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

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Question

(D) Let $	heta = \frac{\pi}{20}$. The expression is $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 \c

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $	heta = \frac{\pi}{20}$. The expression is $E = (4 \cos^2 	heta - 1)(4 \cos^2 3	heta - 1)(4 \cos^2 5	heta + 1)(4 \cos^2 7	heta - 1)(4 \cos^2 9	heta - 1)$.Using $4 \cos^2 	heta - 1 = \frac{\sin 3	heta}{\sin 	heta}$,we have:$E = \left(\frac{\sin 3	heta}{\sin 	heta}ight) \left(\frac{\sin 9	heta}{\sin 3	heta}ight) (4 \cos^2 5	heta + 1) \left(\frac{\sin 21	heta}{\sin 7	heta}ight) \left(\frac{\sin 27	heta}{\sin 9	heta}ight)$.Since $	heta = \frac{\pi}{20}$,$5	heta = \frac{\pi}{4}$,so $4 \cos^2 5	heta + 1 = 4(\frac{1}{2}) + 1 = 3$.Also,$\sin 21	heta = \sin(\pi + 	heta) = -\sin 	heta$ and $\sin 27	heta = \sin(\frac{3\pi}{2} + 3	heta) = -\cos 3	heta$.Simplifying the product: $E = \frac{\sin 9	heta}{\sin 	heta} \cdot 3 \cdot \frac{-\sin 	heta}{\sin 7	heta} \cdot \frac{-\cos 3	heta}{\sin 9	heta} = 3 \cdot \frac{\cos 3	heta}{\sin 7	heta}$.Since $7	heta = \frac{7\pi}{20}$ and $3	heta = \frac{3\pi}{20}$,$7	heta + 3	heta = \frac{10\pi}{20} = \frac{\pi}{2}$,so $\sin 7	heta = \cos 3	heta$.Thus,$E = 3 \cdot \frac{\cos 3	heta}{\cos 3	heta} = 3$.

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

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