question evaluate the expression below (for angles measured in radians). arccos (sin (4π/3))

Explanation:

Step1: Find the value of $\sin(\frac{4\pi}{3})$

We know that $\sin(A + B)=\sin A\cos B+\cos A\sin B$. Also, $\frac{4\pi}{3}=\pi+\frac{\pi}{3}$. So, $\sin(\frac{4\pi}{3})=\sin(\pi + \frac{\pi}{3})=-\sin(\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$.

Step2: Evaluate $\arccos(-\frac{\sqrt{3}}{2})$

The range of the inverse - cosine function $y = \arccos(x)$ is $[0,\pi]$. We know that $\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}$ and $\frac{5\pi}{6}\in[0,\pi]$. So, $\arccos(-\frac{\sqrt{3}}{2})=\frac{5\pi}{6}$.

Answer:

$\frac{5\pi}{6}$