How do you find the exact value of $arctan - \sqrt{3}$ $Arctan -sqrt3$ ?

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Answer 1 · 2016-02-06T00:20:49

$arctan (- \sqrt{3}) = - \frac{π}{3}$ $arctan(-sqrt3)=-pi/3$

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

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

$arctan (x)$ $arctan(x)$ and $tan (x)$ $tan(x)$ are inverse functions.

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

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

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

We see that:

So, back to the original question:

What is the value of $arctan (- \sqrt{3})$ $arctan(-sqrt3)$ ?

This is essentially asking, the tangent of what angle gives $- \sqrt{3}$ $mathbf(-sqrt3)$ ?

Since $tan (\frac{π}{3}) = \sqrt{3}$ $tan(pi/3)=sqrt3$ , we know that $tan (- \frac{π}{3}) = - \sqrt{3}$ $tan(-pi/3)=-sqrt3$ .

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

$arctan (- \sqrt{3}) = - \frac{π}{3}$ $arctan(-sqrt3)=-pi/3$

How do you find the exact value of arctan−√3 Arctan -sqrt3?