# How do you find the exact value of  Arctan -sqrt3?

Feb 6, 2016

$\arctan \left(- \sqrt{3}\right) = - \frac{\pi}{3}$

#### Explanation:

First, recognize that the domain of the function $\arctan \left(x\right)$ is $- \frac{\pi}{2} < x < \frac{\pi}{2}$.

Now, a little bit about what $\arctan \left(x\right)$ means:

$\arctan \left(x\right)$ and $\tan \left(x\right)$ are inverse functions.

This means that mathbf(arctan(tan(x))=x and mathbf(tan(arctan(x))=x.

We can see tangent and arctangent as "undoing" one another. Another way we can see $\arctan \left(x\right)$ is as $\tan \left(x\right)$ in reverse.

We know that $\tan \left(\frac{\pi}{4}\right) = 1$. This means that $\arctan \left(1\right) = \frac{\pi}{4}$.

We see that:

• in $\tan \left(x\right)$, $x$ is an angle
• in $\arctan \left(x\right)$, $x$ is the value of the tangent function

So, back to the original question:

What is the value of $\arctan \left(- \sqrt{3}\right)$?

This is essentially asking, the tangent of what angle gives $m a t h b f \left(- \sqrt{3}\right)$ ?

Since $\tan \left(\frac{\pi}{3}\right) = \sqrt{3}$, we know that $\tan \left(- \frac{\pi}{3}\right) = - \sqrt{3}$.

We can reverse this with the $\arctan$ function to see that

$\arctan \left(- \sqrt{3}\right) = - \frac{\pi}{3}$