If theta = sin^{-1}[sin(-600^circ)],then one | vedclass

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

Question

(A) Given $	heta = \sin^{-1}[\sin(-600^\circ)]$. Since $\sin(-	heta) = -\sin(	heta)$,we have $	heta = \sin^{-1}[-\sin(600^\circ)]$. Now,express $6

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Given $	heta = \sin^{-1}[\sin(-600^\circ)]$. Since $\sin(-	heta) = -\sin(	heta)$,we have $	heta = \sin^{-1}[-\sin(600^\circ)]$. Now,express $600^\circ$ as a multiple of $180^\circ$ or $360^\circ$: $600^\circ = 360^\circ + 240^\circ$. So,$\sin(600^\circ) = \sin(360^\circ + 240^\circ) = \sin(240^\circ)$. Also,$\sin(240^\circ) = \sin(180^\circ + 60^\circ) = -\sin(60^\circ)$. Substituting this back: $	heta = \sin^{-1}[-(-\sin(60^\circ))] = \sin^{-1}[\sin(60^\circ)]$. Since $60^\circ$ is in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $	heta = 60^\circ = \frac{\pi}{3}$.

If $\theta = \sin^{-1}[\sin(-600^\circ)]$,then one of the possible values of $\theta$ is

Similar Questions

If $2 \cos \left(2 \tan ^{-1} x\right)=1$,then $x=$ . . . . . .

$\cos^{-1}\left(x^2 + \frac{1}{x^2} - 1\right) + \sin^{-1}\left(x^2 - \frac{1}{x^2}\right) + \tan^{-1}(x^2)$ is equal to (where $x \in R - \{0\}$)

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is equal to . . . . . . .

If the area of the region bounded by the parabola $y^2 = 4kx$ and the line $x = k$, (where $k > 0$) is $\frac{128}{3}$ sq. units, then the value of $\sin^{-1} \left( \frac{2}{k} \right)$ is equal to...

Evaluate: $\cos^{-1}\sqrt{1-x} + \sin^{-1}\sqrt{1-x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module