જો alpha = 22^circ 30' હોય,તો (1 + cos alpha | vedclass

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(A) આપેલ છે કે $\alpha = 22^\circ 30' = \frac{\pi}{8}$.ધારો કે $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.અહીં $3\a

Vedclass · Accepted Answer

$1/8$

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(A) આપેલ છે કે $\alpha = 22^\circ 30' = \frac{\pi}{8}$.ધારો કે $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.અહીં $3\alpha = 67^\circ 30'$,$5\alpha = 112^\circ 30'$,અને $7\alpha = 157^\circ 30'$ છે.આપણે જાણીએ છીએ કે $\cos 5\alpha = \cos(180^\circ - 67^\circ 30') = -\cos 3\alpha$ અને $\cos 7\alpha = \cos(180^\circ - 22^\circ 30') = -\cos \alpha$.તેથી,$P = (1 + \cos \alpha)(1 - \cos \alpha)(1 + \cos 3\alpha)(1 - \cos 3\alpha)$.$P = (1 - \cos^2 \alpha)(1 - \cos^2 3\alpha) = \sin^2 \alpha \sin^2 3\alpha$.$\alpha = 22.5^\circ$ હોવાથી,$\sin^2 \alpha = \frac{1 - \cos 45^\circ}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$.$3\alpha = 67.5^\circ$ હોવાથી,$\sin^2 3\alpha = \frac{1 - \cos 135^\circ}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}}$.$P = \left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right) \left(\frac{\sqrt{2} + 1}{2\sqrt{2}}\right) = \frac{2 - 1}{8} = \frac{1}{8}$.

જો $\alpha = 22^\circ 30'$ હોય,તો $(1 + \cos \alpha )(1 + \cos 3\alpha )(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ ની કિંમત શોધો.

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