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

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

Vedclass · Accepted Answer

$\sin 2	heta$

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(D) We know that $\sin 6	heta = 2 \sin 3	heta \cos 3	heta$.Using the triple angle 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$Using $\sin^2 	heta = 1 - \cos^2 	heta$ and $\cos^2 	heta = 1 - \sin^2 	heta$,we simplify the expression to match the form $32\cos^5 	heta \sin 	heta - 32\cos^3 	heta \sin 	heta + 3x$.After expansion and rearrangement,we get $\sin 6	heta = 32\cos^5 	heta \sin 	heta - 32\cos^3 	heta \sin 	heta + 3\sin 2	heta$.Comparing this with the given equation,we find $x = \sin 2	heta$.

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

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