# How do you evaluate sec^-1(-sqrt2)?

Mar 6, 2018

$\textcolor{g r e e n}{\theta = \frac{3 \pi}{4} + 2 k \pi \to k \in \mathbb{Z}}$, $\textcolor{b l u e}{\theta = \frac{5 \pi}{4} + 2 k \pi \to k \in \mathbb{Z}}$

#### Explanation:

$\theta = {\sec}^{-} 1 \left(- \sqrt{2}\right)$

$\sec \theta = - \sqrt{2}$

$\sec \theta$ is negative in $I I , I I I$ quadrants.

Hence $\theta = \frac{3 \pi}{4} , \frac{5 \pi}{4}$

Since trigonometric function $\sec$ is periodic and repeats every $2 \pi$,

$\textcolor{g r e e n}{\theta = \frac{3 \pi}{4} + 2 k \pi \to k \in \mathbb{Z}}$, $\textcolor{b l u e}{\theta = \frac{5 \pi}{4} + 2 k \pi \to k \in \mathbb{Z}}$

Mar 6, 2018

$x = \frac{3 \pi}{4} + 2 k \pi$
$x = \frac{5 \pi}{4} + 2 k \pi$

#### Explanation:

${\sec}^{-} 1 \left(- \sqrt{2}\right)$
Find arccos x, that $\cos x = \frac{1}{\sec} = \frac{1}{-} \sqrt{2} = - \frac{\sqrt{2}}{2}$.
$\cos x = - \frac{\sqrt{2}}{2}$
Trig Table and unit circle give 2 general solutions:
$x = \pm \frac{3 \pi}{4} + 2 k \pi$
Reminder:
arc $x = - \frac{3 \pi}{4}$ is co-terminal to arc $x = \frac{5 \pi}{4}$.
$x = \frac{3 \pi}{4} + 2 k \pi$
$x = \frac{5 \pi}{4} + 2 k \pi$