Q. 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
A) $x = \frac{\pi}{12}$
B) $x = \frac{\pi}{6}$
C) $x = \frac{\pi}{3}$
D) $x = \frac{\pi}{2}$
Explanation
(a)
$\cos\left( x + \frac{\pi}{6} \right) + \sin\left( x + \frac{\pi}{6} \right)$
$= \sqrt{2}\left\lbrack \frac{1}{\sqrt{2}}\cos{\left( x + \frac{\pi}{6} \right) + \frac{1}{\sqrt{2}}\sin\left( x + \frac{\pi}{6} \right)} \right\rbrack$
$= \sqrt{2}\cos\left( x + \frac{\pi}{6} - \frac{\pi}{4} \right) = \sqrt{2}\cos\left( x - \frac{\pi}{12} \right)$
$\therefore$ For maximum value $x = \frac{\pi}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.