The greatest and the least value of {(sin ^{ | vedclass

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(C) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$. We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$. Let $u = \sin^{-1} x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$. We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$. Let $u = \sin^{-1} x$. Then $\cos^{-1} x = \frac{\pi}{2} - u$. Since $x \in [-1, 1]$,$u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Now,$f(u) = u^3 + (\frac{\pi}{2} - u)^3 = u^3 + \frac{\pi^3}{8} - 3u^2(\frac{\pi}{2}) + 3u(\frac{\pi^2}{4}) - u^3 = \frac{3\pi^2}{4}u - \frac{3\pi}{2}u^2 + \frac{\pi^3}{8}$. Completing the square: $f(u) = \frac{3\pi}{2} [ -u^2 + \frac{\pi}{2}u ] + \frac{\pi^3}{8} = \frac{3\pi}{2} [ -(u - \frac{\pi}{4})^2 + \frac{\pi^2}{16} ] + \frac{\pi^3}{8} = \frac{3\pi^3}{32} - \frac{3\pi}{2}(u - \frac{\pi}{4})^2 + \frac{\pi^3}{8} = \frac{7\pi^3}{32} - \frac{3\pi}{2}(u - \frac{\pi}{4})^2$. Wait,let's re-evaluate: $f(u) = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8} = \frac{3\pi}{2}(u - \frac{\pi}{4})^2 + \frac{\pi^3}{32}$. For $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum occurs at $u = \frac{\pi}{4}$,giving $f_{min} = \frac{\pi^3}{32}$. The maximum occurs at the boundary $u = -\frac{\pi}{2}$,giving $f_{max} = \frac{3\pi}{2}(-\frac{\pi}{2} - \frac{\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(-\frac{3\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(\frac{9\pi^2}{16}) + \frac{\pi^3}{32} = \frac{27\pi^3}{32} + \frac{\pi^3}{32} = \frac{28\pi^3}{32} = \frac{7\pi^3}{8}$.

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

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