A quite compact maxima problem

$\begin{aligned} f(x) & = \cos^2 (\cos x) + \sin^2 (\sin x) & \small \color{#3D99F6} \text{Differentiat both sides w.r.t. }x. \\ f'(x) & = 2\sin (\cos x) \cos (\cos x) \sin x + 2\sin (\sin x) \cos (\sin x) \cos x \end{aligned}$

We note that $f'(x) = 0$ , when $\sin x = 0 \implies x=0$ or $\cos x = 0 \implies x = \frac \pi 2$ . Now let us check the values of $f''(x)$ , when $x=0$ and $x = \frac \pi 2$ .

$\begin{aligned} f'(x) & = 2\sin (\cos x) \cos (\cos x) \sin x + 2\sin (\sin x) \cos (\sin x) \cos x \\ & = \sin (2\cos x)\sin x + \sin (2\sin x) \cos x \\ f''(x) & = -2 \sin^2 x \cos (2\cos x) + \cos x \sin (2\cos x) + 2\cos^2 x \cos (2\sin x) - \sin x \sin (2\sin x) \end{aligned}$

$\begin{cases} f''(0) = \sin 2 + 2 > 0 & \small \color{#3D99F6} \implies f(0) \text{ is a minimum.} \\ f''\left(\frac \pi 2\right) = -2 - \sin 2 < 0 & \small \color{#3D99F6} \implies f\left(\frac \pi 2\right) \text{ is a maximum.} \end{cases}$

Therefore, $\max (f(x)) = f\left(\frac \pi 2\right) = \cos^2 \left(\cos \frac \pi 2\right) + \sin^2 \left(\sin \frac \pi 2\right) = \cos^2 0 + \sin^2 1 \approx \boxed{1.708}$ .