How do you solve the inverse trig function $sin ({sin}^{- 1} (\frac{1}{3}))$ $sin(sin^-1 (1/3))$ ?

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How do you solve the inverse trig function $sin ({sin}^{- 1} (\frac{1}{3}))$ $sin(sin^-1 (1/3))$ ?

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Answer 1 · 2015-05-25T16:58:58

${sin}^{- 1} (\frac{1}{3})$ $sin^-1(1/3)$ or $arcsin (\frac{1}{3})$ $arcsin(1/3)$ is the angle $θ$ $theta$ for which $sin (θ) = \frac{1}{3}$ $sin(theta) = 1/3$

Therefore $sin ({sin}^{- 1} (\frac{1}{3}))$ $sin(sin^-1(1/3))$
$= sin (θ)$ $= sin(theta)$ for the value of $θ$ $theta$ for which $sin (θ) = \frac{1}{3}$ $sin(theta) = 1/3$

That is
$sin ({sin}^{- 1} (\frac{1}{3})) = \frac{1}{3}$ $sin(sin^-1(1/3)) = 1/3$

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How do you solve the inverse trig function $sin ({sin}^{- 1} (\frac{1}{3}))$ $sin(sin^-1 (1/3))$ ?

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