If 1 + cos alpha + cos^2 alpha + dots infty = | vedclass

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(D) The given series is an infinite geometric series with the first term $a = 1$ and common ratio $r = \cos \alpha$.For an infinite geometric series,t

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$3\pi / 4$

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(D) The given series is an infinite geometric series with the first term $a = 1$ and common ratio $r = \cos \alpha$.For an infinite geometric series,the sum $S = \frac{a}{1 - r}$,provided $|r| Given $S = 2 - \sqrt{2}$,we have $\frac{1}{1 - \cos \alpha} = 2 - \sqrt{2}$.Taking the reciprocal,$1 - \cos \alpha = \frac{1}{2 - \sqrt{2}}$.Rationalizing the denominator: $\frac{1}{2 - \sqrt{2}} 	imes \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.So,$1 - \cos \alpha = 1 + \frac{1}{\sqrt{2}}$,which implies $-\cos \alpha = \frac{1}{\sqrt{2}}$,or $\cos \alpha = -\frac{1}{\sqrt{2}}$.Since $0 $\cos \alpha = -\cos(\pi/4) = \cos(\pi - \pi/4) = \cos(3\pi/4)$.Therefore,$\alpha = 3\pi/4$.

If $1 + \cos \alpha + \cos^2 \alpha + \dots \infty = 2 - \sqrt{2}$,then $\alpha$ $(0 < \alpha < \pi)$ is:

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