How do you evaluate ${sin}^{- 1} (sin (\frac{π}{7}))$ $sin^-1(sin(pi/7))$ ?

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How do you evaluate ${sin}^{- 1} (sin (\frac{π}{7}))$ $sin^-1(sin(pi/7))$ ?

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Answer 1 · 2016-10-13T16:06:34

${sin}^{- 1} (sin (\frac{π}{7})) = \frac{π}{7}$ $sin^-1(sin(pi/7))=pi/7$

Explanation:

Since $\frac{π}{7}$ $pi/7$ is within the restriction for arcsin which is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ we use the property $f^{- 1} (f (x) = x$ $f^-1(f(x)=x$ to evaluate and hence ${sin}^{- 1} (sin (\frac{π}{7})) = \frac{π}{7}$ $sin^-1(sin(pi/7))=pi/7$

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How do you evaluate ${sin}^{- 1} (sin (\frac{π}{7}))$ $sin^-1(sin(pi/7))$ ?

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