Proof of the tangent angle sum and difference identities

In this video, I'm going to assume that you already know a few things, and we've covered this. We've proved this in other videos that sine of x plus y is equal to sine of x cosine y plus, and then you swap the cosines and the sines: cosine of x sine y.

That cosine of x plus y is equal to cosine x cosine y minus sine x sine y. Once again, we've proven this in these in other videos. And then there's some other properties we know of cosine and sine that we have looked at in other videos: cosine of negative x is equal to cosine of x, and that sine of negative x is equal to negative sine of x; and that, of course, the tangent of something is defined as sine over cosine of that something.

Now, with that out of the way, I want to come up with a formula for tangent of x plus y expressed just in terms of tangent of x and tangent of y. You can view it as the analog for what we did up here for sine and cosine. Well, the immediate thing that you might recognize is that tangent of x plus y, based on the definition of tangent, is the same thing as sine of x plus y over cosine of x plus y.

And what's that going to be equal to? Well, we know that sine of x plus y can be expressed this way, so let me write that down. So that's going to be sine of x cosine y plus cosine of x sine of y. And then, actually, so that we can save a little bit of writing, I'm going to awkwardly make the line here down here because we're going to put something here in a second, but I think you'll get the idea.

So that's going to be that over cosine of x plus y, which is this expression when you just express it in terms of cosines of x and cosines of y and sines of x and sines of y. So let me write it here: so you're going to have cosine of x cosine y minus sine of x sine y.

Now we want to express everything in terms of tangents of x's and y's, and so it might make sense here to say, "All right, well, we know tangent is sine over cosine." So what if we were to divide both the numerator and the denominator by some expression that can start to make the numerator and denominator express in terms of tangents?

And I will cut a little bit to the chase here. So in the numerator, what I can do is—and I'm going to do this just in the numerator and then I'm going to do it in the denominator as well—I'm going to divide the numerator by cosine of x cosine y. And, of course, I can't just divide the numerator by cosine of x cosine y; that would change the value of this rational expression.

I have to do that to the denominator as well, so I know this is a very complex-looking fraction here, but it's going to simplify in a second. So I'm also going to divide the denominator by cosine of x cosine of y, and now let's see if we can simplify this in certain ways.

In the numerator, we can see that this cosine y cancels with this cosine y, and so the first term becomes—write in another color here—so this sine of x over cosine of x. And so in the numerator, I can say this is going to be equal to sine of x over cosine of x, which is tangent of x; and then this second term here, we can see that this cosine of x cancels with this cosine of x, and so we're left with sine of y over cosine of y, which is, of course, tangent of y.

So, plus tangent of y, and then all of that is going to be over—now we can look at the denominator. So this first term here, we can see the cosine of x cancels with the cosine of x, and the cosine of y cancels out with the cosine of y. So you could view this first term here, when you divide by this cosine of x cosine y, it just becomes one.

And then we're going to have the minus, and now this second term is interesting. We have sine of x over cosine of x, sine of y over cosine of y. So sine of x over cosine of x—that over there is tangent of x—and then sine of y over cosine of y—that's tangent of y. So this is going to be tangent of x times tangent of y.

And just like that, we have come up with an expression for tangent of x plus y that just deals with tangent of x's and tangent of y's. Now, the next question you might say, "Well, all right, that's great for tangent of x plus y, but what about tangent of x minus y?"

Well, here we just have to recognize a little bit of what we've seen before. Let me write it over here: tangent of negative x is equal to sine of negative x over cosine of negative x. And what's that going to be equal to? And I know I'm running out of space; this is going to be equal to sine of negative x is the same thing as negative sine of x. Negative sine of x; and then cosine of negative x is just cosine of x.

Well, this is just the negative of the tangent of x, so this is negative tangent of x. And the reason why that is useful is I can rewrite this as being—let me write it here—this is the same thing as tangent of x plus negative y. So everywhere we saw a y here, we can replace it with a negative y.

So this is going to be equal to tangent of x plus the tangent of negative y, all of that over 1 minus the tangent of x times the tangent of negative y. Well, we know the tangent of negative y is the same thing as the negative tangent of y, and we know that over here as well. So this could just be—we could write the tangent of y here, and then the negative would turn this into a plus.

And so just to write everything neatly, we know that also the tangent of x minus y can be rewritten as tangent of x minus tangent of y, all of that over 1 plus tangent of x and tangent of y.