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

Identify the correct order.

The solution of the inequality $(\cot^{-1}x)^2 - 5\cot^{-1}x + 6 > 0$ is:

If $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$, then $x=$

Solve $\sin(\tan^{-1} x)$,where $|x| < 1$,is equal to:

$A$ positive acute angle is divided into two parts whose tangents are $\frac{1}{2}$ and $\frac{1}{3}$. Then the angle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module