मान ज्ञात कीजिए: left(1+cos frac{pi}{8}right) | vedclass

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Question

(C) माना व्यंजक $E = \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(C) माना व्यंजक $E = \left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{3 \pi}{8}ight)\left(1+\cos \frac{5 \pi}{8}ight)\left(1+\cos \frac{7 \pi}{8}ight)$ है।सर्वसमिका $\cos(\pi - 	heta) = -\cos 	heta$ का उपयोग करते हुए,हमें $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$ और $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$ प्राप्त होता है।इन मानों को प्रतिस्थापित करने पर:$E = \left(1+\cos \frac{\pi}{8}ight)\left(1+\cos \frac{3 \pi}{8}ight)\left(1-\cos \frac{3 \pi}{8}ight)\left(1-\cos \frac{\pi}{8}ight)$$E = \left(1-\cos^2 \frac{\pi}{8}ight)\left(1-\cos^2 \frac{3 \pi}{8}ight)$$E = \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$चूंकि $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,इसलिए $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$ होता है।$E = \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = \left(\sin \frac{\pi}{8} \cos \frac{\pi}{8}ight)^2$$2 \sin 	heta \cos 	heta = \sin 2	heta$ का उपयोग करते हुए,$\sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{1}{2} \sin \frac{\pi}{4} = \frac{1}{2} 	imes \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$ प्राप्त होता है।$E = \left(\frac{1}{2\sqrt{2}}ight)^2 = \frac{1}{8}$.

मान ज्ञात कीजिए: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

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