cos ^2left(frac{7 pi}{8}right)+cos ^2left(fra | vedclass

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Question

(C) આપણે જાણીએ છીએ કે $\cos(\pi - 	heta) = -\cos 	heta$,તેથી $\cos^2(\pi - 	heta) = \cos^2 	heta$. વળી,$\cos(\frac{\pi}{2} - 	heta) = \sin 	heta

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $\cos(\pi - 	heta) = -\cos 	heta$,તેથી $\cos^2(\pi - 	heta) = \cos^2 	heta$. વળી,$\cos(\frac{\pi}{2} - 	heta) = \sin 	heta$,તેથી $\cos^2(\frac{\pi}{2} - 	heta) = \sin^2 	heta$. આપેલ પદાવલિ: $E = \cos^2(\frac{7\pi}{8}) + \cos^2(\frac{5\pi}{8}) + \cos^2(\frac{3\pi}{8}) + \cos^2(\frac{\pi}{8})$. અહીં $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$,તેથી $\cos^2(\frac{7\pi}{8}) = \cos^2(\frac{\pi}{8})$. અહીં $\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$,તેથી $\cos^2(\frac{5\pi}{8}) = \cos^2(\frac{3\pi}{8})$. તેથી,$E = 2\cos^2(\frac{\pi}{8}) + 2\cos^2(\frac{3\pi}{8})$. કારણ કે $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,તેથી $\cos^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8})$. આ કિંમત $E$ માં મૂકતા: $E = 2\cos^2(\frac{\pi}{8}) + 2\sin^2(\frac{\pi}{8})$. $E = 2(\cos^2(\frac{\pi}{8}) + \sin^2(\frac{\pi}{8}))$. નિત્યસમ $\sin^2 	heta + \cos^2 	heta = 1$ નો ઉપયોગ કરતા,આપણને $E = 2(1) = 2$ મળે છે.

$\cos ^2\left(\frac{7 \pi}{8}\right)+\cos ^2\left(\frac{5 \pi}{8}\right)+\cos ^2\left(\frac{3 \pi}{8}\right)+\cos ^2\left(\frac{\pi}{8}\right)=$

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