UrbanPro

Your Worksheet is Ready

CBSE - Class 11 Mathematics Trigonometric Functions Worksheet

EXERCISE 3.3

1.
Prove the following: $\cos 6x = 32 \cos^6 x - 48\cos^4 x + 18 \cos^2 x - 1$
2.
Prove the following: $\sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x$
3.
Prove the following: $\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x$
4.
Prove the following: $\cos(\frac{3\pi}{2} + x) \cos(2\pi + x) [\cot(\frac{3\pi}{2} - x) + \cot(2\pi + x)] = 1$
5.
Prove that: $\cot^2 \frac{\pi}{6} + \text{cosec} \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6$
6.
Prove the following: $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$
7.
Prove that: $2\sin^2 \frac{\pi}{6} + \text{cosec}^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2}$
8.
Prove the following: $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$
9.
Prove the following: $\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x$
10.
Prove the following: $\cos (\frac{\pi}{4} - x) \cos (\frac{\pi}{4} - y) - \sin (\frac{\pi}{4} - x) \sin (\frac{\pi}{4} - y) = \sin (x+y)$
11.
Prove the following: $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$
12.
Prove the following: $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$
13.
Prove that: $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}$
14.
Prove the following: $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x)$
15.
Prove that: $2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$
16.
Prove the following: $\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x-y}{2}$
17.
Prove the following: $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$
18.
Find the value of: (ii) $\tan 15^{\circ}$
19.
Prove the following: $\cos(\frac{3\pi}{4} + x) - \cos(\frac{3\pi}{4} - x) = -\sqrt{2} \sin x$
20.
Find the value of: (i) $\sin 75^{\circ}$

Worksheet Answers

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All