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

$\arccos \left(\cos \left(\frac{5 \pi}{4}\right)\right) = \arccos \left(\cos \left(\frac{3 \pi}{4}\right)\right) = \frac{3 \pi}{4}$
$\arccos \left(\cos \left(\frac{5 \pi}{4}\right)\right)$
Since the restriction for arccos is $\left[0 , \pi\right]$ we see that the argument is in quadrant III so we need to find the reference angle which is $\frac{\pi}{4}$. 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 $\pi - \frac{\pi}{4} = \frac{3 \pi}{4}$
Now using the property ${f}^{-} 1 \left(f \left(x\right)\right) = x$ we have $\arccos \left(\cos \left(\frac{5 \pi}{4}\right)\right) = \arccos \left(\cos \left(\frac{3 \pi}{4}\right)\right) = \frac{3 \pi}{4}$