Maximum value + Minimum value in trigonometry

Let $f(x) = \sin x + \cos x = \sqrt 2\left( \frac 1{\sqrt 2} \sin x + \frac 1{\sqrt 2} \cos x \right) = \sqrt 2 \sin \left( x + \frac \pi 4 \right)$ .

$\implies \begin{cases} \max(f(x)) = \sqrt 2 \max \left(\sin \left( x + \frac \pi 4 \right)\right) = \sqrt 2 \\ \min(f(x)) = \sqrt 2 \min \left(\sin \left( x + \frac \pi 4 \right)\right) = - \sqrt 2 \end{cases}$

Therefore $\max(f(x) + \min(f(x) = \boxed 0$ .