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

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(C) આપેલ પદાવલિ: $E = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8}$$\cos(\pi - 	heta) = -\cos 	heta$

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) આપેલ પદાવલિ: $E = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8}$$\cos(\pi - 	heta) = -\cos 	heta$ હોવાથી,$\cos \frac{7\pi}{8} = \cos(\pi - \frac{\pi}{8}) = -\cos \frac{\pi}{8}$ અને $\cos \frac{5\pi}{8} = \cos(\pi - \frac{3\pi}{8}) = -\cos \frac{3\pi}{8}$ થાય.તેમની $4$ ઘાત લેતા,$\cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8}$ અને $\cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8}$ મળે.આ કિંમતો પદાવલિમાં મૂકતા: $E = 2(\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8})$.નિત્યસમ $a^2 + b^2 = (a+b)^2 - 2ab$ નો ઉપયોગ કરતા,જ્યાં $a = \cos^2 \frac{\pi}{8}$ અને $b = \cos^2 \frac{3\pi}{8}$:$E = 2[(\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8})^2 - 2\cos^2 \frac{\pi}{8} \cos^2 \frac{3\pi}{8}]$અહીં $\cos^2 \frac{3\pi}{8} = \sin^2 \frac{\pi}{8}$ હોવાથી,$\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} = \cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} = 1$ થાય.$E = 2[1^2 - 2(\cos \frac{\pi}{8} \cos \frac{3\pi}{8})^2] = 2[1 - 2(\frac{1}{2} (2 \cos \frac{\pi}{8} \cos \frac{3\pi}{8}))^2]$$2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ સૂત્ર વાપરતા:$2 \cos \frac{\pi}{8} \cos \frac{3\pi}{8} = \cos(\frac{4\pi}{8}) + \cos(-\frac{2\pi}{8}) = \cos \frac{\pi}{2} + \cos \frac{\pi}{4} = 0 + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.$E = 2[1 - 2(\frac{1}{2} \cdot \frac{1}{\sqrt{2}})^2] = 2[1 - 2(\frac{1}{2\sqrt{2}})^2] = 2[1 - 2(\frac{1}{8})] = 2[1 - \frac{1}{4}] = 2(\frac{3}{4}) = \frac{3}{2}$.

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

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