4 cos frac{7 theta}{2} cos frac{3 theta}{2} s | vedclass

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(C) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.Applying this to $4 \cos \frac{7 	heta}{2} \cos \frac{3 	heta}{2} \sin 5 	heta$:$=

Vedclass · Accepted Answer

$\sin 10 	heta + \sin 7 	heta + \sin 3 	heta$

Vedclass · Answer

(C) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.Applying this to $4 \cos \frac{7 	heta}{2} \cos \frac{3 	heta}{2} \sin 5 	heta$:$= 2 \left( 2 \cos \frac{7 	heta}{2} \cos \frac{3 	heta}{2} ight) \sin 5 	heta$$= 2 \left( \cos(\frac{7 	heta}{2} + \frac{3 	heta}{2}) + \cos(\frac{7 	heta}{2} - \frac{3 	heta}{2}) ight) \sin 5 	heta$$= 2 (\cos 5 	heta + \cos 2 	heta) \sin 5 	heta$$= 2 \cos 5 	heta \sin 5 	heta + 2 \cos 2 	heta \sin 5 	heta$Using $2 \sin A \cos A = \sin 2A$ and $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$= \sin 10 	heta + (\sin(5 	heta + 2 	heta) + \sin(5 	heta - 2 	heta))$$= \sin 10 	heta + \sin 7 	heta + \sin 3 	heta$.

$4 \cos \frac{7 \theta}{2} \cos \frac{3 \theta}{2} \sin 5 \theta = $

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