The value of $\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)$ is
The value of $\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)$ is
$\frac{8 \pi}{3}$
$\frac{\pi}{3}$
$\frac{2 \pi}{3}$
$\frac{3 \pi}{2}$
Solution
Step 1: Reduce $\frac{8 \pi}{3}$ within the principal range of cosine.
The principal range of $\cos ^{-1}$ is $[0, \pi]$, so we first reduce $\frac{8 \pi}{3}$ to an equivalent angle in the range $[0,2 \pi]$.
Since $2 \pi=\frac{6 \pi}{3}$, we subtract $2 \pi$ from $\frac{8 \pi}{3}$ :
$\frac{8 \pi}{3}-2 \pi=\frac{8 \pi}{3}-\frac{6 \pi}{3}=\frac{2 \pi}{3}$
Thus, $\cos \frac{8 \pi}{3}=\cos \frac{2 \pi}{3}$.
Step 2: Find the inverse cosine.
Since $\cos \frac{2 \pi}{3}=-\frac{1}{2}$ and the angle $\frac{2 \pi}{3}$ lies in the range $[0, \pi]$, we have:
$\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)=\frac{2 \pi}{3}$
Final Answer:
$\frac{2 \pi}{3}$