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: $2\sin X \cos Y = \sin(X+Y) + \sin(X-Y)$.$32\sin \frac{A}{2}\cos \frac{5A}{2} = 16

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$-\sqrt{7}$

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