Question

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

• A. $\cos \theta$
• B. $\cos 2\theta$
• C. $\sin \theta$
• D. $\sin 2\theta$

Answer 1

The correct option is D

$\sin 2\theta$

Explanation for the correct option.

Given that, $\sin 6\theta =32{\cos }^5\theta \sin \theta -32{\cos }^3\theta \sin \theta +3x....\left(1\right)$

We know that, $\sin 2\theta =2\sin \theta \cos \theta$ consider the LHS

$\begin{array}{rcl}2\sin 3\theta \cos 3\theta & =& 2\left(3\sin \theta -4{\sin }^3\theta \right)\left(4{\cos }^3\theta -3\cos \theta \right)\\ & =& 2\left(12\sin \theta {\cos }^3\theta -9\sin \theta \cos \theta -16{\sin }^3\theta {\cos }^3\theta +12{\sin }^3\theta \cos \theta \right)\\ & =& 24\sin \theta {\cos }^3\theta -18\sin \theta \cos \theta -32{\sin }^3\theta {\cos }^3\theta +24{\sin }^3\theta \cos \theta \\ & =& 24\sin \theta \cos \theta \left(1-{\sin }^2\theta \right)-18\sin \theta \cos \theta -32\sin \theta \left(1-{\cos }^2\theta \right){\cos }^3\theta +24{\sin }^3\theta \cos \theta \\ & =& 24\sin \theta \cos \theta -24{\sin }^3\theta \cos \theta -18\sin \theta \cos \theta -32\sin \theta {\cos }^3\theta +32{\cos }^5\theta \sin \theta +24{\sin }^3\cos \theta \\ & =& 6\sin \theta \cos \theta -32\sin \theta {\cos }^3\theta +32{\cos }^5\theta \sin \theta \\ & =& 32{\cos }^5\theta \sin \theta -32{\cos }^3\theta \sin \theta +3\sin 2\theta ....\left(2\right)\end{array}$

Comparing (1) & (2) we have:

$x=\sin 2\theta$

Hence, option D is correct.