# How do you find the arcsin(sin((7pi)/6))?

Feb 10, 2015

$\arcsin \left(\sin \left(7 \frac{\pi}{6}\right)\right) = - \frac{\pi}{6}$.

The range of a function $\arcsin \left(x\right)$ is, by definition ,
$- \frac{\pi}{2} \le \arcsin \left(x\right) \le \frac{\pi}{2}$
It means that we have to find an angle $\alpha$ that lies between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ and whose $\sin \left(\alpha\right)$ equals to a $\sin \left(7 \frac{\pi}{6}\right)$.

From trigonometry we know that
$\sin \left(\phi + \pi\right) = - \sin \left(\phi\right)$
for any angle $\phi$.
This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle $\phi$ with the X-axis (counterclockwise from the X-axis to a radius).
We also know that sine is an odd function, that is
$\sin \left(- \phi\right) = - \sin \left(\phi\right)$.

We will use both properties as follows:
$\sin \left(7 \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6} + \pi\right) = - \sin \left(\frac{\pi}{6}\right) = \sin \left(- \frac{\pi}{6}\right)$

As we see, the angle $\alpha = - \frac{\pi}{6}$ fits our conditions. It is in the range from $- \frac{\pi}{2}$ to $\frac{\pi}{2}$ and its sine equals to $\sin \left(7 \frac{\pi}{6}\right)$. Therefore, it's a correct answer to a problem.