यदि cos ^4 frac{pi}{8}+cos ^4 frac{3 pi}{8}+c | vedclass

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(A) दिया गया व्यंजक: $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} = k$. $\cos(\frac{\pi}{2} + 	heta

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(A) दिया गया व्यंजक: $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} = k$. $\cos(\frac{\pi}{2} + 	heta) = -\sin 	heta$ का उपयोग करते हुए,$\cos^4(\frac{\pi}{2} + 	heta) = \sin^4 	heta$ प्राप्त होता है। अतः,$\cos^4 \frac{5 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{\pi}{8}) = \sin^4 \frac{\pi}{8}$ और $\cos^4 \frac{7 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{3 \pi}{8}) = \sin^4 \frac{3 \pi}{8}$. इन मानों को प्रतिस्थापित करने पर,$k = (\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}) + (\cos^4 \frac{3 \pi}{8} + \sin^4 \frac{3 \pi}{8})$. $a^2 + b^2 = (a+b)^2 - 2ab$ का उपयोग करते हुए,$k = [(\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}] + [(\cos^2 \frac{3 \pi}{8} + \sin^2 \frac{3 \pi}{8})^2 - 2 \sin^2 \frac{3 \pi}{8} \cos^2 \frac{3 \pi}{8}]$. चूंकि $\sin^2 	heta \cos^2 	heta = \frac{1}{4} \sin^2(2 	heta)$,इसलिए $k = [1 - \frac{1}{2} \sin^2 \frac{\pi}{4}] + [1 - \frac{1}{2} \sin^2 \frac{3 \pi}{4}]$. $k = 2 - \frac{1}{2}(\frac{1}{2}) - \frac{1}{2}(\frac{1}{2}) = 2 - \frac{1}{4} - \frac{1}{4} = 2 - \frac{1}{2} = \frac{3}{2}$. अब,$\sin^{-1}(\sqrt{\frac{k}{2}}) + \cos^{-1}(\frac{k}{3}) = \sin^{-1}(\sqrt{\frac{3/2}{2}}) + \cos^{-1}(\frac{3/2}{3}) = \sin^{-1}(\frac{\sqrt{3}}{2}) + \cos^{-1}(\frac{1}{2})$. $= \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

यदि $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ है,तो $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$

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