<s>[INST] Minimum value of inverse trigo function (cubic) 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 minimum value of $(\sin^{-1}x)^3 = (\frac{-\pi}{2})^3$= -$\frac{\pi}{8}$
Please guide..
 [/INST] Hint: Use the fact that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$
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