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# The Tangent Function - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I want to complete a table of values for the tangent function, tangent theta. I have some angles here, they're all angles in the first quadrant so I want to focus on those today. Remember the unit circle definition for tangent. Tangent theta equals y over x where theta is the angle, say this angle, and y over x, y and x come from the coordinates of this point.

Now there are these five angles in the first quadrant that you have to be really familiar with and the nice thing about them is they all have coordinates that come from these set of numbers. For example this point has coordinates 1,0 and this point has coordinates 0,1 and for the rest of them you just have to think about the order of these numbers. They're actually written in order from smallest to largest.

So as we work our way up, the x values go down, so this will have an x value of root 3 over 2, root 2 over 2 and a one-half. Also as we go up, the y values increase, so starting at zero then going to a half, root 2 over 2 and root 3 over 2 and that gives us all five of our special points in the first quadrant. With these points you can find the sine, cosine, tangent of any angle in this quadrant.

Back to our definition; tangent theta equals y over x, so let's look at this point. This point represents the tangent of 0. The tangent of 0 would be 0 over 1 or 0. Pi over 6 is represented by this point. Tangent of pi over 6 would be one-half divided by root 3 over 2. One-half divided by root 3 over 2 that's the same as one-half times 2 over root 3 which is 1 over root 3 and if we rationalize the denominator that's the same as root 3 over 3.

So the tangent of pi over 6 is root 3 over 3. This one is pretty easy the numbers aren't so nice, but the tangent of this angle is going to be 1 because the x and y coordinates are exactly the same, this angle is pi over 4 or 45 degrees. So the tangent of pi over 4 is 1, y over x. This angle is pi over 3, the tangent of this angle is going to be root 3 over 2 divided by one-half, root 3 over 2 divided by one-half that's the same as root over 2 times 2 which is root 3, so the tangent of pi over 3 is root 3.

Finally pi over 2, it's 90 degrees, the tangent would be 1 over 0 that's undefined, and so it turns out that the tangent is undefined at pi over 2.

These are the basic values of tangent and we'll use these whenever we're trying to find the tangent of any special angle using reference angles in a later episode, but tangent of 0 was 0, tangent of pi over 6 is root 3 over 3, tangent of pi over 4 is 1, tangent of pi over 3 is root 3 and very important tangent of pi over 2 is undefined.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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