Question #2712b

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Question #2712b

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Answer 1 · 2017-03-31T08:25:03

$\frac{1}{\sqrt{2}}$ $1/sqrt 2$

Explanation:

$cos x = cos (\frac{x}{2} + \frac{x}{2}) = {cos}^{2} (\frac{x}{2}) - {sin}^{2} (\frac{x}{2}) = 2 {cos}^{2} (\frac{x}{2}) - 1$ $cos x = cos(x/2 + x/2) = cos^2 (x/2) -sin^2 (x/2) =2 cos^2 (x/2) -1$

$cos x = 2 {cos}^{2} (\frac{x}{2}) - 1$ $cos x = 2 cos^2 (x/2) -1$

$cos x + 1 = 2 {cos}^{2} (\frac{x}{2})$ $cos x + 1 = 2 cos^2 (x/2)$

$\frac{1 + cos x}{2} = {cos}^{2} (\frac{x}{2})$ $(1 + cos x)/2 = cos^2 (x/2)$

$\pm \sqrt{\frac{1 + cos x}{2}} = cos (\frac{x}{2}) \to$ $+- sqrt((1 + cos x)/2) = cos (x/2)->$ proved for all $x$ $x$ .

To find their value when $x = \frac{π}{2}$ $x =pi/2$

A . $cos (\frac{x}{2}) = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $cos (x/2) = cos (pi/4) = 1/sqrt 2$

B. $\pm \sqrt{\frac{1 + cos x}{2}} = \pm \sqrt{\frac{1 + cos (\frac{π}{2})}{2}}$ $+- sqrt((1 + cos x)/2) = +- sqrt((1 + cos (pi/2))/2)$

$= \pm \sqrt{\frac{1 + 0}{2}} = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{1}}{\sqrt{2}} = \pm \frac{1}{\sqrt{2}}$ $= +- sqrt((1 + 0)/2) = +- sqrt(1/2) = +- sqrt1/sqrt 2 = +- 1/sqrt 2$
since $x$ $x$ in quadrant I, $\sqrt{\frac{1 + cos x}{2}} = \frac{1}{\sqrt{2}}$ $sqrt((1 + cos x)/2) = 1/sqrt 2$

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Question #2712b

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