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Question
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
Solution
\[\because \frac{\pi}{2} < x < \frac{3\pi}{2}\]
\[ \therefore \sqrt{\frac{1 + \cos2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = \left| \text{ cos } x \right|\]
\[\text { In second quadrant } \text{ cos } x \text{ is negative } . \]
\[ \therefore \sqrt{\frac{1 + \cos2x}{2}} = - \text{ cos } x\]
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