# How do you find the exact value of cos^-1(1/2)?

May 1, 2018

${\cos}^{- 1} \left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 n \pi$, $\forall n \in \mathbb{Z}$

#### Explanation:

Let $\theta = {\cos}^{- 1} \left(\frac{1}{2}\right)$. In other words,
$\cos \left(\theta\right) = \frac{1}{2}$
But we know that $\theta = \frac{\pi}{3}$ is a solution to this equality;
$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

Now, since $\cos$ is a periodic function with period $2 n \pi$, for integer $n$, we can rewrite this as:

$\cos \left(\frac{\pi}{3} + 2 n \pi\right) = \frac{1}{2}$
Or
$\frac{\pi}{3} + 2 n \pi = {\cos}^{- 1} \left(\frac{1}{2}\right)$, $\forall n \in \mathbb{Z}$.

May 1, 2018

$\frac{\pi}{3} + 2 k \pi$
$\frac{5 \pi}{3} + 2 k \pi$

#### Explanation:

Find arc x knowing cos x = 1/2
Trig table and unit circle give 2 solutions:
$x = \pm \frac{\pi}{3} + 2 k \pi$
Note that $\left(- \frac{\pi}{3}\right)$ is co-terminal to $\frac{5 \pi}{3}$.
Answers for $\left(0 , 2 \pi\right)$:
$\frac{\pi}{3} , \mathmr{and} \frac{5 \pi}{3}$
$x = \frac{\pi}{3} + 2 k \pi$
$x = \frac{5 \pi}{3} + 2 k \pi$