The least and greatest values of (sin^{-1} x) | vedclass

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(C) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$. We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.Using the identity $a^3 + b^3 = (a+b)^3 -

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$\frac{\pi^3}{32}, \frac{7\pi^3}{8}$

Vedclass · Answer

(C) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$. We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$,we get:$f(x) = (\sin^{-1} x + \cos^{-1} x)^3 - 3(\sin^{-1} x)(\cos^{-1} x)(\sin^{-1} x + \cos^{-1} x)$$f(x) = (\frac{\pi}{2})^3 - 3(\sin^{-1} x)(\frac{\pi}{2} - \sin^{-1} x)(\frac{\pi}{2})$$f(x) = \frac{\pi^3}{8} - \frac{3\pi}{2}(\sin^{-1} x)(\frac{\pi}{2} - \sin^{-1} x)$$f(x) = \frac{\pi^3}{8} - \frac{3\pi^2}{4}(\sin^{-1} x) + \frac{3\pi}{2}(\sin^{-1} x)^2$Completing the square for the quadratic in $\sin^{-1} x$:$f(x) = \frac{3\pi}{2} [(\sin^{-1} x)^2 - \frac{\pi}{2}(\sin^{-1} x) + \frac{\pi^2}{16} - \frac{\pi^2}{16}] + \frac{\pi^3}{8}$$f(x) = \frac{3\pi}{2}(\sin^{-1} x - \frac{\pi}{4})^2 - \frac{3\pi^3}{32} + \frac{4\pi^3}{32} = \frac{3\pi}{2}(\sin^{-1} x - \frac{\pi}{4})^2 + \frac{\pi^3}{32}$Since $\sin^{-1} x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $\sin^{-1} x = \frac{\pi}{4}$,which is $\frac{\pi^3}{32}$.The maximum value occurs at the boundary $\sin^{-1} x = -\frac{\pi}{2}$:$f(-\frac{\pi}{2}) = \frac{3\pi}{2}(-\frac{\pi}{2} - \frac{\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(-\frac{3\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(\frac{9\pi^2}{16}) + \frac{\pi^3}{32} = \frac{27\pi^3}{32} + \frac{\pi^3}{32} = \frac{28\pi^3}{32} = \frac{7\pi^3}{8}$.

The least and greatest values of $(\sin^{-1} x)^3 + (\cos^{-1} x)^3$ are:

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