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

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Answer 1 · 2016-12-16T06:56:42

$sin ({sin}^{- 1} (\frac{1}{2})) = \frac{1}{2}$ $sin(sin^(-1)(1/2))=1/2$

${sin}^{- 1} (\frac{1}{2})$ $sin^(-1)(1/2)$ means an angle $θ$ $theta$ so that $sin θ = \frac{1}{2}$ $sintheta=1/2$

As we have $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$ , ${sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$ $sin^(-1)(1/2)=pi/6$

and $sin ({sin}^{- 1} (\frac{1}{2})) = sin (\frac{π}{6}) = \frac{1}{2}$ $sin(sin^(-1)(1/2))=sin(pi/6)=1/2$

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