How do you calculate ${sin}^{- 1} (- \frac{\sqrt{2}}{2})$ $sin^-1(-sqrt2/2)$ ?

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How do you calculate ${sin}^{- 1} (- \frac{\sqrt{2}}{2})$ $sin^-1(-sqrt2/2)$ ?

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Answer 1 · 2015-04-30T15:31:48

225, 315 degrees

It is known that sin 45= $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ . Hence, ${sin}^{- 1} (- \frac{\sqrt{2}}{2})$ $sin^-1 (-sqrt(2)/2)$ 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 $2 π$ $2pi$ .

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How do you calculate ${sin}^{- 1} (- \frac{\sqrt{2}}{2})$ $sin^-1(-sqrt2/2)$ ?

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