# How do you evaluate  arccos(-1/sqrt2)?

Apr 15, 2016

As the range is not specified, the general value is given. The principal value of $a r c \cos \left(- \frac{1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.
The general value is $2 k \pi \pm \frac{3 \pi}{4}$, k = 0. 1. 2. 3.,...

#### Explanation:

Cosine is positive in the 1st quadrant [0, pi/2] and negative in the second $\left[\frac{\pi}{2} , \pi\right]$

The principal value for arc cos is defined to be in $\left[0 , \pi\right]$.

The general value is $2 k \pi \pm$(principal value), k = 0. 1. 2. 3.,...

Here, the principal value of $a r c \cos \left(- \frac{1}{\sqrt{2}}\right) = \frac{3 \pi}{4}$.