Question

if q=sin^-1(sin(-600 degree)) then principal value of q is?

Answer 1

we know,

sin^-1(sinx) = x when -π/2≤ x ≤ π/2

here x = -600°

sin(-600°) = -sin(600°)
= -sin(720° -120°)
= sin120°
=sin(180-60°)
=sin(60°)
hence, sin (-600°) = sin(60°) =sin(π/3)

use this

q =sin^-1(sin(π/3))
q =π/3

Answer 2

Answer:

$q=\frac{\pi}{3}$

Step-by-step explanation:

Given : If $q=\sin^{-1}(\sin(-600^\circ))$

To find : The principal value of q is ?

Solution :

$q=\sin^{-1}(\sin(-600^\circ))$

We can re-write, $\sin(-\theta)=-\sin \theta$

$q=\sin^{-1}(-\sin(600^\circ))$

$q=\sin^{-1}(-\sin((720-120)^\circ))$

$q=\sin^{-1}(\sin(120)^\circ)$

$q=\sin^{-1}(\sin(180-60)^\circ)$

$q=\sin^{-1}(\sin(60)^\circ)$

$q=\sin^{-1}(\sin(\frac{\pi}{3}))$

Using inverse trigonometry identity, $\sin^{-1}\sin x=x$

$q=\frac{\pi}{3}$

Therefore, The principal value of q is $q=\frac{\pi}{3}$