If sin 6theta = 32cos^5theta sintheta - 32cos | vedclass

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

Question

(D) We know that $\sin 6	heta = 2\sin 3	heta \cos 3	heta$. Using the identities $\sin 3	heta = 3\sin	heta - 4\sin^3	heta$ and $\cos 3	heta = 4\

Vedclass · Accepted Answer

$\sin 2	heta$

Vedclass · Answer

(D) We know that $\sin 6	heta = 2\sin 3	heta \cos 3	heta$. Using the identities $\sin 3	heta = 3\sin	heta - 4\sin^3	heta$ and $\cos 3	heta = 4\cos^3	heta - 3\cos	heta$: $\sin 6	heta = 2(3\sin	heta - 4\sin^3	heta)(4\cos^3	heta - 3\cos	heta)$ $= 2(12\sin	heta \cos^3	heta - 9\sin	heta \cos	heta - 16\sin^3	heta \cos^3	heta + 12\sin^3	heta \cos	heta)$ $= 24\sin	heta \cos^3	heta - 18\sin	heta \cos	heta - 32\sin^3	heta \cos^3	heta + 24\sin^3	heta \cos	heta$ By rearranging and using $\sin^2	heta = 1 - \cos^2	heta$,we simplify the expression to match the form $32\cos^5	heta \sin	heta - 32\cos^3	heta \sin	heta + 3x$. After algebraic simplification,we find that $3x = 3\sin 2	heta$. Therefore,$x = \sin 2	heta$.

If $\sin 6\theta = 32\cos^5\theta \sin\theta - 32\cos^3\theta \sin\theta + 3x$,then $x = $

Similar Questions

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

$\cot \theta - \tan \theta - 2 \tan 2 \theta - 4 \tan 4 \theta = $

If the identity $\cos^4 \theta = a \cos 4\theta + b \cos 2\theta + c$ holds for some $a, b, c \in \mathbb{Q}$,then $(a, b, c) =$

The value of $\tan \left(\frac{\pi}{8}\right)$ is

$3 \tan^6 10^{\circ} - 27 \tan^4 10^{\circ} + 33 \tan^2 10^{\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module