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

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Answer 1 · 2016-07-27T03:11:17

$\frac{π}{3}$ $pi/3$

In the first quadrant, $sin (\frac{π}{6}) = cos (\frac{π}{2} - \frac{π}{6}) = cos (\frac{π}{3})$ $sin (pi/6) = cos(pi/2-pi/6)=cos(pi/3)$

The given expression = ${cos}^{- 1} cos (\frac{π}{3}) = \frac{π}{3}$ $cos^(-1)cos(pi/3)=pi/3$ .

Note: For general values also, $sin a = cos (\frac{π}{2} - a)$ $sin a = cos (pi/2-a)$ .

How do you find the exact value of cos^-1(sin(pi/6))cos−1(sin(π6))cos^-1(sin(pi/6))?