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  4. The maximum value of sin (x+(π/6))+cos(x+(π /6)) is in the interval (0,+(π/6)) if the value of x is

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Q. The maximum value of sin $\left(x+\frac{\pi}{6}\right)+cos\left(x+\frac{\pi }{6}\right)$ is in the interval $\left(0,+\frac{\pi}{6}\right)$ if the value of x is

Trigonometric Functions

Solution:

Let, y = $sin \left(x+\frac{\pi}{6}\right)+cos\left(x+\frac{\pi }{6}\right)$
$= \sqrt{2} \left[\frac{1}{\sqrt{2}}sin \left(x+\frac{\pi }{6}\right)+\frac{1}{\sqrt{2}}cos\left(x+\frac{\pi }{6}\right)\right]$
$= \sqrt{2}\left[sin\frac{\pi}{4}sin \left(x+\frac{\pi }{6}\right)+cos\frac{\pi }{4}cos\left(x+\frac{\pi }{6}\right)\right]$
$= \sqrt{2}\left[cos\left(x+\frac{\pi }{6}-\frac{\pi }{4}\right)\right] = \sqrt{2}\left[cos\left(x-\frac{\pi }{12}\right)\right]$
$\Rightarrow \quad x-\frac{\pi }{12} = 0\quad$ [$\because$ y to be max.]
$\Rightarrow \quad x = \frac{\pi }{12}$