The principal value of {sin ^{ - 1}}left[ {si | vedclass

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

Question

(D) The principal value branch of ${\sin^{-1}}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\frac{2\pi}{3}$ does not lie in the interval $[-\frac{\

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(D) The principal value branch of ${\sin^{-1}}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $\frac{2\pi}{3}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we simplify the expression.We know that $\sin(\pi - 	heta) = \sin(	heta)$.Therefore,$\sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3})$.Now,${\sin^{-1}}[\sin(\frac{2\pi}{3})] = {\sin^{-1}}[\sin(\frac{\pi}{3})]$.Since $\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the value is $\frac{\pi}{3}$.

The principal value of ${\sin ^{ - 1}}\left[ {\sin \left( {\frac{{2\pi }}{3}} \right)} \right]$ is

Similar Questions

If $x + \frac{1}{x} = 2$,the principal value of $\sin^{-1} x$ is

$\tan ^{-1}(-\sqrt{3})-\sec ^{-1}(-2)=$ . . . . . . .

$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2)$ is equal to

If $y = \cos^{-1}(\cos(|x| - f(x)))$,where $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \end{cases}$,then $\left. \frac{dy}{dx} \right|_{x = \frac{5\pi}{4}}$ is

Write the function in the simplest form: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), x < \pi$

Vedclass Test Series

Exam Paper Generator

Online Exam Module