The maximum value of $\sin x+\cos x, x \in R$ is:

$\sqrt{2}$

$y=\sin x+\cos x$

we know that value of sinusoidal functions are maximum 1 multiplying and dividing equation by $\sqrt{2}$

we get $y=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)$

$y=\sqrt{2}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right)$

as  $\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

$y=\sqrt{2} \sin \underbrace{\left(x+\frac{\pi}{4}\right)}_{\text {let it be } \theta}$

$y=\sqrt{2} \sin \theta$

as $sin (A+B)$

$=\sin A \cos B+\sin B \cos A$

$-1 \leq \sin \theta \leq 1$

$\Rightarrow -\sqrt{2} \leq \sqrt{2} \sin \theta \leq \sqrt{2}$

$\Rightarrow -\sqrt{2} \leq y \leq \sqrt{2}$

⇒  y_max = $\sqrt{2}$