Question

How do you evaluate `cos^-1(cos ((9pi)/8))`?

Answer 1

`cos^-1(cos((9pi)/8))=cos^-1(cos((7pi)/8))=(7pi)/8`

Explanation:

Since the restriction for `cos^-1x` is `[0,pi], (9pi)/8` is in quadrant three and cosine is negative in quadrant three. So we need to find the reference angle which is `(9pi)/8-pi=pi/8`. Hence the argument x in quadrant II is`(7pi)/8` since cosine is negative in quadrant two from the restriction. Lastly, use the property `f^-1(f(x)=x` so

`cos^-1(cos((9pi)/8))=cos^-1(cos((7pi)/8))=(7pi)/8`