Write the value of `sin^-1(sin1550^@)`

Explanation

`1550^@ = 1550 xx pi/180`

`= (155 pi)/18`

`=> sin^-1(sin ((155 pi)/18))`

`= sin^-1 (sin (8 pi + (11 pi)/18))`

`=> sin^-1(sin((11pi)/18))`

`= sin^-1[sin(pi- (7pi)/18)]`

`=> sin^-1(sin((7pi)/18))= (7pi)/18`

Answer