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

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

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Answer 1 · 2016-08-25T18:34:34

${sin}^{- 1} (- \frac{\sqrt{2}}{2}) = - \frac{π}{4}$ $sin^-1(-sqrt2/2)=-pi/4$ .

Explanation:

The Defn. of ${sin}^{- 1}$ $sin^-1$ function is :

${sin}^{- 1} x = θ, | x | \leq 1 \Leftrightarrow sin θ = x, θ \in [- \frac{π}{2}, \frac{π}{2}]$ $sin^-1x=theta, |x|<=1 iff sin theta=x, theta in [-pi/2,pi/2]$ .

Knowing that, $sin (- \frac{π}{4}) = - sin (\frac{π}{4}) = - \frac{\sqrt{2}}{2}, &, - \frac{π}{4} \in [- \frac{π}{2}, \frac{π}{2}]$ $sin(-pi/4)=-sin(pi/4)=-sqrt2/2, &, -pi/4 in [-pi/2,pi/2]$ , we have,

${sin}^{- 1} (- \frac{\sqrt{2}}{2}) = - \frac{π}{4}$ $sin^-1(-sqrt2/2)=-pi/4$ .

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

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