For some a, b, c in R,if sin 5 theta = a cos^ | vedclass

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(D) We know that $\sin 5 	heta = \sin(3 	heta + 2 	heta) = \sin 3 	heta \cos 2 	heta + \cos 3 	heta \sin 2 	heta$. Using the identities $\sin 3

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-$50$

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(D) We know that $\sin 5 	heta = \sin(3 	heta + 2 	heta) = \sin 3 	heta \cos 2 	heta + \cos 3 	heta \sin 2 	heta$. Using the identities $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$,$\cos 2 	heta = 1 - 2 \sin^2 	heta$,$\cos 3 	heta = 4 \cos^3 	heta - 3 \cos 	heta$,and $\sin 2 	heta = 2 \sin 	heta \cos 	heta$: $\sin 5 	heta = (3 \sin 	heta - 4 \sin^3 	heta)(1 - 2 \sin^2 	heta) + (4 \cos^3 	heta - 3 \cos 	heta)(2 \sin 	heta \cos 	heta)$ $= 3 \sin 	heta - 6 \sin^3 	heta - 4 \sin^3 	heta + 8 \sin^5 	heta + 8 \sin 	heta \cos^4 	heta - 6 \sin 	heta \cos^2 	heta$ $= 8 \sin 	heta \cos^4 	heta - 6 \sin 	heta \cos^2 	heta + 10 \sin^3 	heta - 10 \sin^5 	heta + 10 \sin^5 	heta$ Alternatively,expanding $\sin 5 	heta$ in terms of $\sin 	heta$ and $\cos 	heta$: $\sin 5 	heta = 5 \sin 	heta \cos^4 	heta - 10 \sin^3 	heta \cos^2 	heta + \sin^5 	heta$. Comparing this with $a \cos^4 	heta \sin 	heta + b \cos^2 	heta \sin^3 	heta + c \sin^5 	heta$,we get $a = 5$,$b = -10$,and $c = 1$. Therefore,$abc = 5 	imes (-10) 	imes 1 = -50$.

For some $a, b, c \in R$,if $\sin 5 \theta = a \cos^4 \theta \sin \theta + b \cos^2 \theta \sin^3 \theta + c \sin^5 \theta$,then $abc =$

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