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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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)

Vedclass · Accepted Answer

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

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:

Similar Questions

If $y = \sin^{-1}\left(\frac{19}{20}x\right) + \cos^{-1}\left(\frac{19}{20}x\right)$,then $\frac{dy}{dx} = $

If $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha$, then $\alpha=$

The value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ is

Evaluate: $\cos^{-1}(\cos \frac{4\pi}{3}) + \sin^{-1}(\sin \frac{4\pi}{3}) = \dots$

Show that $\sin ^{-1}(2 x \sqrt{1-x^{2}})=2 \cos ^{-1} x$ for $\frac{1}{\sqrt{2}} \leq x \leq 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module