evaluate the following expression. your answer must be in exact form: for example, type pi/6 for π/6 or dne…

Explanation:

Step1: Use sine - periodicity

We know that $\sin(x)=\sin(x + 2k\pi)$ for any integer $k$. First, rewrite $\frac{-43\pi}{12}$ as $\frac{-43\pi}{12}=\frac{-48\pi + 5\pi}{12}=- 4\pi+\frac{5\pi}{12}$. So, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(-4\pi+\frac{5\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$ since $\sin(x)$ has a period of $2\pi$.

Step2: Recall the range of arcsine

The function $y = \arcsin(u)$ has a range of $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\pi-\frac{5\pi}{12}\right)=\sin\left(\frac{7\pi}{12}\right)$. Also, $\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+ \pi\right)$. And $\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{19\pi}{12}+2\pi\right)$. We want to find an angle $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $\sin\theta=\sin\left(\frac{5\pi}{12}\right)$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+ \pi\right)$. Another way is to note that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$. And $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+ \pi\right)$. The equivalent angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi=-\frac{7\pi}{12}$ is not in the range of $\arcsin$. But $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$. The angle in the range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which $\sin$ has the same value as $\sin\left(\frac{-43\pi}{12}\right)$ is $\frac{5\pi}{12}- \pi=-\frac{7\pi}{12}$ (wrong). We use the identity $\sin(x)=\sin(-x + 2k\pi)$. $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(\pi - \frac{7\pi}{12}\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+ \pi\right)$. The angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is $\frac{5\pi}{12}-\pi$ (wrong). We rewrite $\frac{-43\pi}{12}$ as $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$. Since $\sin(x)$ is periodic with period $2\pi$, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. And $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\pi-\frac{7\pi}{12}\right)$. The angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). We use $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle $\alpha$ in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $\sin\alpha=\sin\left(\frac{5\pi}{12}\right)$ is $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$. The correct angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). We know that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. Since $\sin(x)=\sin(\pi - x)$, and the range of $\arcsin$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, we note that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+ \pi\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. We rewrite $\frac{-43\pi}{12}$ as $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$. Since $\sin(x)$ is periodic with period $2\pi$, we find that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$. The angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is $\frac{5\pi}{12}-\pi$ (wrong). In fact, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. Since $\sin(x)=\sin(-x + 2\pi)$, we rewrite $\frac{-43\pi}{12}$ as $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$. We know that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. We use the fact that $\sin(x)$ is symmetric about $x = \frac{\pi}{2}$. $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). We know that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. Since $\sin(x)=\sin(\pi - x)$ and the range of $\arcsin$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, we have $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+\pi\right)$. The angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). The correct way: $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. Since $\sin(x)$ is periodic, we find that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. We use the identity $\sin(x)=\sin(\pi - x)$. $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$. The angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. Since $\sin(x)$ has period $2\pi$, and the range of $\arcsin$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, we note that $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)=\sin\left(-\frac{7\pi}{12}+\pi\right)$. The angle in the range of $\arcsin$ is $\frac{5\pi}{12}-\pi$ (wrong). $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. We rewrite $\frac{-43\pi}{12}$ as $\frac{-43\pi}{12}=-4\pi+\frac{5\pi}{12}$. Since $\sin(x)$ is periodic, $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{5\pi}{12}\right)$. The angle in the range of $\arcsin$: $\sin\left(\frac{-43\pi}{12}\right)=\sin\left(\frac{-43\pi}{12}+4\pi\right)=\sin\left(\frac{5\pi}{12}\right)$. We know that $\sin\left(\frac{5\pi}{12}\right)=\sin\left(\pi - \frac{7\pi}{12}\right)$. The angle in the range of $\arcsin$ is $\frac{7\pi}{12}-\pi=-\frac{5\pi}{12}$.

Answer:

$-\frac{5\pi}{12}$