question evaluate the expression below (for angles measured in radians). arcsin(sin(-5π/3)) answer attempt 1…

Explanation:

Step1: Find the value of $\sin(-\frac{5\pi}{3})$

We know that $\sin(-\alpha)=-\sin\alpha$ and $\sin(x + 2k\pi)=\sin x,k\in\mathbb{Z}$. So, $\sin(-\frac{5\pi}{3})=-\sin\frac{5\pi}{3}=-\sin(2\pi-\frac{\pi}{3})$. Since $\sin(2\pi - \theta)=-\sin\theta$, then $-\sin(2\pi-\frac{\pi}{3})=-(-\sin\frac{\pi}{3})=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$.

Step2: Evaluate $\arcsin(\sin(-\frac{5\pi}{3}))$

We found that $\sin(-\frac{5\pi}{3})=\frac{\sqrt{3}}{2}$, and we need to find $\arcsin(\frac{\sqrt{3}}{2})$. The range of the inverse - sine function $y = \arcsin x$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$. And $\arcsin(\frac{\sqrt{3}}{2})=\frac{\pi}{3}$ because $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\frac{\pi}{3}\in[-\frac{\pi}{2},\frac{\pi}{2}]$.

Answer:

$\frac{\pi}{3}$