If cos A = frac{3}{4},then 32 sin left(frac{A | vedclass

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(D) Given expression: $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right)$$= 16 \left[ 2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(D) Given expression: $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right)$$= 16 \left[ 2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) \right]$Using the formula $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$:$= 16 \left[ \cos \left( \frac{5A}{2} - \frac{A}{2} \right) - \cos \left( \frac{5A}{2} + \frac{A}{2} \right) \right]$$= 16 [ \cos 2A - \cos 3A ]$$= 16 [ (2 \cos^2 A - 1) - (4 \cos^3 A - 3 \cos A) ]$Substitute $\cos A = \frac{3}{4}$:$= 16 \left[ 2 \left( \frac{3}{4} \right)^2 - 1 - 4 \left( \frac{3}{4} \right)^3 + 3 \left( \frac{3}{4} \right) \right]$$= 16 \left[ 2 \left( \frac{9}{16} \right) - 1 - 4 \left( \frac{27}{64} \right) + \frac{9}{4} \right]$$= 16 \left[ \frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4} \right]$$= 16 \left[ \frac{18 - 16 - 27 + 36}{16} \right]$$= 18 - 16 - 27 + 36 = 11$

If $\cos A = \frac{3}{4}$,then $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) = $

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