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

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Question

(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 ]$$= 16 [ (2 \cos^2 A - 1) - (4 \cos^3 A - 3 \cos A) ]$$\cos A = \frac{3}{4}$ મુકતા:$= 16 \left[ 2 \left( \frac{3}{4} \right)^2 - 1 - 4 \left( \frac{3}{4} \right)^3 + 3 \left( \frac{3}{4} \right) \right]$$= 16 \left[ 2 \left( \frac{9}{16} \right) - 1 - 4 \left( \frac{27}{64} \right) + \frac{9}{4} \right]$$= 16 \left[ \frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4} \right]$$= 16 \left[ \frac{18 - 16 - 27 + 36}{16} \right]$$= 18 - 16 - 27 + 36 = 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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