[Math] Minimum and maximum of $ \sin^2(\sin x) + \cos^2(\cos x) $

calculustrigonometry

I want to find the maximum and minimum value of this expression:
$$ \sin^2(\sin x) + \cos^2(\cos x) $$

Best Answer

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 $0+\cos^21\approx 0.29$