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The maximum value of $\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$ in the
interval $\left(0, \frac{\pi}{2}\right)$ is attained at
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interval $\left(0, \frac{\pi}{2}\right)$ is attained at
Solution:
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Verified Answer
The correct answer is:
$\frac{\pi}{12}$
$\quad \sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$
$=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \left(x+\frac{\pi}{6}\right)+\frac{1}{\sqrt{2}} \cos \left(x+\frac{\pi}{6}\right)\right]$
$=\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}\right) \cos \frac{\pi}{4}+\cos \left(x+\frac{\pi}{6}\right) \sin \frac{\pi}{4}\right]$
$=\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}+\frac{\pi}{4}\right)\right]$
$[\because \sin (A+B)=\sin A \cos B+\cos A \operatorname{Sin} B]$
$=\sqrt{2}\left[\sin \left(x+\frac{5 \pi}{12}\right)\right]$
Given interval is $\left(0, \frac{\pi}{2}\right)$ For, maximum value $x+\frac{5 \pi}{12}=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{2}-\frac{5 \pi}{12}=\frac{6 \pi-5 \pi}{12}=\frac{\pi}{12}$
$=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \left(x+\frac{\pi}{6}\right)+\frac{1}{\sqrt{2}} \cos \left(x+\frac{\pi}{6}\right)\right]$
$=\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}\right) \cos \frac{\pi}{4}+\cos \left(x+\frac{\pi}{6}\right) \sin \frac{\pi}{4}\right]$
$=\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}+\frac{\pi}{4}\right)\right]$
$[\because \sin (A+B)=\sin A \cos B+\cos A \operatorname{Sin} B]$
$=\sqrt{2}\left[\sin \left(x+\frac{5 \pi}{12}\right)\right]$
Given interval is $\left(0, \frac{\pi}{2}\right)$ For, maximum value $x+\frac{5 \pi}{12}=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{2}-\frac{5 \pi}{12}=\frac{6 \pi-5 \pi}{12}=\frac{\pi}{12}$
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