# How do you evaluate cos^-1 (cos ((17pi)/6))?

Sep 16, 2017

$\frac{17 \pi}{6}$

#### Explanation:

The arc cosine function, often abbreviated arccos or ${\cos}^{-} 1$, is the inverse of the cosine function. It is defined in such a way that ${\cos}^{-} 1 \left(\cos \left(x\right)\right) = x$. Thus, ${\cos}^{-} 1 \left(\cos \left(\frac{17 \pi}{6}\right)\right) = \frac{17 \pi}{6}$

Sep 17, 2017

$\frac{5 \pi}{6}$

#### Explanation:

${\cos}^{-} 1 x \text{ is defined as the angle "theta" such that } 0 \le \theta \le \pi$

$\Rightarrow {\cos}^{-} 1 \left(\cos \theta\right) = \theta$

$\Rightarrow {\cos}^{-} 1 \left(\cos \left(\frac{17 \pi}{6}\right)\right)$

$= {\cos}^{-} 1 \left(\cos \left(\frac{5 \pi}{6}\right)\right) = \frac{5 \pi}{6}$