sin ^4 frac{pi}{8}+sin ^4 frac{3 pi}{8}+sin ^ | vedclass

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Question

(C) આપણી પાસે પદાવલિ છે: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$ $\sin(\pi - 	heta) = \si

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) આપણી પાસે પદાવલિ છે: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$ $\sin(\pi - 	heta) = \sin 	heta$ હોવાથી,$\sin \frac{5 \pi}{8} = \sin \frac{3 \pi}{8}$ અને $\sin \frac{7 \pi}{8} = \sin \frac{\pi}{8}$ થાય. તેથી,$S = 2 \left[ \sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} ight]$. $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ નો ઉપયોગ કરતા,$\sin^4 	heta = \left( \frac{1 - \cos 2	heta}{2} ight)^2$ મળે. $S = 2 \left[ \left( \frac{1 - \cos(\pi/4)}{2} ight)^2 + \left( \frac{1 - \cos(3\pi/4)}{2} ight)^2 ight]$ $S = 2 \left[ \left( \frac{1 - 1/\sqrt{2}}{2} ight)^2 + \left( \frac{1 - (-1/\sqrt{2})}{2} ight)^2 ight]$ $S = \frac{2}{4} \left[ (1 - 1/\sqrt{2})^2 + (1 + 1/\sqrt{2})^2 ight] = \frac{1}{2} \left[ (1 + 1/2 - \sqrt{2}) + (1 + 1/2 + \sqrt{2}) ight]$ $S = \frac{1}{2} \left[ 3 ight] = \frac{3}{2}$.

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

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જો $\tan \theta - \cot \theta = a$ અને $\sin \theta + \cos \theta = b$ હોય,તો ${({b^2} - 1)^2}({a^2} + 4)$ ની કિંમત શોધો.

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

ધારો કે $E = \left( 1 - \frac{\cos 61^\circ}{\cos 1^\circ} \right) \left( 1 - \frac{\cos 62^\circ}{\cos 2^\circ} \right) \dots \left( 1 - \frac{\cos 119^\circ}{\cos 59^\circ} \right)$,તો $E$ ની કિંમત શોધો.

જો $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$ હોય,તો $x$ ની કિંમત શોધો.

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