The value of left( 1 + cos frac{pi }{9} right | vedclass

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Question

(A) Let $E = \left( 1 + \cos \frac{\pi}{9} \right) \left( 1 + \cos \frac{3\pi}{9} \right) \left( 1 + \cos \frac{5\pi}{9} \right) \left( 1 + \cos \frac

Vedclass · Accepted Answer

$\frac{9}{16}$

Vedclass · Answer

(A) Let $E = \left( 1 + \cos \frac{\pi}{9} ight) \left( 1 + \cos \frac{3\pi}{9} ight) \left( 1 + \cos \frac{5\pi}{9} ight) \left( 1 + \cos \frac{7\pi}{9} ight)$.Since $\cos \frac{3\pi}{9} = \cos \frac{\pi}{3} = \frac{1}{2}$,we have $1 + \cos \frac{3\pi}{9} = \frac{3}{2}$.Thus,$E = \frac{3}{2} \left( 1 + \cos 20^\circ ight) \left( 1 + \cos 100^\circ ight) \left( 1 + \cos 140^\circ ight)$.Using $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,we get:$E = \frac{3}{2} \left( 2 \cos^2 10^\circ ight) \left( 2 \cos^2 50^\circ ight) \left( 2 \cos^2 70^\circ ight) = 12 \left( \cos 10^\circ \cos 50^\circ \cos 70^\circ ight)^2$.Using the identity $\cos 	heta \cos(60^\circ - 	heta) \cos(60^\circ + 	heta) = \frac{1}{4} \cos 3	heta$ with $	heta = 10^\circ$:$\cos 10^\circ \cos 50^\circ \cos 70^\circ = \frac{1}{4} \cos 30^\circ = \frac{1}{4} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.Therefore,$E = 12 	imes \left( \frac{\sqrt{3}}{8} ight)^2 = 12 	imes \frac{3}{64} = \frac{36}{64} = \frac{9}{16}$.

The value of $\left( 1 + \cos \frac{\pi }{9} \right) \left( 1 + \cos \frac{3\pi }{9} \right) \left( 1 + \cos \frac{5\pi }{9} \right) \left( 1 + \cos \frac{7\pi }{9} \right)$ is

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