If cos A = frac{3}{4},then 32sin frac{A}{2}co | vedclass

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(B) Given $\cos A = \frac{3}{4}$.We use the product-to-sum formula: $2\sin X \cos Y = \sin(X+Y) + \sin(X-Y)$.Here,$X = \frac{A}{2}$ and $Y = \frac{5A}

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$-\sqrt{7}$

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(B) Given $\cos A = \frac{3}{4}$.We use the product-to-sum formula: $2\sin X \cos Y = \sin(X+Y) + \sin(X-Y)$.Here,$X = \frac{A}{2}$ and $Y = \frac{5A}{2}$.$32\sin \frac{A}{2}\cos \frac{5A}{2} = 16 	imes (2\sin \frac{A}{2}\cos \frac{5A}{2})$$= 16 	imes [\sin(\frac{A}{2} + \frac{5A}{2}) + \sin(\frac{A}{2} - \frac{5A}{2})]$$= 16 	imes [\sin(3A) + \sin(-2A)]$$= 16 	imes [\sin(3A) - \sin(2A)]$Using $\sin(3A) = 3\sin A - 4\sin^3 A$ and $\sin(2A) = 2\sin A \cos A$:$= 16 	imes [3\sin A - 4\sin^3 A - 2\sin A \cos A]$$= 16\sin A [3 - 4\sin^2 A - 2\cos A]$Since $\cos A = \frac{3}{4}$,$\sin^2 A = 1 - (\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$,so $\sin A = \frac{\sqrt{7}}{4}$.$= 16 	imes \frac{\sqrt{7}}{4} [3 - 4(\frac{7}{16}) - 2(\frac{3}{4})]$$= 4\sqrt{7} [3 - \frac{7}{4} - \frac{3}{2}]$$= 4\sqrt{7} [\frac{12 - 7 - 6}{4}] = 4\sqrt{7} [-\frac{1}{4}] = -\sqrt{7}$.

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

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