Can We Compound The Range?

Relevant wiki: Trigonometric Equations - R method

$\sin(x) + \cos(x) = \sqrt{2} \sin\left(x+\dfrac{\pi}{4}\right)$
$-\sqrt{2} \leq\sqrt{2} \sin\left(x+\dfrac{\pi}{4}\right) \leq \sqrt{2}$

Since we want to maximize $\sin(x)+\cos(x)$ we can take its derivative and set it to zero to find the value of x. $\frac{d}{dx}(\sin(x)+\cos(x))$ $=\cos(x)-\sin(x)$ $=0$ $\cos(x)=\sin(x)$ $x=\frac{\pi}{4}$ $\sin(\frac{\pi}{4})+\cos(\frac{\pi}{4})$ $=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}$ $=\boxed{\sqrt{2}}$

$1=\sin^2x+\cos^2x=(\sin x+\cos x)^2-2\sin x\cos x=(\sin x+\cos x)^2-\sin 2x$

$(\sin x+\cos x)^2=1+\sin 2x\leq2$

$|\sin x+\cos x|\leq\sqrt2$

The equality holds when $x=\frac\pi4+\pi k$ , $k \in \mathbb{Z}$ .