How do you evaluate ${sin}^{- 1} (sin (- \frac{π}{10}))$ $sin^-1(sin(-(pi)/10))$ ?

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How do you evaluate ${sin}^{- 1} (sin (- \frac{π}{10}))$ $sin^-1(sin(-(pi)/10))$ ?

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Answer 1 · 2016-10-13T21:43:38

${sin}^{- 1} (sin (- \frac{π}{10})) = - \frac{π}{10}$ $sin^-1 (sin(-pi/10))=-pi/10$

Explanation:

${sin}^{- 1} (sin (- \frac{π}{10}))$ $sin^-1 (sin(-pi/10))$

The restriction for the range of ${sin}^{- 1} x$ $sin^-1x$ is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ and since $- \frac{π}{10}$ $-pi/10$ is in the range we can apply the property $f^{- 1} (f (x) = x$ $f^-1(f(x)=x$ . That is,
${sin}^{- 1} (sin (- \frac{π}{10})) = - \frac{π}{10}$ $sin^-1 (sin(-pi/10))=-pi/10$

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How do you evaluate ${sin}^{- 1} (sin (- \frac{π}{10}))$ $sin^-1(sin(-(pi)/10))$ ?

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How do you evaluate ${sin}^{- 1} (sin (- \frac{π}{10}))$ $sin^-1(sin(-(pi)/10))$ ?