Question

$\sin 6\theta =32{\cos }^5\theta \sin \theta -32{\cos }^3\theta \sin \theta +3x$.Find the value of $x$

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

Answer 1

The correct option is D $\sin 2\theta$

$\sin 6\theta =2\sin 3\theta \cos 3\theta =2(3\sin \theta -4{\sin }^3\theta )(4{\cos }^3\theta -3\cos \theta )=24\sin \theta {\cos }^3\theta -18\sin \theta \cos \theta -32{\sin }^3\theta {\cos }^3\theta +24{\sin }^3\theta \cos \theta =24\sin \theta \cos \theta (1-{\sin }^2\theta )-18\sin \theta \cos \theta -32\sin \theta (1-{\cos }^2\theta ){\cos }^3\theta +24{\sin }^3\theta \cos \theta =24\sin \theta \cos \theta -24{\sin }^3\theta \cos \theta -18\sin \theta \cos \theta -32\sin \theta {\cos }^3\theta +32{\cos }^5\theta \sin \theta +24{\sin }^3\theta \cos \theta =32{\cos }^5\theta \sin \theta -32\sin \theta {\cos }^3\theta +6\sin \theta \cos \theta =32{\cos }^5\theta \sin \theta -32\sin \theta {\cos }^3\theta +3\sin 2\theta$

Therefore $x=\sin 2\theta$