# How do you calculate sin^-1(-sqrt2/2)?

It is known that sin 45=$\frac{\sqrt{2}}{2}$. Hence, ${\sin}^{-} 1 \left(- \frac{\sqrt{2}}{2}\right)$ would be either in the IIIrd quadrant that is 180+45= 225 degrees or in the IVth quadrant, that is 360-45=315 degrees . These are the two values of the angle between 0 to $2 \pi$.