જો cos A = frac{3}{4} હોય,તો 32sin left( frac | vedclass

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Question

(C) આપણે સૂત્ર $2\sin X \sin Y = \cos(X - Y) - \cos(X + Y)$ નો ઉપયોગ કરીએ છીએ.આપેલ પદાવલિ: $32\sin \left( \frac{A}{2} \right)\sin \left( \frac{5A}{2}

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) આપણે સૂત્ર $2\sin X \sin Y = \cos(X - Y) - \cos(X + Y)$ નો ઉપયોગ કરીએ છીએ.આપેલ પદાવલિ: $32\sin \left( \frac{A}{2} ight)\sin \left( \frac{5A}{2} ight) = 16 	imes 2\sin \left( \frac{A}{2} ight)\sin \left( \frac{5A}{2} ight)$.$= 16 \left[ \cos\left( \frac{5A}{2} - \frac{A}{2} ight) - \cos\left( \frac{5A}{2} + \frac{A}{2} ight) ight]$.$= 16 (\cos 2A - \cos 3A)$.$\cos 2A = 2\cos^2 A - 1$ અને $\cos 3A = 4\cos^3 A - 3\cos A$ નો ઉપયોગ કરતા:$= 16 [ (2\cos^2 A - 1) - (4\cos^3 A - 3\cos A) ]$.આપેલ છે કે $\cos A = \frac{3}{4}$,તેથી $\cos^2 A = \frac{9}{16}$ અને $\cos^3 A = \frac{27}{64}$.$= 16 \left[ \left( 2 	imes \frac{9}{16} - 1 ight) - \left( 4 	imes \frac{27}{64} - 3 	imes \frac{3}{4} ight) ight]$.$= 16 \left[ \left( \frac{9}{8} - 1 ight) - \left( \frac{27}{16} - \frac{9}{4} ight) ight]$.$= 16 \left[ \frac{1}{8} - \left( \frac{27 - 36}{16} ight) ight] = 16 \left[ \frac{1}{8} - \left( -\frac{9}{16} ight) ight]$.$= 16 \left( \frac{2 + 9}{16} ight) = 11$.

જો $\cos A = \frac{3}{4}$ હોય,તો $32\sin \left( \frac{A}{2} \right)\sin \left( \frac{5A}{2} \right) = $

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