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Question
If \[\frac{\pi}{2}\] then find \[\frac{d}{dx}\left( \sqrt{\frac{1 + \cos 2x}{2}} \right)\]
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Solution
\[\sqrt{\frac{1 + \cos 2x}{2}}\]
\[ = \sqrt{\frac{2 \cos^2 x}{2}}\]
\[ = \sqrt{\cos^2 x}\]
\[ = - \cos x (\because\frac{\pi}{2}<x<\pi)\]
\[\frac{d}{dx}\left( \sqrt{\frac{1 + \cos 2x}{2}} \right)\]
\[ = \frac{d}{dx}\left( - \cos x \right)\]
\[ = - \left( - \sin x \right)\]
\[ = \sin x\]
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