Which of the following identities is true?

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $\cos 5 	heta = \cos (4 	heta + 	heta)$.Using the formula $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we have:$\cos 5 	heta =

Vedclass · Accepted Answer

$\cos 5 	heta = 16 \cos ^5 	heta - 20 \cos ^3 	heta + 5 \cos 	heta$

Vedclass · Answer

(D) We know that $\cos 5 	heta = \cos (4 	heta + 	heta)$.Using the formula $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we have:$\cos 5 	heta = \cos 4 	heta \cos 	heta - \sin 4 	heta \sin 	heta$$= (2 \cos ^2 2 	heta - 1) \cos 	heta - (2 \sin 2 	heta \cos 2 	heta) \sin 	heta$$= \{2(2 \cos ^2 	heta - 1)^2 - 1\} \cos 	heta - 2(2 \sin 	heta \cos 	heta) \cos 2 	heta \sin 	heta$$= \{2(4 \cos ^4 	heta - 4 \cos ^2 	heta + 1) - 1\} \cos 	heta - 4 \sin ^2 	heta \cos 	heta (2 \cos ^2 	heta - 1)$$= (8 \cos ^4 	heta - 8 \cos ^2 	heta + 1) \cos 	heta - 4(1 - \cos ^2 	heta) \cos 	heta (2 \cos ^2 	heta - 1)$$= (8 \cos ^5 	heta - 8 \cos ^3 	heta + \cos 	heta) - 4 \cos 	heta (2 \cos ^2 	heta - 1 - 2 \cos ^4 	heta + \cos ^2 	heta)$$= 8 \cos ^5 	heta - 8 \cos ^3 	heta + \cos 	heta - 4 \cos 	heta (3 \cos ^2 	heta - 1 - 2 \cos ^4 	heta)$$= 8 \cos ^5 	heta - 8 \cos ^3 	heta + \cos 	heta - 12 \cos ^3 	heta + 4 \cos 	heta + 8 \cos ^5 	heta$$= 16 \cos ^5 	heta - 20 \cos ^3 	heta + 5 \cos 	heta$

Similar Questions

If $\sin x - \sin y = \frac{1}{2}$ and $\cos x - \cos y = 1$,then $\tan(x + y)$ is equal to

If $\sin(\frac{\pi}{18}) \sin(\frac{5\pi}{18}) \sin(\frac{7\pi}{18}) = K$, then the value of $\sin(\frac{10K\pi}{3})$ is :

If $\tan \frac{\theta}{2} = \operatorname{cosec} \theta - \sin \theta$,then $\tan^2 \frac{\theta}{2} =$

$\sin 4\theta$ can be written as

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module