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

Oct 12, 2016

${\cos}^{-} 1 \left(\cos \left(\frac{9 \pi}{8}\right)\right) = {\cos}^{-} 1 \left(\cos \left(\frac{7 \pi}{8}\right)\right) = \frac{7 \pi}{8}$

#### Explanation:

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

${\cos}^{-} 1 \left(\cos \left(\frac{9 \pi}{8}\right)\right) = {\cos}^{-} 1 \left(\cos \left(\frac{7 \pi}{8}\right)\right) = \frac{7 \pi}{8}$