Minimum and maximum of $ \sin^2(\sin x) + \cos^2(\cos x) $ I want to find the maximum and minimum value of this expression:
$$ \sin^2(\sin x) + \cos^2(\cos x) $$
 A: Following George's hint,
Because $-1\le \sin x \le 1$, and $\sin x$ is strictly increasing on $-1\le x\le 1$, we see that $\sin (\sin x)$ (and hence $\sin^2(\sin x)$) is maximized when $\sin x=1$, e.g. at $x=\pi/2$.
On the other hand, $\cos x$ is maximized when $x=0$, so $\cos (\cos x)$ (and hence $\cos^2(\cos x)$ is maximized when $\cos x=0$, e.g. at $x=\pi/2$.
Hence the combined function is maximal at $\pi/2$, when it is $1+\sin^21\approx 1.7$.
For the other direction, $x=0$ gives $\sin^2(\sin x)=0$, clearly minimal.  Because $-1\le \cos x\le 1$, and $\cos x$ is increasing on $[-1,0)$ and decreasing on $(0,1]$, we minimize $\cos x$ for $x=\pm 1$.  Hence in particular $x=0$ minimizes $\cos(\cos x)$ and thus $\cos^2(\cos x)$.  Combining, the minimum value is $0+\cos^21\approx 0.29$
A: Your expression simplifies to 
$$1+\cos(2\cos x)-\cos (2\sin x).$$
We optimize of $1+\cos u-\cos v$ under the constraint $u^2+v^2=4$.
$\cos$ is an even function, so we can say that we optimize $1+\cos 2u-\cos(2\sqrt{1-u^2})$, $u\in [0,1]$, which should be doable.
