If alpha = 22^circ 30',then (1 + cos alpha )( | vedclass

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(A) Given $\alpha = 22^\circ 30' = \frac{\pi}{8}$.Let $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.Note that $3\alpha

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$1/8$

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(A) Given $\alpha = 22^\circ 30' = \frac{\pi}{8}$.Let $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.Note that $3\alpha = 67^\circ 30'$,$5\alpha = 112^\circ 30'$,and $7\alpha = 157^\circ 30'$.We know $\cos 5\alpha = \cos(180^\circ - 67^\circ 30') = -\cos 3\alpha$ and $\cos 7\alpha = \cos(180^\circ - 22^\circ 30') = -\cos \alpha$.So,$P = (1 + \cos \alpha)(1 - \cos \alpha)(1 + \cos 3\alpha)(1 - \cos 3\alpha)$.$P = (1 - \cos^2 \alpha)(1 - \cos^2 3\alpha) = \sin^2 \alpha \sin^2 3\alpha$.Since $\alpha = 22.5^\circ$,$\sin^2 \alpha = \frac{1 - \cos 45^\circ}{2} = \frac{1 - 1/\sqrt{2}}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$.Since $3\alpha = 67.5^\circ$,$\sin^2 3\alpha = \frac{1 - \cos 135^\circ}{2} = \frac{1 - (-1/\sqrt{2})}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}}$.$P = \left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right) \left(\frac{\sqrt{2} + 1}{2\sqrt{2}}\right) = \frac{2 - 1}{8} = \frac{1}{8}$.

If $\alpha = 22^\circ 30'$,then $(1 + \cos \alpha )(1 + \cos 3\alpha )(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ equals

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