left(4 cos ^2 frac{pi}{20}-1right)left(4 cos | vedclass

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(D) ધારો કે $	heta = \frac{\pi}{20}$. પદાવલિ $E = (4 \cos^2 	heta - 1)(4 \cos^2 3	heta - 1)(4 \cos^2 5	heta + 1)(4 \cos^2 7	heta - 1)(4 \cos^2 9\

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) ધારો કે $	heta = \frac{\pi}{20}$. પદાવલિ $E = (4 \cos^2 	heta - 1)(4 \cos^2 3	heta - 1)(4 \cos^2 5	heta + 1)(4 \cos^2 7	heta - 1)(4 \cos^2 9	heta - 1)$ છે.$4 \cos^2 	heta - 1 = \frac{\sin 3	heta}{\sin 	heta}$ નો ઉપયોગ કરતા:$E = \left(\frac{\sin 3	heta}{\sin 	heta}ight) \left(\frac{\sin 9	heta}{\sin 3	heta}ight) (4 \cos^2 5	heta + 1) \left(\frac{\sin 21	heta}{\sin 7	heta}ight) \left(\frac{\sin 27	heta}{\sin 9	heta}ight)$.$	heta = \frac{\pi}{20}$ હોવાથી,$5	heta = \frac{\pi}{4}$,તેથી $4 \cos^2 5	heta + 1 = 4(\frac{1}{2}) + 1 = 3$.વળી,$\sin 21	heta = \sin(\pi + 	heta) = -\sin 	heta$ અને $\sin 27	heta = \sin(\frac{3\pi}{2} + 3	heta) = -\cos 3	heta$.ગુણાકારનું સાદું રૂપ આપતા: $E = \frac{\sin 9	heta}{\sin 	heta} \cdot 3 \cdot \frac{-\sin 	heta}{\sin 7	heta} \cdot \frac{-\cos 3	heta}{\sin 9	heta} = 3 \cdot \frac{\cos 3	heta}{\sin 7	heta}$.$7	heta = \frac{7\pi}{20}$ અને $3	heta = \frac{3\pi}{20}$ હોવાથી,$7	heta + 3	heta = \frac{10\pi}{20} = \frac{\pi}{2}$,તેથી $\sin 7	heta = \cos 3	heta$.આમ,$E = 3 \cdot \frac{\cos 3	heta}{\cos 3	heta} = 3$.

$\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \cos ^2 \frac{5 \pi}{20}+1\right)\left(4 \cos ^2 \frac{7 \pi}{20}-1\right)\left(4 \cos ^2 \frac{9 \pi}{20}-1\right)=$

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