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

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(C) We use the product-to-sum formula: $2\sin x \sin y = \cos(x - y) - \cos(x + y)$.Applying this to the expression $32\sin \left( \frac{A}{2} \right)

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) We use the product-to-sum formula: $2\sin x \sin y = \cos(x - y) - \cos(x + y)$.Applying this to the expression $32\sin \left( \frac{A}{2} ight)\sin \left( \frac{5A}{2} ight)$:$= 16 	imes [2\sin \left( \frac{A}{2} ight)\sin \left( \frac{5A}{2} ight)]$$= 16 	imes [\cos \left( \frac{5A}{2} - \frac{A}{2} ight) - \cos \left( \frac{5A}{2} + \frac{A}{2} ight)]$$= 16 	imes [\cos(2A) - \cos(3A)]$Using the identities $\cos(2A) = 2\cos^2 A - 1$ and $\cos(3A) = 4\cos^3 A - 3\cos A$:$= 16 	imes [(2\cos^2 A - 1) - (4\cos^3 A - 3\cos A)]$Given $\cos A = \frac{3}{4}$,substitute the value:$= 16 	imes [2(\frac{3}{4})^2 - 1 - 4(\frac{3}{4})^3 + 3(\frac{3}{4})]$$= 16 	imes [2(\frac{9}{16}) - 1 - 4(\frac{27}{64}) + \frac{9}{4}]$$= 16 	imes [\frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4}]$$= 16 	imes [\frac{18 - 16 - 27 + 36}{16}]$$= 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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