Minimum value of $\sin^{-1}x \cos^{-1}x$

Question

Let us consider the function $~~f(x)= \sin^{-1}x \cos^{-1}x.$ I want to find the minimum value of $f$.
So, first I find the derivative $df$ of $f$ and put $df=0$ to get $\cos^{-1}x = \sin^{-1}x$ and from this critical point I got the maximum value of $f$ which is $\frac{\pi^2}{16}.$ But I want to find the minimum value of $f$.

Please help me to find this.

Answer 1

As $-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2$

and $0\le\cos^{-1}x\le\pi$

Clearly $x=-1$ yields the minimum value

Alternatively,

$$f(x)=\sin^{-1}x\cos^{-1}x=\left(\dfrac\pi2-\cos^{-1}x\right)\cos^{-1}x=\left(\dfrac\pi4\right)^2-\left(\cos^{-1}x-\dfrac\pi4\right)^2$$

Now $f(x)$ will be minimum if $\left(\cos^{-1}x-\dfrac\pi4\right)^2$ is maximum

$0\le\cos^{-1}x\le\pi\implies0-\dfrac\pi4\le\cos^{-1}x-\dfrac\pi4\le\pi-\dfrac\pi4$

$\implies\left(\cos^{-1}x-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$

Answer 2

As you can see by the fact that at $\frac{\pi^2}{16}$ it is the only maximum (or stationary point of any kind), that means it is strictly decreasing in both directions. Taking the range of $\arcsin(x)$ and $\arccos(x)$, you see that the function has a range of [-1,1]. Checking both -1 and 1 (because you are strictly decreasing in both directions from $\frac{\pi^2}{16}$ but at different rates), you find that f(-1) is smaller and thus the minimum.