The value of sin^{-1}(sin 10) is | vedclass

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

Question

(C) We know that $\sin^{-1}(\sin x) = x$ only when $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $10$ radians is not in the interval $[-\frac{\pi}{2},

Vedclass · Accepted Answer

$3\pi - 10$

Vedclass · Answer

(C) We know that $\sin^{-1}(\sin x) = x$ only when $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $10$ radians is not in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we need to find an angle $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(y) = \sin(10)$.We know that $\sin(x) = \sin(n\pi + (-1)^n x)$.For $n = 3$,$\sin(10) = \sin(3\pi - 10)$.Since $3\pi \approx 9.42$,we have $3\pi Specifically,$3\pi Subtracting $3\pi$ from all sides,we get $0 Multiplying by $-1$,we get $-\frac{\pi}{2} Since $3\pi - 10$ lies in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin^{-1}(\sin 10) = \sin^{-1}(\sin(3\pi - 10)) = 3\pi - 10$.

The value of $\sin^{-1}(\sin 10)$ is

Similar Questions

$\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)$ is equal to $.........$

The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}(-\frac{1}{2})$ is

The value of $\cos ^{-1}\left(\tan \left(\frac{7 \pi}{4}\right)\right)$ is

$\sin^{-1}(\sin \frac{7\pi}{6}) = $

$A$ value of $\tan^{-1} \left( \sin \left( \cos^{-1} \sqrt{\frac{2}{3}} \right) \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module