How do you find the exact value of ${sin}^{- 1} (cos (\frac{π}{3}))$ $sin^-1(cos(pi/3))$ ?

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How do you find the exact value of ${sin}^{- 1} (cos (\frac{π}{3}))$ $sin^-1(cos(pi/3))$ ?

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Answer 1 · 2016-07-26T01:27:33

Use ${sin}^{- 1} sin x = x$ $sin^(-1)sin x = x$

Here, ${sin}^{- 1} cos (\frac{π}{3}) = {sin}^{- 1} sin (\frac{π}{2} - \frac{π}{3}) = \frac{π}{2} - \frac{π}{3} = \frac{π}{6}$ $sin^(-1)cos(pi/3)=sin^(-1)sin(pi/2-pi/3)=pi/2-pi/3=pi/6$ .

This is the conventional principal value.

The general value for the angle whose sine = $cos (\frac{π}{3}) = sin (\frac{π}{6})$ $cos (pi/3)=sin(pi/6)$ is

$npi+(-1)^npi/6, n = 0, +-1, +-2, +-3, ...$

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How do you find the exact value of ${sin}^{- 1} (cos (\frac{π}{3}))$ $sin^-1(cos(pi/3))$ ?

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