How do you evaluate ${cos}^{- 1} (sin (\frac{5 π}{6}))$ $cos^-1(sin((5pi)/6))$ without a calculator?

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Answer 1 · 2016-09-15T09:44:31

${cos}^{- 1} (sin (\frac{5 π}{6})) = \frac{π}{3}$ $cos^(-1)(sin((5pi)/6))=pi/3$

We know that $sin (\frac{π}{2} + θ) = cos θ$ $sin(pi/2+theta)=costheta$

Hence $sin (\frac{5 π}{6}) = sin (\frac{3 π + 2 π}{6}) = sin (\frac{3 π}{6} + \frac{2 π}{6})$ $sin((5pi)/6)=sin((3pi+2pi)/6)=sin((3pi)/6+(2pi)/6)$

= $sin (\frac{π}{2} + \frac{π}{3}) = cos (\frac{π}{3}) = \frac{1}{2}$ $sin(pi/2+pi/3)=cos(pi/3)=1/2$

Hence ${cos}^{- 1} (sin (\frac{5 π}{6})) = {cos}^{- 1} (\frac{1}{2}) = \frac{π}{3}$ $cos^(-1)(sin((5pi)/6))=cos^(-1)(1/2)=pi/3$

How do you evaluate cos−1(sin(5π6))cos^-1(sin((5pi)/6)) without a calculator?