How do you find the exact value of ${cos}^{- 1} (cos (\frac{π}{7}))$ $cos^-1(cos(pi/7))$ ?

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Answer 1 · 2017-02-02T17:10:38

${cos}^{- 1} (cos (\frac{π}{7})) = \frac{π}{7}$ $cos^(-1)(cos(pi/7))=pi/7$

As per definition of ${cos}^{- 1}$ $cos^(-1)$ , if $cos a = x$ $cosa=x$ , $a = {cos}^{- 1} x$ $a=cos^(-1)x$

In the given problem, let $cos (\frac{π}{7}) = v$ $cos(pi/7)=v$

then as per definition ${cos}^{- 1} v = \frac{π}{7}$ $cos^(-1)v=pi/7$ .................(1)

and putting value of $v = cos (\frac{π}{7})$ $v=cos(pi/7)$ in (1), we get

${cos}^{- 1} (cos (\frac{π}{7})) = \frac{π}{7}$ $cos^(-1)(cos(pi/7))=pi/7$

How do you find the exact value of cos^-1(cos(pi/7))cos−1(cos(π7))cos^-1(cos(pi/7))?