The minimum value of (sin^{-1} x)^2 + (cos^{- | vedclass

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Question

(A) Let $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$. We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Substituting this, $f(x) = (\sin^{-1} x)

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$\frac{\pi^2}{8}$

Vedclass · Answer

(A) Let $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$. We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Substituting this, $f(x) = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$. Let $u = \sin^{-1} x$. Since $x \in [-1, 1]$, $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. $f(u) = u^2 + (\frac{\pi}{2} - u)^2 = u^2 + \frac{\pi^2}{4} - \pi u + u^2 = 2u^2 - \pi u + \frac{\pi^2}{4}$. This is a parabola opening upwards. The minimum occurs at $u = -\frac{b}{2a} = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$. Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the minimum value is $f(\frac{\pi}{4}) = 2(\frac{\pi}{4})^2 - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = 2(\frac{\pi^2}{16}) - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$.

The minimum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in [-1, 1]$ is:

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