A true statement among the following identiti | vedclass

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(A) We have,$\sin (5 	heta) = \sin (3 	heta + 2 	heta)$.Using the identity $\sin (A + B) = \sin A \cos B + \cos A \sin B$,we get:$\sin (5 	heta) =

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$\sin 5 	heta = 16 \cos^4 	heta \sin 	heta - 12 \cos^2 	heta \sin 	heta + \sin 	heta$

Vedclass · Answer

(A) We have,$\sin (5 	heta) = \sin (3 	heta + 2 	heta)$.Using the identity $\sin (A + B) = \sin A \cos B + \cos A \sin B$,we get:$\sin (5 	heta) = \sin 3 	heta \cos 2 	heta + \cos 3 	heta \sin 2 	heta$.Substituting the multiple angle formulas $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$,$\cos 2 	heta = 2 \cos^2 	heta - 1$,$\cos 3 	heta = 4 \cos^3 	heta - 3 \cos 	heta$,and $\sin 2 	heta = 2 \sin 	heta \cos 	heta$:$\sin (5 	heta) = (3 \sin 	heta - 4 \sin^3 	heta)(2 \cos^2 	heta - 1) + (4 \cos^3 	heta - 3 \cos 	heta)(2 \sin 	heta \cos 	heta)$.Expanding the terms:$\sin (5 	heta) = (6 \sin 	heta \cos^2 	heta - 3 \sin 	heta - 8 \sin^3 	heta \cos^2 	heta + 4 \sin^3 	heta) + (8 \sin 	heta \cos^4 	heta - 6 \sin 	heta \cos^2 	heta)$.Simplifying by canceling $6 \sin 	heta \cos^2 	heta$:$\sin (5 	heta) = 8 \sin 	heta \cos^4 	heta - 8 \sin^3 	heta \cos^2 	heta + 4 \sin^3 	heta - 3 \sin 	heta$.Using $\sin^2 	heta = 1 - \cos^2 	heta$:$\sin (5 	heta) = 8 \sin 	heta \cos^4 	heta - 8 \sin 	heta (1 - \cos^2 	heta) \cos^2 	heta + 4 \sin 	heta (1 - \cos^2 	heta) - 3 \sin 	heta$.$\sin (5 	heta) = 8 \sin 	heta \cos^4 	heta - 8 \sin 	heta \cos^2 	heta + 8 \sin 	heta \cos^4 	heta + 4 \sin 	heta - 4 \sin 	heta \cos^2 	heta - 3 \sin 	heta$.$\sin (5 	heta) = 16 \sin 	heta \cos^4 	heta - 12 \sin 	heta \cos^2 	heta + \sin 	heta$.

$A$ true statement among the following identities is

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