Question
If $\frac{\pi}{2}<\text{x}<\pi,$ then find $\frac{\text{d}}{\text{dx}}\Bigg(\sqrt{\frac{1+\cos\text{2x}}{2}}\Bigg)$

Answer

$\frac{\text{d}}{\text{dx}}\sqrt{\frac{1+\cos\text{2x}}{2}}=\frac{\text{d}}{\text{dx}}\sqrt{\frac{\sin^2\text{x}+\cos^2\text{x}+\cos^2\text{x}-\sin^2\text{x}}{2}}$$=\frac{\text{d}}{\text{dx}}\cos\text{x}$
$=-\sin\text{x}$
$=\sin\text{x}\ \Big[\because\frac{\pi}{2}<\text{x}<\pi\Big]$

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