If $\frac{\pi}{2}
$\because \frac{\pi}{2} $\therefore \sqrt{\frac{1+\cos 2 x}{2}}=\sqrt{\frac{2 \cos ^{2} x}{2}}=|\cos x|$ In second quadrant $\cos x$ is negative. $\therefore \sqrt{\frac{1+\cos 2 x}{2}}=-\cos x$
$\therefore \sqrt{\frac{1+\cos 2 x}{2}}=\sqrt{\frac{2 \cos ^{2} x}{2}}=|\cos x|$
In second quadrant $\cos x$ is negative.
$\therefore \sqrt{\frac{1+\cos 2 x}{2}}=-\cos x$
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