Using the tangent angle addition identity | Trigonometry | Precalculus | Khan Academy

In this video, we're going to try to compute what tangent of 13 pi over 12 is without using a calculator. But I will give you a few hints.

First of all, you can rewrite tangent of 13 pi over 12 as tangent of... instead of 13 pi over 12, we can express that in terms of angles where we might be able to figure out the tangents just based on other things we know about the unit circle.

13 pi over 12 is the same thing as 15 pi over 12 minus 2 pi over 12, which is the same thing as the tangent of 5 pi over 4 minus pi over 6. Or we could even view it as plus negative pi over 6. So that's my hint right over there. Pause this video and see if you can keep going with this train of reasoning to evaluate what tangent of 13 pi over 12 is without using a calculator.

All right, now let's keep on going together. Well, we already know what the tangent of the sum of two angles are; we've proven that in another video. We know that this is going to be equal to the tangent of the first of these angles, 5 pi over 4, plus the tangent of the second angle, tangent of negative pi over 6. All of that is going to be over 1 minus tangent of the first angle, 5 pi over 4, times the tangent of the second angle, negative pi over 6.

And so now we can break out our unit circles to figure out what these things are. So I have pre-put some unit circles here, and so let's first think about what 5 pi over 4 looks like. Pi over 4, you might already associate it with 45 degrees; that's pi over 4 right over there. 2 pi over 4 would get you here, 3 pi over 4 would get you there, 4 pi over 4, which is the same thing as pi, gets us over there. 5 pi over 4 would get us right about there.

Now you might already recognize the tangent of an angle as the slope of the radius, and so this you might already be able to intuit that the tangent here is going to be one. But we can also break out our knowledge of triangles and the unit circle to figure this out if you didn't realize that. So what we need to do is figure out the coordinates of that point right over there. And to help us do that, we can set up a little bit of a right triangle, which you might immediately recognize as a 45-45-90 triangle.

How do I know that? Well, 5 pi over 4, remember we go 4 pi over 4 to get here, then we have one more pi over 4 to go down here. So this angle right over here is pi over 4, or you could view it as 45 degrees. And of course, if that's 45 degrees, that's 90, then this has to be 45 degrees because they all add up to 180 degrees.

We know a triangle like this by the Pythagorean theorem. If our hypotenuse is 1, each of the other two sides is square root of 2 over 2 times the hypotenuse. So this is square root of 2 over 2, and then this is square root of 2 over 2. Now if we think about the coordinates, our x coordinate is square root of 2 over 2 in the negative direction, so our x coordinate is negative square root of 2 over 2, and our y coordinate is square root of 2 over 2 going down, so that's also negative square root of 2 over 2.

And the tangent is just the y-coordinate over the x-coordinate here, so the tangent is just going to be negative square root of 2 over 2 over negative square root of 2 over 2, which is once again 1, which was our intuition. So we can write that tangent of 5 pi over 4 is equal to 1.

And then what about negative pi over 6? Well, negative pi over 6, you might recognize pi over 6 as being a 30-degree angle. Pi is 180 degrees, so divided by 6 is 30 degrees, and negative pi over 6 would be going 30 degrees below the positive x-axis, so it would look just like that. As we said, this angle right over here is pi over 6, which you can also view as a 30-degree angle.

If we were to drop a perpendicular right over here, you might immediately recognize this as a 30-60-90 triangle. We know if the hypotenuse is of length 1, the side opposite the 30-degree side is one half the hypotenuse, and then the longer non-hypotenuse side is going to be square root of three times the shorter side, so square root of three over two.

And so our coordinates right over here, we are moving square root of 3 over 2 in the positive x-direction, square root of 3 over 2, and then we are going negative one half in the y-direction, so we put negative one half right over there. And so now we know that the tangent of negative pi over 6 is going to be equal to negative one-half over square root of 3 over 2, which is the same thing as negative one-half times... let me write this way: negative one-half times two over the square root of three, which is equal to negative one over the square root of three.

This right over here is one we saw that there, this right over here is 1, and then this right over here is negative 1 over the square root of 3, and then this is negative 1 over the square root of 3.

And so I can rewrite this entire expression as being equal to 1 minus 1 over the square root of 3, all of that over 1. And then I have a negative here, negative here, so then that becomes a negative times a negative, so positive 1 plus 1 over the square root of 3.

And if we multiply both the numerator and the denominator by square root of 3, what we're going to get in the numerator is square root of 3 minus 1, and then our denominator is going to be square root of 3 plus 1. And we are done, that's tangent of 13 pi over 12 without using a calculator.