# How can I prove that tan(angle) = opposite/adjacent

• May 7th 2011, 11:52 PM
jlzuri
How can I prove that tan(angle) = opposite/adjacent
Hello,

Given the following naming convention:

It doesn't seam trivial to me to prove that the length of the segment named "tan \theta " is in fact equal to opposite / adjacent, where opposite and adjacent refer to the sides of the triangle inside the circle.

Can any help with this?

Thanks
JL
• May 8th 2011, 12:38 AM
HappyJoe
What are you allowed to use here? For instance, are you allowed to use that tan is defined as sin/cos, and that sin is opposite/hypotenuse and cos is adjacent/hypotenuse?
• May 8th 2011, 02:49 PM
jlzuri
The title of the thread might be a little confusing. What I am trying to prove here is that the length of the segment named "tan \theta" is actually the same as the value of the trigonometric function tangent(angle).

By now I actually think I solved the problem though:

Given the tangent definition "tan = sin / cos" and known that sin = oppos / hyp and cos = adj / hyp -> tan = opposite / adjacent. Now my real question was "How do we know that the length of the segment is equal to oppos / adj" The answer is straight forward if we give a meaning to each side of the triangle. the adjacent side can be thought as the distance covered along the x axis, the opposite side is a difference of height between the ends of the hyp segment, and hypotenuse is therefore the slope which is deffined as height / distance.

Finally since the radius of the circle is 1 we can assert that the the length of the segment named "tan \theta" is actually equal to the trigonometric function tangent(\theta )

Does it make sense?

JL
• May 8th 2011, 10:37 PM
HappyJoe
Quote:

Originally Posted by jlzuri
Does it make sense?

Possibly. :) I can't see the pictures you've posted in the OP though, so not sure. Don't you also have this problem?
• May 8th 2011, 10:43 PM
Prove It
Using the unit circle, you explain that the right-angle triangle that is formed by the lengths $\displaystyle \sin{\theta}, \cos{\theta}$ is similar to the right-angle triangle formed by the lengths $\displaystyle \tan{\theta}, 1$.

That means $\displaystyle k\sin{\theta} = \tan{\theta}$ and $\displaystyle k\cos{\theta} = 1$, where $\displaystyle k$ is some constant.

From the second equation, we can see $\displaystyle k = \frac{1}{\cos{\theta}}$ and so $\displaystyle \frac{1}{\cos{\theta}}\sin{\theta} = \tan{\theta}$, or $\displaystyle \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$.

Now remembering that $\displaystyle \sin{\theta} = \frac{O}{H}$ and $\displaystyle \cos{\theta} = \frac{A}{H}$, that means

$\displaystyle \tan{\theta} = \frac{\frac{O}{H}}{\frac{A}{H}} = \frac{O}{A}$.

Q.E.D.
• May 10th 2011, 10:09 AM
HallsofIvy
I can't see the pictures but I suspect you have a unit circle (center at (0, 0), radius 1) with a line through the origin at angle $\theta$ and a vertical line tangent to that circle at (1, 0). Yes, the length of the segment from (1, 0) to the original line is $tan(\theta)$.

The point at which the line crosses the circle has coordinates $(cos(\theta), sin(\theta))$ (that's one common way of defining sine and cosine). That is, the slope of the line is $(sin(\theta)- 0)/(cos(\theta)- 0)= tan(\theta)$. But then, of course, the slope as calculated by using (1, y) (the point where the vertical line at (1, 0) crosses that line) and (0, 0) is $(y- 0)/(1- 0)= y= tan(\theta)$.