Find the minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$.

Question

The minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$ is equal to ( following options)

a) $\displaystyle \frac{\pi^3}{32}$ b) $\displaystyle\frac{5\pi^3}{32}$ c) $\displaystyle\frac{9\pi^3}{32}$ d) $\displaystyle\frac{11\pi^3}{32}$

Can we go like this :

$$ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$$

Therefore the minimum value of $(\sin^{-1}x)^3$ is$$(\frac{-\pi}{2})^3= -\frac{\pi}{8}$$

Please guide..

Answer 1

As $\sin^{-1}x+\cos^{-1}x=\frac\pi2$

$$(\sin^{-1}x)^3+(\cos^{-1}x)^3$$ $$=(\sin^{-1}x)^3+\left(\frac\pi2-\sin^{-1}x\right)^3$$ $$=\left(\frac\pi2\right)^3-3\left(\frac\pi2\right)^2\sin^{-1}x+3\left(\frac\pi2\right)(\sin^{-1}x)^2$$

$$=\left(\frac\pi2\right)^3+3\left(\frac\pi2\right) \{(\sin^{-1}x)^2-\frac\pi2\sin^{-1}x\}$$

$$=\left(\frac\pi2\right)^3+3\left(\frac\pi2\right) \{\left(\sin^{-1}x-\frac\pi4\right)^2-\left(\frac\pi4\right)^2\}$$

$$\ge \left(\frac\pi2\right)^3-3\left(\frac\pi2\right) \left(\frac\pi4\right)^2$$

Answer 2

Hint: Use the fact that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$