How do you find ${sin}^{- 1} (cos (\frac{π}{6}))$ $sin^-1 (cos(pi/6))$ ?

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How do you find ${sin}^{- 1} (cos (\frac{π}{6}))$ $sin^-1 (cos(pi/6))$ ?

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Answer 1 · 2018-04-18T04:58:36

${sin}^{- 1} (cos (\frac{π}{6})) = \frac{π}{3} or 60^{\circ}$ $color(blue)(sin^-1 (cos (pi/6)) = pi/3 " or " 60^@$

Explanation:

![www.math-only-math.com/images/http://trigonometrical-ratios-table.png](https://useruploads.socratic.org/Rod2GhDAR8qhxzUwIaKk_trigonometry%20values.png)

$Let θ = {sin}^{- 1} (cos (\frac{π}{6}))$ $"Let " theta = sin^-1 (cos (pi/6))$

From the table above, $cos (\frac{π}{6}) = (\frac{\sqrt{3}}{2})$ $cos (pi/6) = (sqrt3/2)$

$:. theta = sin^-1 (sqrt3/2)$

$sin theta = sqrt3/2$

But $sin 60 = sqrt3/2$ as seen in the table.

Hence $color(blue)(theta = sin ^-1 (sin (60)) = 60^@ " or " pi/3$

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How do you find ${sin}^{- 1} (cos (\frac{π}{6}))$ $sin^-1 (cos(pi/6))$ ?

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