Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (B)

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