# Math Help - Evaluate cos(arctan (-1))

1. ## Evaluate cos(arctan (-1))

so i know that you start with the arctan (-1) .. and for that you just do inverse (tan (-1)) right?
i entered that in to my calcultor and got -.7853981634
and when i did cos(ans) i got .7071067812
but my teacher wants it in exact value? how do i get that?

2. Originally Posted by DarkAngelxXxXx
so i know that you start with the arctan (-1) .. and for that you just do inverse (tan (-1)) right?
i entered that in to my calcultor and got -.7853981634
and when i did cos(ans) i got .7071067812
but my teacher wants it in exact value? how do i get that?
Hi DarkAngelxXxXx,

For angles ranging from 0 to 360 degrees, tan(angle) gives the slope of a line passing through the origin.

Hence, we can represent it with 2 angles.

$tan^{-1}(-1)$ calculates the angle of a line with slope -1.

This is a line going through the origin at 45 degrees downward from left to right.

Hence, it's angle is 90+45=135 or -45 degrees.

You used radian mode on your calculator and you worked with the -45 degree angle.

It helps to understand that cos(angle) gives a horizontal co-ordinate in the unit-radius circle, centre (0,0).

Now, for this line, which co-ordinate will you find ?
Because, in the second quadrant, where the line cuts the circle,
the horizontal co-ordinate is negative.

For the angle you worked with (in the 4th quadrant), the horizontal co-ordinate is positive.

The one you quoted is correct, the other is the negative of that.

I'm not sure what you mean by "your teacher wants it in exact value". Is it surd form?
In that case, use the surd forms for

$cos\left(\frac{{\pi}}{4}\right),\ -cos\left(\frac{{\pi}}{4}\right)$

3. Originally Posted by DarkAngelxXxXx
so i know that you start with the arctan (-1) .. and for that you just do inverse (tan (-1)) right?
i entered that in to my calcultor and got -.7853981634
and when i did cos(ans) i got .7071067812
but my teacher wants it in exact value? how do i get that?
you are looking for $\frac{\sqrt{2}}{2}$

4. Here's a slightly different and more general method: Let $\theta= arctan(x)$. Then $x= tan(\theta)$. Represent that as a right triangle having one angle of measure $\theta$, opposite side of length x, and near side of length 1 so that $tan(\theta)= x/1$= "opposite side/near side". By the Pythagorean theorem the hypotenuse has length $\sqrt{x^2+ 1}$ and so $cos(\theta)= cos(arctan(x))$ is "near side over hypotenuse" or $cos(arctan(x))= \frac{1}{\sqrt{x^2+ 1}}$.

When x= -1, that is, of course, $\frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$.