CBSE Class 10 Chapter 8 Introduction of Trigonometry - Trigonometric Identities

 Trigonometric Identities of Opposite Angles

• $\sin(-\theta) = -\sin \theta$
• $\cos(-\theta) = \cos \theta$
• $\tan(-\theta) = -\tan \theta$
• $\cot(-\theta) = -\cot \theta$
• $\sec(-\theta) = \sec \theta$
• $\csc(-\theta) = -\csc \theta$

Trigonometric Identities of Complementary Angles

When the sum of two angle are equal to $90^{0}$ , they are known as complementary angles.

• $\sin(90 - \theta) = \cos \theta$

Proof:

Using the sum-to-product identity for sine:

$\sin(90 - \theta) = \sin 90 \cos \theta - \cos 90 \sin \theta$

Simplifying with $\sin 90 = 1$ and $\cos 90 = 0$:

$\sin(90 - \theta) = \cos \theta$

Therefore, $\sin (90-\theta)=\cos\theta$ is proven.

• $\cos(90 - \theta) = \sin \theta$

Proof:

Using the sum-to-product identity for cosine:

$\cos(90 - \theta) = \cos 90 \cos \theta + \sin 90 \sin \theta$

Simplifying with $\cos 90 = 0$ and $\sin 90 = 1$:

$\cos(90 - \theta) = \sin \theta$

Therefore, $\cos(90 - \theta) = \sin \theta$ is proven.

• $\tan(90 - \theta) = \cot \theta$

Proof:

Using the relationship $\tan\theta=\frac {\sin\theta}{\cos\theta}$:

We know that  $\sin(90 - \theta) = \cos \theta$ and $\cos(90 - \theta) = \sin \theta$

$\tan(90 - \theta) = \frac{\sin(90 - \theta)}{\cos(90 - \theta)}$

Therefore, $\tan(90 - \theta) = \cot \theta$ is proven.

• $\cot(90 - \theta) = \tan \theta$

Proof:

Using the relationship $cot\theta = \frac {1}{\tan\theta}$:

We know that  $\tan(90 - \theta) = \cot \theta$ ;

$\cot(90 - \theta) = \frac{1}{\cot \theta} = \tan \theta$

Therefore, $\cot(90 - \theta) = \tan \theta$ is proven.

$\sec(90 - \theta) = \csc \theta$

$\csc(90 - \theta) = \sec \theta$

Trigonometric Identities of Supplementary Angles

Two angles are supplementary if their sum is equal to 90 degrees. Similarly, when we can learn here the trigonometric identities for supplementary angles.

$\sin(180^\circ - \theta) = \sin \theta$

$\cos(180^\circ - \theta) = -\cos \theta$

$\csc(180^\circ - \theta) = \csc \theta$

$\sec(180^\circ - \theta) = -\sec \theta$

$\tan(180^\circ - \theta) = -\tan \theta$

$\cot(180^\circ - \theta) = -\cot \theta$

Sum and Difference of Angles Trigonometric Identities

Consider two angles , α and β, the trigonometric sum and difference identities are as follows:

$\sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$

$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta)$

$\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$

$\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$

$\tan (\alpha+\beta)=\frac {\tan\alpha+\tan\beta}{1-\tan\alpha.\tan\beta}$

$\tan (\alpha-\beta)=\frac {\tan\alpha-\tan\beta}{1+\tan\alpha.\tan\beta}$

Double Angle Trigonometric Identities

If the angles are doubled, then the trigonometric identities for sin, cos and tan are:

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

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

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

Half Angle Identities

If the angles are halved, then the trigonometric identities for sin, cos and tan are:

$\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{2}}$

$\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos\theta}{2}}$

$\tan\left(\frac{\theta}{2}\right) = \pm \sqrt{(1 - \cos\theta)(1 + \cos\theta)}$

Product-Sum Trigonometric Identities

The product-sum trigonometric identities change the sum or difference of sines or cosines into a product of sines and cosines.

$\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$

$\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$

$\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$

$\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$

Trigonometric Identities of Products

These identities are:

$\sin A \cdot \sin B = \frac{\cos(A - B) - \cos(A + B)}{2}$

$\sin A \cdot \cos B = \frac{\sin(A + B) + \sin(A - B)}{2}$

$\cos A \cdot \cos B = \frac{\cos(A + B) + \cos(A - B)}{2}$