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

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(A) Given $\alpha = 22^\circ 30' = 22.5^\circ$.We need to evaluate $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.Note

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$1/8$

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(A) Given $\alpha = 22^\circ 30' = 22.5^\circ$.We need to evaluate $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.Note that $3\alpha = 67.5^\circ$,$5\alpha = 112.5^\circ = 180^\circ - 67.5^\circ$,and $7\alpha = 157.5^\circ = 180^\circ - 22.5^\circ$.Using $\cos(180^\circ - 	heta) = -\cos 	heta$,we have:$1 + \cos 5\alpha = 1 - \cos 67.5^\circ = 1 - \cos 3\alpha$$1 + \cos 7\alpha = 1 - \cos 22.5^\circ = 1 - \cos \alpha$Substituting these into the expression:$P = (1 + \cos \alpha)(1 - \cos \alpha)(1 + \cos 3\alpha)(1 - \cos 3\alpha)$$P = (1 - \cos^2 \alpha)(1 - \cos^2 3\alpha) = \sin^2 \alpha \cdot \sin^2 3\alpha$Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$:$\sin^2 22.5^\circ = \frac{1 - \cos 45^\circ}{2} = \frac{1 - 1/\sqrt{2}}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$$\sin^2 67.5^\circ = \frac{1 - \cos 135^\circ}{2} = \frac{1 - (-1/\sqrt{2})}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}}$$P = \left( \frac{\sqrt{2} - 1}{2\sqrt{2}} ight) \left( \frac{\sqrt{2} + 1}{2\sqrt{2}} ight) = \frac{(\sqrt{2})^2 - 1^2}{8} = \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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