# How do you find the exact value of sin^-1(sin((7pi)/6))?

Oct 19, 2016

$- \frac{\pi}{6}$

#### Explanation:

${\sin}^{-} 1 \left(\sin \left(\frac{7 \pi}{6}\right)\right)$

Start inside the parentheses by finding $\sin \left(\frac{7 \pi}{6}\right)$.

According to the unit circle at $\frac{7 \pi}{6}$,
the $y$ coordinate or sine of $\frac{7 \pi}{6}$ is equal to $- \frac{1}{2}$.

Next, substitute $- \frac{1}{2}$ into the original problem.

${\sin}^{-} 1 \left(- \frac{1}{2}\right)$

Recall that the range of ${\sin}^{-} 1$ is $- \frac{\pi}{2}$ to $\frac{\pi}{2}$.

If you are finding ${\sin}^{-} 1$ of a positive value, the answer will be in the first quadrant between $0$ and $\frac{\pi}{2}$.

If you are finding ${\sin}^{-} 1$ of a negative value, the answer will be in the fourth quadrant between $- \frac{\pi}{2}$ and $0$.

Again using the unit circle, the fourth quadrant angle with a sine of $- \frac{1}{2}$ is $\frac{11 \pi}{6}$. But this is NOT the answer! Because of the restriction on the range, you need to find an angle between $- \frac{\pi}{2}$ and $0$. The angle is then $- \frac{\pi}{6}$.

Also note that ${\sin}^{-} 1 \left(\sin x\right)$ does not automatically "cancel out" and yield x.