left( 1 + cos frac{pi }{9} right) left( 1 + c | vedclass

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(A) माना $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 \fra

Vedclass · Accepted Answer

$\frac{9}{16}$

Vedclass · Answer

(A) माना $E = \left( 1 + \cos \frac{\pi}{9} ight) \left( 1 + \cos \frac{3\pi}{9} ight) \left( 1 + \cos \frac{5\pi}{9} ight) \left( 1 + \cos \frac{7\pi}{9} ight)$.चूंकि $\cos \frac{3\pi}{9} = \cos \frac{\pi}{3} = \frac{1}{2}$,इसलिए $1 + \cos \frac{3\pi}{9} = \frac{3}{2}$ है।अतः,$E = \frac{3}{2} \left( 1 + \cos 20^\circ ight) \left( 1 + \cos 100^\circ ight) \left( 1 + \cos 140^\circ ight)$.$1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$ का उपयोग करने पर:$E = \frac{3}{2} \left( 2 \cos^2 10^\circ ight) \left( 2 \cos^2 50^\circ ight) \left( 2 \cos^2 70^\circ ight) = 12 \left( \cos 10^\circ \cos 50^\circ \cos 70^\circ ight)^2$.सर्वसमिका $\cos 	heta \cos(60^\circ - 	heta) \cos(60^\circ + 	heta) = \frac{1}{4} \cos 3	heta$ का उपयोग करने पर $(	heta = 10^\circ)$:$\cos 10^\circ \cos 50^\circ \cos 70^\circ = \frac{1}{4} \cos 30^\circ = \frac{\sqrt{3}}{8}$.अतः,$E = 12 	imes \left( \frac{\sqrt{3}}{8} ight)^2 = 12 	imes \frac{3}{64} = \frac{9}{16}$.

$\left( 1 + \cos \frac{\pi }{9} \right) \left( 1 + \cos \frac{3\pi }{9} \right) \left( 1 + \cos \frac{5\pi }{9} \right) \left( 1 + \cos \frac{7\pi }{9} \right)$ का मान है

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