# How do you evaluate cos^-1(cos((19pi)/10))?

Aug 31, 2016

$\frac{19}{10} \pi$

#### Explanation:

Use that, if f is a single valued function,

${f}^{- 1} f \left(x\right) = x$

Here, it is ${\cos}^{- 1} \cos \left(\frac{19}{10} \pi\right) = \frac{19}{10} \pi$

Disambiguation note on uniqueness, despite the convention

that the inverse trigonometric function values are defined as

principal values:

If f is a single-valued function operator, in $y = f \left(x\right)$, the ordered and

coupled operation

$f {f}^{- 1} \left(y\right)$ returns y

and the ordered couple operation

${f}^{- 1} f \left(x\right)$ returns x.

Despite that f might not be bijective, being single-valued, it is

bijective in an infinitesimal neighborhood

$\left(x - \in , x + \in\right)$.

So there is unique answer for single-valued function under these

twin operations.

This note is important for applications wherein

$x = c$(time t), $t \in \left(- \infty , \infty\right)$ .