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

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(D) આપેલ પદાવલિ: $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}\

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(D) આપેલ પદાવલિ: $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) = 16 \left[ 2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) \right]$ સૂત્ર $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$ નો ઉપયોગ કરતા: $= 16 \left[ \cos\left(\frac{5A}{2} - \frac{A}{2}\right) - \cos\left(\frac{5A}{2} + \frac{A}{2}\right) \right]$ $= 16 [ \cos(2A) - \cos(3A) ]$ $\cos 2A = 2 \cos^2 A - 1$ અને $\cos 3A = 4 \cos^3 A - 3 \cos A$ નો ઉપયોગ કરતા: $= 16 [ (2 \cos^2 A - 1) - (4 \cos^3 A - 3 \cos A) ]$ $\cos A = \frac{3}{4}$ મૂકતા: $= 16 \left[ 2\left(\frac{3}{4}\right)^2 - 1 - 4\left(\frac{3}{4}\right)^3 + 3\left(\frac{3}{4}\right) \right]$ $= 16 \left[ \frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4} \right]$ $= 16 \left[ \frac{18 - 16 - 27 + 36}{16} \right] = 11$

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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