# Prove that $f(x)=\sin x+\sqrt 3\cos x$ has maximum value at $x=\Large\frac{1}{6}$.

Toolbox:
• Let $f(x)$ be a function with domain $D \subset R$. Then $f(x)$ is said to attain the maximum value at a point, $a \in D,$ if $f'(x) \leq f(a)$ for all $x \in D$.
Step 1
$f(x)= \sin x+ \sqrt 3 \cos x$
differentiating w.r.t $x$ we get,
$f'(x)= \cos x+ \sqrt 3(-\sin x)$
$= \cos x-\sqrt 3 \sin x$
at $x = \large\frac{\pi}{6}$
$f'(x)= \cos \large\frac{\pi}{6}-\sqrt 3 \times \sin \large\frac{\pi}{6}$
But $\cos \large\frac{\pi}{6}=\large\frac{\sqrt 3}{2} \: and \: \sin \large\frac{\pi}{6}=\large\frac{1}{2}$
$\therefore f'(x)=\large\frac{\sqrt 3}{2}-\large\frac{\sqrt 3}{2}=0$
$f(x)=\sin\large\frac{\pi}{6}+\sqrt 3 \times \cos \large\frac{\pi}{6}$
$= \large\frac{1}{2}+\large\frac{\sqrt 3}{2}$
$= \large\frac{\sqrt 3+1}{2}$
Step 2
Again differentiating $f(x)$ w.r.t $x$ we get,
$f''(x)-\sin x - \sqrt 3 \cos x$
$f''(x)=-( \sin x+ \sqrt 3 \cos x)$
$f''(x)=-f(x)$
Since $f''(x)<0,$ the given function is maximum.
Hence the maximum value is $\large\frac{\sqrt 3+1}{2}$

edited Aug 6, 2013