How do you find the exact value of ${sin}^{- 1} (\frac{\sqrt{2}}{2})$ $sin^-1 (sqrt2/2)$ ?

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How do you find the exact value of ${sin}^{- 1} (\frac{\sqrt{2}}{2})$ $sin^-1 (sqrt2/2)$ ?

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Answer 1 · 2017-04-14T11:19:15

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

Explanation:

Let ${sin}^{- 1} (\frac{\sqrt{2}}{2}) = θ$ $sin^-1 (sqrt2/2) = theta$ , then $sin θ = \frac{\sqrt{2}}{2}$ $sin theta = sqrt2/2$ . We know $sin (pi/4) = sqrt2/2 :. theta = pi/4 or sin^-1 (sqrt2/2) = pi/4$ [Ans]

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How do you find the exact value of ${sin}^{- 1} (\frac{\sqrt{2}}{2})$ $sin^-1 (sqrt2/2)$ ?

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How do you find the exact value of ${sin}^{- 1} (\frac{\sqrt{2}}{2})$ $sin^-1 (sqrt2/2)$ ?