# Inverse Trig Functions

• September 20th 2009, 12:49 PM
horsepower.850
Inverse Trig Functions
I'm having trouble understanding this:
tan[arc cos (square root of 3 / 2) = square root of 3 / 3
How does this answer come to be?
• September 20th 2009, 01:17 PM
artvandalay11
This question is saying what is the Tangent of the angle who's cosine is $\frac{\sqrt{3}}{2}$

So what is the angle who's cosine equals that. Then take the tangent of your answer....

But even more cleverly, $\tan x=\frac{\sin x}{\cos x}$

When cosine equals $\frac{\sqrt{3}}{2}$, sin = $\frac{1}{2}$

So now compute sin over cos
• September 20th 2009, 10:08 PM
macosxnerd101
RE: Inverse Trig Functions
Let's look at the unit circle: File:Unit circle angles.svg - Wikipedia, the free encyclopedia

As artvandalay said, the first thing the question is asking for is the angle whose cosine is sqrt(3)/2. Knowing that the unit circle coordinates are (cos x, sin x), we can determine that the angle measure whose cosine is sqrt(3)/2 is pi/3. Now from here, the question is asking for the tangent of this angle, which is the sine value over the cosine value. The sine value here is 1/2, and we know the cosine value is sqrt(3)/2. So:

tan pi/3 = (1/2)/(sqrt(3)/2)
The twos cancel out, leaving 1/sqrt(3). When we rationalize the function, we get sqrt(3)/3 (multiply the numerator and denominator by sqrt(3) to remove the radical).

Hope this helps some.