derived directly from the sum and difference formulas.

The core formulas we will use for the derivations are:


1. Angles related to $\frac{\pi}{2}$

Functions of $(\frac{\pi}{2} - x)$ (Co-function Identities)

These identities show the relationship between a function and its "co-function" (sine and cosine, tangent and cotangent, secant and cosecant).

Derivations:

(i) $\sin(\frac{\pi}{2} - x) = \cos x$

Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$ with $A = \frac{\pi}{2}$ and $B=x$:

$\sin(\frac{\pi}{2} - x) = \sin(\frac{\pi}{2})\cos x - \cos(\frac{\pi}{2})\sin x$

Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$:

$ = (1)\cos x - (0)\sin x = \cos x$

(ii) $\cos(\frac{\pi}{2} - x) = \sin x$

Using the formula $\cos(A-B) = \cos A \cos B + \sin A \sin B$:

$\cos(\frac{\pi}{2} - x) = \cos(\frac{\pi}{2})\cos x + \sin(\frac{\pi}{2})\sin x$

$ = (0)\cos x + (1)\sin x = \sin x$

(iii) $\tan(\frac{\pi}{2} - x) = \cot x$

$\tan(\frac{\pi}{2} - x) = \frac{\sin(\frac{\pi}{2} - x)}{\cos(\frac{\pi}{2} - x)} = \frac{\cos x}{\sin x} = \cot x$

The remaining co-function identities follow directly: $\cot(\frac{\pi}{2} - x) = \tan x$, $\sec(\frac{\pi}{2} - x) = \text{cosec } x$, and $\text{cosec}(\frac{\pi}{2} - x) = \sec x$.

Functions of $(\frac{\pi}{2} + x)$

Derivations:

(i) $\sin(\frac{\pi}{2} + x) = \cos x$

Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

$\sin(\frac{\pi}{2} + x) = \sin(\frac{\pi}{2})\cos x + \cos(\frac{\pi}{2})\sin x$

$ = (1)\cos x + (0)\sin x = \cos x$

(ii) $\cos(\frac{\pi}{2} + x) = -\sin x$

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

$\cos(\frac{\pi}{2} + x) = \cos(\frac{\pi}{2})\cos x - \sin(\frac{\pi}{2})\sin x$

$ = (0)\cos x - (1)\sin x = -\sin x$

(iii) $\tan(\frac{\pi}{2} + x) = -\cot x$

$\tan(\frac{\pi}{2} + x) = \frac{\sin(\frac{\pi}{2} + x)}{\cos(\frac{\pi}{2} + x)} = \frac{\cos x}{-\sin x} = -\cot x$


2. Angles related to $\pi$

Functions of $(\pi - x)$

Derivations:

We use the values $\sin(\pi) = 0$ and $\cos(\pi) = -1$.

(i) $\sin(\pi - x) = \sin x$

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

$ = (0)\cos x - (-1)\sin x = \sin x$

(ii) $\cos(\pi - x) = -\cos x$

$\cos(\pi - x) = \cos(\pi)\cos x + \sin(\pi)\sin x$

$ = (-1)\cos x + (0)\sin x = -\cos x$

(iii) $\tan(\pi - x) = -\tan x$

$\tan(\pi - x) = \frac{\sin(\pi - x)}{\cos(\pi - x)} = \frac{\sin x}{-\cos x} = -\tan x$

Functions of $(\pi + x)$

Derivations:

(i) $\sin(\pi + x) = -\sin x$

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

$ = (0)\cos x + (-1)\sin x = -\sin x$

(ii) $\cos(\pi + x) = -\cos x$

$\cos(\pi + x) = \cos(\pi)\cos x - \sin(\pi)\sin x$

$ = (-1)\cos x - (0)\sin x = -\cos x$

(iii) $\tan(\pi + x) = \tan x$

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


3. Angles related to $\frac{3\pi}{2}$

Derivations:

We use the values $\sin(\frac{3\pi}{2}) = -1$ and $\cos(\frac{3\pi}{2}) = 0$.

Functions of $(\frac{3\pi}{2} - x)$

(i) $\sin(\frac{3\pi}{2} - x) = -\cos x$

$\sin(\frac{3\pi}{2} - x) = \sin(\frac{3\pi}{2})\cos x - \cos(\frac{3\pi}{2})\sin x$

$ = (-1)\cos x - (0)\sin x$

$ = -\cos x$

(ii) $\cos(\frac{3\pi}{2} - x) = -\sin x$

$\cos(\frac{3\pi}{2} - x) = \cos(\frac{3\pi}{2})\cos x + \sin(\frac{3\pi}{2})\sin x$

$ = (0)\cos x + (-1)\sin x$

$ = -\sin x$

Functions of $(\frac{3\pi}{2} + x)$

(i) $\sin(\frac{3\pi}{2} + x) = -\cos x$

$\sin(\frac{3\pi}{2} + x) = \sin(\frac{3\pi}{2})\cos x + \cos(\frac{3\pi}{2})\sin x$

$ = (-1)\cos x + (0)\sin x$

$ = -\cos x$

(ii) $\cos(\frac{3\pi}{2} + x) = \sin x$

$\cos(\frac{3\pi}{2} + x) = \cos(\frac{3\pi}{2})\cos x - \sin(\frac{3\pi}{2})\sin x$

$ = (0)\cos x - (-1)\sin x$

$ = \sin x$


4. Angles related to $2\pi$

Functions of $(2\pi - x)$ (or $-x$)

These are the even/odd function identities. We use $\sin(2\pi) = 0$ and $\cos(2\pi) = 1$.

Derivations:

(i) $\sin(2\pi - x) = -\sin x$

$\sin(2\pi - x) = \sin(2\pi)\cos x - \cos(2\pi)\sin x$

$ = (0)\cos x - (1)\sin x$

$ = -\sin x$

(ii) $\cos(2\pi - x) = \cos x$

$\cos(2\pi - x) = \cos(2\pi)\cos x + \sin(2\pi)\sin x$

$ = (1)\cos x + (0)\sin x$

$ = \cos x$

Functions of $(2\pi + x)$ (Periodicity)

These identities demonstrate the periodic nature of trigonometric functions.

Derivations:

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

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

$ = (0)\cos x + (1)\sin x$

$ = \sin x$

(ii) $\cos(2\pi + x) = \cos x$

$\cos(2\pi + x) = \cos(2\pi)\cos x - \sin(2\pi)\sin x$

$ = (1)\cos x - (0)\sin x$

$ = \cos x$


Summary and Mnemonics

A summary table helps to visualize the patterns. A common mnemonic to remember the signs of functions in the four quadrants is "All Silver Tea Cups" or "All Students Take Calculus".

Another key rule:

Combining these two rules allows for the quick determination of any allied angle identity.

Angle $\sin$ $\cos$ $\tan$ $\cot$ $\sec$ $\text{cosec}$
$\frac{\pi}{2} - x$ $\cos x$ $\sin x$ $\cot x$ $\tan x$ $\text{cosec } x$ $\sec x$
$\frac{\pi}{2} + x$ $\cos x$ $-\sin x$ $-\cot x$ $-\tan x$ $-\text{cosec } x$ $\sec x$
$\pi - x$ $\sin x$ $-\cos x$ $-\tan x$ $-\cot x$ $-\sec x$ $\text{cosec } x$
$\pi + x$ $-\sin x$ $-\cos x$ $\tan x$ $\cot x$ $-\sec x$ $-\text{cosec } x$
$\frac{3\pi}{2} - x$ $-\cos x$ $-\sin x$ $\cot x$ $\tan x$ $-\text{cosec } x$ $-\sec x$
$\frac{3\pi}{2} + x$ $-\cos x$ $\sin x$ $-\cot x$ $-\tan x$ $\text{cosec } x$ $-\sec x$
$2\pi - x$ (or $-x$) $-\sin x$ $\cos x$ $-\tan x$ $-\cot x$ $\sec x$ $-\text{cosec } x$
$2\pi + x$ $\sin x$ $\cos x$ $\tan x$ $\cot x$ $\sec x$ $\text{cosec } x$


Trigonometric Equations

A trigonometric equation is an equation that involves one or more trigonometric functions of a variable angle. The goal of solving a trigonometric equation is to find all possible values of the angle that satisfy the equation.

For example, $\sin x = \frac{1}{2}$ or $2\cos^2 \theta + 3\sin \theta = 0$ are trigonometric equations.

Unlike algebraic equations which typically have a finite number of solutions, trigonometric equations often have an infinite number of solutions. This is because trigonometric functions are periodic. They repeat their values at regular intervals. For example, if $\sin x = \frac{1}{2}$, then $x$ could be $\frac{\pi}{6}$, but it could also be $\frac{\pi}{6} + 2\pi$, $\frac{\pi}{6} + 4\pi$, etc., as adding a full circle ($2\pi$ radians) brings us back to the same point on the unit circle.

A sine wave showing its periodic nature. The line y=0.5 intersects the sine wave at infinitely many points, illustrating the infinite solutions to sin(x)=0.5.

Principal Solution and General Solution

Due to the infinite nature of solutions, we categorize them into two types:

Principal Solution

The principal solution(s) of a trigonometric equation are the solutions that lie within a specific interval, typically the interval $[0, 2\pi)$. This interval represents one full counter-clockwise rotation around the unit circle. For most basic trigonometric equations, there are usually two principal solutions (unless the value is ±1 or 0 for sine and cosine).

General Solution

The general solution of a trigonometric equation is an expression that represents the set of all possible solutions. It is expressed using an integer variable, typically denoted by 'n' (where $n \in \mathbb{Z}$, the set of all integers), to account for all the periodic repetitions of the solutions.


General Solutions of Basic Trigonometric Equations

The general solutions for the most fundamental trigonometric equations are derived from the properties of the unit circle and the periodicity of the functions.

1. General Solution of $\sin x = \sin y$

If $\sin x = \sin y$, where $y$ is a known angle, the general solution for $x$ is given by the formula:

$\sin x = \sin y \implies x = n\pi + (-1)^n y$

where $n \in \mathbb{Z}$           ... (1)

Explanation: The sine of an angle is represented by the y-coordinate on the unit circle. For a given value of sine (say, positive), there are two angles in one rotation ($[0, 2\pi)$): one in the first quadrant ($y$) and one in the second quadrant ($\pi - y$).

Unit circle showing two angles, y and (pi - y), that have the same sine value. This illustrates the basis for the general solution of sin(x) = sin(y).

Derivation of Formula (1)

Given: The trigonometric equation $\sin x = \sin y$.

To Prove: The general solution is $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$.

Proof:

We start with the given equation:

$\sin x = \sin y$

Rearrange the equation to set it equal to zero:

$\sin x - \sin y = 0$

Now, we apply the sum-to-product formula: $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$.

$2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = 0$

For this product to be zero, at least one of the trigonometric factors must be zero.

Case 1: $\cos\left(\frac{x+y}{2}\right) = 0$

The general solution for $\cos \theta = 0$ is $\theta = (2k+1)\frac{\pi}{2}$ for any integer $k \in \mathbb{Z}$.

$\frac{x+y}{2} = (2k+1)\frac{\pi}{2}$

Multiplying both sides by 2, we get:

$x + y = (2k+1)\pi$

Solving for $x$:

$x = (2k+1)\pi - y$

Let $n = 2k+1$, which represents any odd integer. For an odd $n$, we have $(-1)^n = -1$. So, this solution can be written as $x = n\pi + (-1)^n y$ for odd $n$.

Case 2: $\sin\left(\frac{x-y}{2}\right) = 0$

The general solution for $\sin \phi = 0$ is $\phi = k\pi$ for any integer $k \in \mathbb{Z}$.

$\frac{x-y}{2} = k\pi$

Multiplying both sides by 2, we get:

$x - y = 2k\pi$

Solving for $x$:

$x = 2k\pi + y$

Let $n = 2k$, which represents any even integer. For an even $n$, we have $(-1)^n = 1$. So, this solution can be written as $x = n\pi + (-1)^n y$ for even $n$.

By combining the results from Case 1 (for odd integers $n$) and Case 2 (for even integers $n$), we arrive at a single, unified general solution for all integers $n \in \mathbb{Z}$:

$\boldsymbol{x = n\pi + (-1)^n y}$


2. General Solution of $\cos x = \cos y$

If $\cos x = \cos y$, where $y$ is a known angle, the general solution for $x$ is given by:

$\cos x = \cos y \implies x = 2n\pi \pm y$

where $n \in \mathbb{Z}$           ... (2)

Explanation: The cosine of an angle is the x-coordinate on the unit circle. For a given x-coordinate, there are two angles in one rotation ($[0, 2\pi)$): one in the upper semi-circle ($y$) and one in the lower semi-circle ($-y$ or $2\pi-y$). The term $2n\pi$ accounts for any number of full rotations, and the $\pm y$ covers both of these possibilities.

Unit circle showing two angles, y and -y, that have the same cosine value. This illustrates the basis for the general solution of cos(x) = cos(y).

Derivation of Formula (2)

Given: The trigonometric equation $\cos x = \cos y$.

To Prove: The general solution is $x = 2n\pi \pm y$, where $n \in \mathbb{Z}$.

Proof:

We start with the equation:

$\cos x - \cos y = 0$

Apply the sum-to-product formula: $\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$.

$-2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = 0$

This implies one of the sine factors must be zero.

Case 1: $\sin\left(\frac{x+y}{2}\right) = 0$

The general solution is $\frac{x+y}{2} = n\pi$ for $n \in \mathbb{Z}$.

$x + y = 2n\pi$

$x = 2n\pi - y$

Case 2: $\sin\left(\frac{x-y}{2}\right) = 0$

The general solution is $\frac{x-y}{2} = n\pi$ for $n \in \mathbb{Z}$.

$x - y = 2n\pi$

$x = 2n\pi + y$

Combining the results from Case 1 ($2n\pi - y$) and Case 2 ($2n\pi + y$), we get the compact general solution:

$\boldsymbol{x = 2n\pi \pm y}$


3. General Solution of $\tan x = \tan y$

If $\tan x = \tan y$, where $y$ is a known angle, the general solution for $x$ is given by:

$\tan x = \tan y \implies x = n\pi + y$

where $n \in \mathbb{Z}$           ... (3)

Explanation: The tangent function has a period of $\pi$. This means its values repeat every $\pi$ radians. If $y$ is a solution, then adding any integer multiple of $\pi$ will also result in a solution. The angles $y$ and $y+\pi$ are diametrically opposite on the unit circle, and they have the same tangent value.

Unit circle showing two angles, y and (pi + y), that are diametrically opposite and have the same tangent value. This illustrates the basis for the general solution of tan(x) = tan(y).

Derivation of Formula (3)

Given: The trigonometric equation $\tan x = \tan y$.

To Prove: The general solution is $x = n\pi + y$, where $n \in \mathbb{Z}$.

Proof:

We assume that $\cos x \neq 0$ and $\cos y \neq 0$ so that the tangent function is defined.

$\tan x - \tan y = 0$

Rewrite in terms of sine and cosine:

$\frac{\sin x}{\cos x} - \frac{\sin y}{\cos y} = 0$

Combine the fractions:

$\frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y} = 0$

For the fraction to be zero, the numerator must be zero.

$\sin x \cos y - \cos x \sin y = 0$

The left side is the angle subtraction formula for sine, $\sin(x-y)$.

$\sin(x - y) = 0$

The general solution for $\sin \theta = 0$ is $\theta = n\pi$ for any integer $n \in \mathbb{Z}$.

$x - y = n\pi$

Solving for $x$ gives the general solution:

$\boldsymbol{x = n\pi + y}$


Summary of General Solutions (Including Special Cases)

Equation Condition General Solution
$\sin x = \sin y$ $x = n\pi + (-1)^n y$
$\cos x = \cos y$ $x = 2n\pi \pm y$
$\tan x = \tan y$ $x = n\pi + y$
Special Cases
$\sin x = 0$ ($y=0$) $x = n\pi$
$\cos x = 0$ ($y=\pi/2$) $x = (2n+1)\frac{\pi}{2}$
$\tan x = 0$ ($y=0$) $x = n\pi$
$\sin x = 1$ ($y=\pi/2$) $x = 2n\pi + \frac{\pi}{2}$
$\cos x = 1$ ($y=0$) $x = 2n\pi$
$\sin x = -1$ ($y=-\pi/2$) $x = 2n\pi - \frac{\pi}{2}$
$\cos x = -1$ ($y=\pi$) $x = (2n+1)\pi$

In all cases, $n \in \mathbb{Z}$.

The solutions for $\text{cosec } x = \text{cosec } y$, $\sec x = \sec y$, and $\cot x = \cot y$ are the same as for their reciprocal functions, with the additional constraint that the original functions must be defined (i.e., their denominators are not zero).


Example 1. Find the principal and general solutions of the equation $\sin x = \frac{\sqrt{3}}{2}$.

Answer:

Given: The trigonometric equation $\sin x = \frac{\sqrt{3}}{2}$.

To Find:

  1. The principal solutions (solutions in the interval $[0, 2\pi)$).
  2. The general solution.

Solution:

First, we identify an angle whose sine is $\frac{\sqrt{3}}{2}$. From standard trigonometric values, we know that $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Finding Principal Solutions:

We need to find angles $x$ in the interval $[0, 2\pi)$ for which $\sin x$ is positive.

  • Quadrant I: The sine function is positive in the first quadrant. The angle is $x = \frac{\pi}{3}$. This is a principal solution.
  • Quadrant II: The sine function is also positive in the second quadrant. The corresponding angle is given by $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$. This is also a principal solution.

The principal solutions are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.

Finding the General Solution:

We write the equation in the form $\sin x = \sin y$. Using $y = \frac{\pi}{3}$, we have:

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

Now, we apply the general solution formula for sine: $x = n\pi + (-1)^n y$.

$x = n\pi + (-1)^n \frac{\pi}{3}$, where $n \in \mathbb{Z}$

This single formula encompasses all solutions, including the principal ones we found. For $n=0$, $x=\frac{\pi}{3}$. For $n=1$, $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Final Answer:

The principal solutions are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.

The general solution is $x = n\pi + (-1)^n \frac{\pi}{3}$, where $n \in \mathbb{Z}$.


Example 2. Find the general solution of $\cos x = \frac{1}{2}$.

Answer:

Given: The equation $\cos x = \frac{1}{2}$.

To Find: The general solution.

Solution:

First, we identify a basic angle $y$ such that $\cos y = \frac{1}{2}$. We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

So, we can write the equation as:

$\cos x = \cos\left(\frac{\pi}{3}\right)$

Now, we apply the general solution formula for cosine: $x = 2n\pi \pm y$.

Substituting $y = \frac{\pi}{3}$:

$x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$

Final Answer: The general solution is $x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.


Example 3. Solve the equation $\tan 2x = -\cot(x + \frac{\pi}{3})$.

Answer:

Given: The equation $\tan 2x = -\cot(x + \frac{\pi}{3})$.

To Find: The general solution for $x$.

Solution:

Our goal is to transform the equation into the form $\tan A = \tan B$. We need to express $-\cot \theta$ in terms of $\tan$. We can use the identity $-\cot \theta = \tan\left(\frac{\pi}{2} + \theta\right)$.

Let $\theta = x + \frac{\pi}{3}$. Applying the identity to the right side of the equation:

$-\cot\left(x + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{2} + \left(x + \frac{\pi}{3}\right)\right)$

Now, simplify the angle inside the tangent function:

$\frac{\pi}{2} + x + \frac{\pi}{3} = x + \frac{3\pi + 2\pi}{6} = x + \frac{5\pi}{6}$

So, the original equation becomes:

$\tan(2x) = \tan\left(x + \frac{5\pi}{6}\right)$

This is now in the form $\tan A = \tan B$, where $A = 2x$ and $B = x + \frac{5\pi}{6}$.

We apply the general solution formula for tangent: $A = n\pi + B$.

$2x = n\pi + \left(x + \frac{5\pi}{6}\right)$, where $n \in \mathbb{Z}$

Now, we solve this algebraic equation for $x$:

$2x - x = n\pi + \frac{5\pi}{6}$

$x = n\pi + \frac{5\pi}{6}$

Final Answer: The general solution is $x = n\pi + \frac{5\pi}{6}$, where $n \in \mathbb{Z}$.

Alternate Solution:

We could also use the identity $\cot \theta = \tan(\frac{\pi}{2} - \theta)$.

$\tan 2x = -\tan\left(\frac{\pi}{2} - \left(x + \frac{\pi}{3}\right)\right)$

Using the identity $-\tan \phi = \tan(-\phi)$:

$\tan 2x = \tan\left(-\left(\frac{\pi}{2} - x - \frac{\pi}{3}\right)\right) = \tan\left(x - \frac{\pi}{6}\right)$

Applying the general solution $2x = n\pi + (x - \frac{\pi}{6})$ gives $x = n\pi - \frac{\pi}{6}$.

Note: The solutions $n\pi + \frac{5\pi}{6}$ and $m\pi - \frac{\pi}{6}$ generate the same set of angles. For instance, for $n=0$, we get $5\pi/6$. For $m=1$, we get $\pi - \pi/6 = 5\pi/6$. Both forms are correct.