Question

Prove that the maximum value of $\sin x+\cos x$ is $\sqrt{2}$.

Answer 1

$f\left(x\right)=\sin x+\cos x$
$f^{\prime }\left(x\right)=f\left(x\right)=\sin x+\cos x$
For maximum value of $f^{\prime }\left(x\right)=0$
$\cos x-\sin x=0$
$\cos x=\sin x$
$\tan x=1\Rightarrow \tan x=\tan \frac{\pi }{4}$
$x=\frac{\pi }{4}$
To find maximum value
We have to calculate $f\left(x\right)$
at $x=\frac{\pi }{4}$

$f\left(\frac{\pi }{4}\right)=\cos \frac{\pi }{4}+\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$
Therefore, maximum value of $\sqrt{2}$.