Trigonometric Ratios of Special Angles and Complementary Angles

Trigonometric Ratios of Some Specific Angles (0°, 30°, 45°, 60°, 90°)

While trigonometric ratios can be calculated for any acute angle using a right-angled triangle, there are certain angles whose trigonometric ratios are frequently used and can be determined geometrically or through logical extension. These specific angles are $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, and $90^\circ$. Knowing the values of trigonometric ratios for these angles is essential.

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Derivation for $45^\circ$

To find the trigonometric ratios for $45^\circ$, we can consider a special type of right-angled triangle: an isosceles right-angled triangle. In such a triangle, one angle is $90^\circ$, and the other two angles are equal, each measuring $\frac{180^\circ - 90^\circ}{2} = 45^\circ$.

Consider a right-angled triangle ABC, where $\angle B = 90^\circ$ and $\angle A = \angle C = 45^\circ$.

Since it's an isosceles right-angled triangle with $\angle A = \angle C$, the sides opposite to these angles must be equal in length. So, AB = BC.

Let's assume the length of AB (and BC) is $a$ units, where $a > 0$.

Now, we find the length of the hypotenuse AC using the Pythagorean Theorem:

$ AC^2 = AB^2 + BC^2 $

Substitute the lengths:

$ AC^2 = a^2 + a^2 $

$ AC^2 = 2a^2 $

Taking the square root of both sides (and considering only the positive length):

$ AC = \sqrt{2a^2} = \sqrt{2} \times \sqrt{a^2} = \sqrt{2}a $

So, the sides of a $45^\circ - 45^\circ - 90^\circ$ triangle are in the ratio $a : a : a\sqrt{2}$, or $1 : 1 : \sqrt{2}$.

Now, let's find the trigonometric ratios for $\theta = 45^\circ$. We can use either $\angle A$ or $\angle C$. Let's use $\angle C$.

• Opposite Side (O) to $\angle C$ = AB = $a$

• Adjacent Side (A) to $\angle C$ = BC = $a$}

• Hypotenuse (H) = AC = $a\sqrt{2}$

Using the definitions of trigonometric ratios:

$ \sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{a}{a\sqrt{2}} = \frac{\cancel{a}}{\cancel{a}\sqrt{2}} = \frac{1}{\sqrt{2}} $

$ \cos 45^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{a}{a\sqrt{2}} = \frac{\cancel{a}}{\cancel{a}\sqrt{2}} = \frac{1}{\sqrt{2}} $

$ \tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{a}{a} = \frac{\cancel{a}}{\cancel{a}} = 1 $

Now, let's find the reciprocal ratios:

$ \text{cosec} \, 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{1/\sqrt{2}} = \sqrt{2} $

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

$ \cot 45^\circ = \frac{1}{\tan 45^\circ} = \frac{1}{1} = 1 $

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Derivation for $30^\circ$ and $60^\circ$

To find the trigonometric ratios for $30^\circ$ and $60^\circ$, we can consider an equilateral triangle. In an equilateral triangle, all sides are equal, and all angles are equal to $60^\circ$.

Consider an equilateral triangle ABC with side length $2a$, where $a > 0$. Thus, AB = BC = CA = $2a$, and $\angle A = \angle B = \angle C = 60^\circ$.

Draw an altitude (perpendicular) AD from vertex A to the side BC. In an equilateral triangle, the altitude also acts as a median and an angle bisector.

Therefore, AD is perpendicular to BC ($\angle ADB = 90^\circ$), D is the midpoint of BC (so BD = DC = $\frac{1}{2} \times 2a = a$), and AD bisects $\angle BAC$ (so $\angle BAD = \angle CAD = \frac{1}{2} \times 60^\circ = 30^\circ$).

Now consider the right-angled triangle ABD.

In $\triangle$ABD:

• $\angle B = 60^\circ$
• $\angle BAD = 30^\circ$
• $\angle ADB = 90^\circ$
• Hypotenuse AB = $2a$
• Side BD = $a$

We need to find the length of the side AD using the Pythagorean Theorem in $\triangle$ABD:

$ AB^2 = AD^2 + BD^2 $

Substitute the known lengths:

$ (2a)^2 = AD^2 + a^2 $

$ 4a^2 = AD^2 + a^2 $

$ AD^2 = 4a^2 - a^2 = 3a^2 $

Taking the square root (positive value):

$ AD = \sqrt{3a^2} = \sqrt{3} \times \sqrt{a^2} = \sqrt{3}a $

So, the sides of a $30^\circ - 60^\circ - 90^\circ$ triangle are in the ratio $a : \sqrt{3}a : 2a$, or $1 : \sqrt{3} : 2$ (corresponding to sides opposite $30^\circ$, $60^\circ$, and $90^\circ$ respectively).

Ratios for $60^\circ$ (using $\angle B$)

In $\triangle$ABD, with respect to $\angle B = 60^\circ$:

• Opposite Side (O) to $\angle B$ = AD = $\sqrt{3}a$

• Adjacent Side (A) to $\angle B$ = BD = $a$

• Hypotenuse (H) = AB = $2a$

Using the definitions:

$ \sin 60^\circ = \frac{O}{H} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}\cancel{a}}{2\cancel{a}} = \frac{\sqrt{3}}{2} $

$ \cos 60^\circ = \frac{A}{H} = \frac{a}{2a} = \frac{\cancel{a}}{2\cancel{a}} = \frac{1}{2} $

$ \tan 60^\circ = \frac{O}{A} = \frac{\sqrt{3}a}{a} = \frac{\sqrt{3}\cancel{a}}{\cancel{a}} = \sqrt{3} $

Reciprocal ratios for $60^\circ$:

$ \text{cosec} \, 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} $

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

$ \cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}} $

Ratios for $30^\circ$ (using $\angle BAD$)

In $\triangle$ABD, with respect to $\angle BAD = 30^\circ$:

• Opposite Side (O) to $\angle BAD$ = BD = $a$

• Adjacent Side (A) to $\angle BAD$ = AD = $\sqrt{3}a$

• Hypotenuse (H) = AB = $2a$

Using the definitions:

$ \sin 30^\circ = \frac{O}{H} = \frac{a}{2a} = \frac{\cancel{a}}{2\cancel{a}} = \frac{1}{2} $

$ \cos 30^\circ = \frac{A}{H} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}\cancel{a}}{2\cancel{a}} = \frac{\sqrt{3}}{2} $

$ \tan 30^\circ = \frac{O}{A} = \frac{a}{\sqrt{3}a} = \frac{\cancel{a}}{\sqrt{3}\cancel{a}} = \frac{1}{\sqrt{3}} $

Reciprocal ratios for $30^\circ$:

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

$ \sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} $

$ \cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{1/\sqrt{3}} = \sqrt{3} $

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Values for $0^\circ$ and $90^\circ$

Defining trigonometric ratios for $0^\circ$ and $90^\circ$ requires extending the concept beyond just acute angles in a right triangle. This can be done using the concept of a unit circle, where an angle is formed by rotating the terminal side from the positive x-axis, and the trigonometric ratios are defined based on the coordinates of the point where the terminal side intersects the unit circle.

Consider a point P(x, y) on the unit circle (a circle with radius 1 centered at the origin). If the angle formed by the positive x-axis and the ray OP is $\theta$, then the coordinates of P are $(\cos \theta, \sin \theta)$.

For $\theta = 0^\circ$, the terminal side lies along the positive x-axis. The point P is (1, 0). Thus:

• $ \cos 0^\circ = x\text{-coordinate of P} = 1 $

• $ \sin 0^\circ = y\text{-coordinate of P} = 0 $

Using quotient and reciprocal identities:

• $ \tan 0^\circ = \frac{\sin 0^\circ}{\cos 0^\circ} = \frac{0}{1} = 0 $

• $ \cot 0^\circ = \frac{1}{\tan 0^\circ} = \frac{1}{0} = \text{Undefined} $

• $ \sec 0^\circ = \frac{1}{\cos 0^\circ} = \frac{1}{1} = 1 $

• $ \text{cosec} \, 0^\circ = \frac{1}{\sin 0^\circ} = \frac{1}{0} = \text{Undefined} $

For $\theta = 90^\circ$, the terminal side lies along the positive y-axis. The point P is (0, 1). Thus:

• $ \cos 90^\circ = x\text{-coordinate of P} = 0 $

• $ \sin 90^\circ = y\text{-coordinate of P} = 1 $

Using quotient and reciprocal identities:

• $ \tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0} = \text{Undefined} $

• $ \cot 90^\circ = \frac{1}{\tan 90^\circ} = \frac{1}{\text{Undefined}} = 0 $ (Alternatively, $\cot 90^\circ = \frac{\cos 90^\circ}{\sin 90^\circ} = \frac{0}{1} = 0$)

• $ \sec 90^\circ = \frac{1}{\cos 90^\circ} = \frac{1}{0} = \text{Undefined} $

• $ \text{cosec} \, 90^\circ = \frac{1}{\sin 90^\circ} = \frac{1}{1} = 1 $

(Note: For $\tan 90^\circ$ and $\sec 90^\circ$, the denominator is zero, hence they are undefined. For $\cot 0^\circ$ and $\text{cosec} \, 0^\circ$, the denominator is zero, hence they are undefined.)

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Table of Trigonometric Ratios for Specific Angles

The values derived above are summarised in the following table, which is crucial to remember for solving many problems.

Angle ($\theta$)	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin \theta$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos \theta$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$
$\tan \theta$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	Undefined
$\text{cosec} \, \theta$	Undefined	$2$	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	$1$
$\sec \theta$	$1$	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$	Undefined
$\cot \theta$	Undefined	$\sqrt{3}$	$1$	$\frac{1}{\sqrt{3}}$	$0$

Tip for remembering the table:

For $\sin \theta$ values (0°, 30°, 45°, 60°, 90°): Write $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$. Simplify these to get $0, \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, 1$.

For $\cos \theta$ values: Write the $\sin \theta$ values in reverse order.

For $\tan \theta$: Use the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

For $\text{cosec} \, \theta, \sec \theta, \cot \theta$: Take the reciprocals of $\sin \theta, \cos \theta, \tan \theta$ respectively.

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Examples

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Trigonometric Ratios of Complementary Angles

Two angles are called complementary if their sum is equal to $90^\circ$. In a right-angled triangle, the sum of all three angles is $180^\circ$. Since one angle is $90^\circ$, the sum of the other two acute angles must be $180^\circ - 90^\circ = 90^\circ$. Therefore, the two acute angles in a right-angled triangle are always complementary to each other.

Consider a right-angled triangle ABC, which is right-angled at B.

In $\triangle$ABC, $\angle A + \angle B + \angle C = 180^\circ$.

$ \angle A + \angle C = 180^\circ - 90^\circ $

This means that $\angle C = 90^\circ - \angle A$, and $\angle A = 90^\circ - \angle C$.

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Deriving the Relationships

Let's derive the relationships between the trigonometric ratios of complementary angles. Consider the right-angled triangle ABC, right-angled at B.

Trigonometric Ratios for Angle C:

With respect to angle C:

• Opposite Side = AB

• Adjacent Side = BC

• Hypotenuse = AC

$ \sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} $

$ \cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} $

$ \tan C = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} $

$ \text{cosec} \, C = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{AC}{AB} $

$ \sec C = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{AC}{BC} $

$ \cot C = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{BC}{AB} $

Trigonometric Ratios for Angle A:

With respect to angle A:

• Opposite Side = BC

• Adjacent Side = AB

• Hypotenuse = AC

Angle A is complementary to angle C, so we can write $\angle A = 90^\circ - \angle C$. Let's express the ratios for angle A:

$ \sin A = \sin (90^\circ - C) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} $

$ \cos A = \cos (90^\circ - C) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} $

$ \tan A = \tan (90^\circ - C) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} $

$ \text{cosec} \, A = \text{cosec} \, (90^\circ - C) = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{AC}{BC} $

$ \sec A = \sec (90^\circ - C) = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{AC}{AB} $

$ \cot A = \cot (90^\circ - C) = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{AB}{BC} $

Comparing the Ratios:

Now, let's compare the ratios for angle C and angle A ($90^\circ - C$):

• We see that $\sin (90^\circ - C) = \frac{BC}{AC}$ and $\cos C = \frac{BC}{AC}$. Therefore, $\mathbf{\sin (90^\circ - C) = \cos C}$.

• We see that $\cos (90^\circ - C) = \frac{AB}{AC}$ and $\sin C = \frac{AB}{AC}$. Therefore, $\mathbf{\cos (90^\circ - C) = \sin C}$.

• We see that $\tan (90^\circ - C) = \frac{BC}{AB}$ and $\cot C = \frac{BC}{AB}$. Therefore, $\mathbf{\tan (90^\circ - C) = \cot C}$.

• We see that $\cot (90^\circ - C) = \frac{AB}{BC}$ and $\tan C = \frac{AB}{BC}$. Therefore, $\mathbf{\cot (90^\circ - C) = \tan C}$.

• We see that $\sec (90^\circ - C) = \frac{AC}{AB}$ and $\text{cosec} \, C = \frac{AC}{AB}$. Therefore, $\mathbf{\sec (90^\circ - C) = \text{cosec} \, C}$.

• We see that $\text{cosec} \, (90^\circ - C) = \frac{AC}{BC}$ and $\sec C = \frac{AC}{BC}$. Therefore, $\mathbf{\text{cosec} \, (90^\circ - C) = \sec C}$.

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Complementary Angle Identities

Replacing the acute angle C with a general acute angle $\theta$ (where $0^\circ < \theta < 90^\circ$), we get the following fundamental identities for trigonometric ratios of complementary angles:

• $\mathbf{\sin (90^\circ - \theta) = \cos \theta}$

• $\mathbf{\cos (90^\circ - \theta) = \sin \theta}$

• $\mathbf{\tan (90^\circ - \theta) = \cot \theta}$

• $\mathbf{\cot (90^\circ - \theta) = \tan \theta}$

• $\mathbf{\sec (90^\circ - \theta) = \text{cosec} \, \theta}$

• $\mathbf{\text{cosec} \, (90^\circ - \theta) = \sec \theta}$

These identities are particularly useful when dealing with angles that are complementary to the standard specific angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$). For example, $72^\circ$ is the complement of $18^\circ$, $50^\circ$ is the complement of $40^\circ$, etc.

While derived using acute angles, these identities hold true for all angles where the respective trigonometric ratios are defined.

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Examples

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