sin ^4 frac{pi}{8}+cos ^4 frac{pi}{8}+sin ^4 | vedclass

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Question

(A) આપણે જાણીએ છીએ કે $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2} \sin^2(2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2} \sin^2(2 	heta)$.ધારો કે $S = \sum_{k=1,3,5,7} (\sin^4 \frac{k \pi}{8} + \cos^4 \frac{k \pi}{8})$.આ નિત્યસમનો ઉપયોગ કરતા,$S = \sum_{k=1,3,5,7} (1 - \frac{1}{2} \sin^2 \frac{k \pi}{4})$.અહીં $4$ પદો હોવાથી,$S = 4 - \frac{1}{2} (\sin^2 \frac{\pi}{4} + \sin^2 \frac{3 \pi}{4} + \sin^2 \frac{5 \pi}{4} + \sin^2 \frac{7 \pi}{4})$.આપણે જાણીએ છીએ કે $\sin^2 \frac{\pi}{4} = \frac{1}{2}$,$\sin^2 \frac{3 \pi}{4} = \frac{1}{2}$,$\sin^2 \frac{5 \pi}{4} = \frac{1}{2}$,અને $\sin^2 \frac{7 \pi}{4} = \frac{1}{2}$.તેથી,$S = 4 - \frac{1}{2} (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}) = 4 - \frac{1}{2} (2) = 4 - 1 = 3$.

$\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=$

Similar Questions

જો $\pi < \alpha < \frac{3\pi}{2}$ હોય,તો $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

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