global max value

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

Then $y=f(\sin x)=\sqrt 2\sin\left (\sin x+\dfrac{\pi}{4} \right )$ .

Consider $\sin\theta$ , which reaches a maximum where $\theta = \dfrac{\pi}{2}+2n\pi\ \left \{n\in\mathbb{Z} \right \}$ .

Clearly $\sin x\in [-1,1]$ , so as $\dfrac{\pi}{4}<\dfrac{\pi}{2}<\dfrac{\pi}{4}+1$ then $f(\sin x)$ acheives its maximum value of $\sqrt 2=\color{#20A900}{\boxed{1.41\ \textnormal{(3 s.f.)}}}$ , at $x=2n\pi+\arcsin \dfrac{\pi}{4}\left \{n\in\mathbb{Z}\right \}$ .

$\begin{aligned} y & = \sin(\sin x) + \cos(\sin x) \\ & = \sqrt 2 \sin \left(\sin x + \frac \pi 4\right) \\ \implies \max(y) & = \sqrt 2 \approx \boxed{1.41} & \small \blue{\text{when }x = 2n\pi + \sin^{-1} \left(\frac \pi 4\right)} \end{aligned}$