Question

What is the maximum value of the function $\sin x+\cos x$?

Answer 1

$f\left(x\right)=\sin \left(x\right)+\cos \left(x\right)$
Dividing and multiplying by $\sqrt{2}$
$=\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)$
$=\sqrt{2}(\cos 45\sin x+\sin 45\cos x)$
$=\sqrt{2}\left(\sin \right(x+45\left)\right)$
$\Rightarrow f\left(x\right)=\sqrt{2}\left(\sin \right(x+45\left)\right)$
We know that
$-1\le \sin \left(x\right)\le 1$
$\Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left(x\right)\le \sqrt{2}$
Here $x$ can take any value
$\Rightarrow -\sqrt{2}\le \sqrt{2}\sin (x+45)\le \sqrt{2}$
Which clearly shows that maximum and minimum value of the function is $\sqrt{2}$ and $-\sqrt{2}$