Question

How do you evaluate `arc cos(cos (5/4pi))`?

Answer 1

`arccos(cos((5pi)/4))=arccos(cos((3pi)/4))=(3pi)/4`

Explanation:

`arccos(cos((5pi)/4))`

Since the restriction for arccos is `[0,pi]` we see that the argument is in quadrant III so we need to find the reference angle which is `pi/4`. And since cosine is negative in quadrant three it means that our argument will be in quadrant two from the restriction. So the argument x in quadrant two is `pi-pi/4=(3pi)/4`

Now using the property `f^-1(f(x))=x` we have `arccos(cos((5pi)/4))=arccos(cos((3pi)/4))=(3pi)/4`