inverse

\sin^{-1} [sin (2010) ]= \sin^{-1} sin ( 360\times 5 + 210 )= \sin^{-1} sin ( 210 )= \sin^{-1} sin ( 180 + 30 )= \sin^{-1} sin ( -30 )= -30

using formula

sin ( n \times 360 + \theta ) = sin \theta
= - \pi /6

Big mistake --- sin^-1 sin -30 ≠-30

Bcoz sin^-1 sin (x) = x only when -1 ≤ x ≤ 1

Hence , you have to convert x into a suitable y such that y lies between -1 to 1.

aN EXAMPLE ---- sin^-1 sin 10 = sin^-1 sin 3pi - 10

value of 3pi - 10 > -1 but < 1 .

hence sin^-1 sin 10 = 3pi - 10 , not 10.

Similarly , solve this one.

@ soumya sin-1 sin (x) = x only when -pi/2 ≤ x ≤ pi/2

And omkar ur input is in degrees but the actual qn is in radians,so writing 2010 in terms of sum of pie's will help us to decide the quadrant.

It is in radians..

Thanks nishant sir and soumya,msp Kastub and omkarsatis.