 In this video, I want to talk about the sum and difference angle formulas for tangent. We've previously talked about the angle sum and angle difference identities for sine and cosine. And given that tangent, of course, tangent of x is just sine of x over cosine of x. We're going to use that to actually help us compute an angle sum and angle difference formula for tangent. So let's look at the angle sum identity. We get tangent of a plus b is equal to tangent of a plus tangent of b over 1 minus tangent of a times tangent of b. And so to prove this identity, we're going to take the left-hand side, tangent of a plus b. All right? Well, tangent, like we saw a moment ago by the ratio identity, is just sine over cosine. So tangent of a plus b becomes sine of a plus b over cosine of a plus b. And now we're going to apply the angle sum identities for sine and cosine. For which sine, because sine lives in an ideal kumbaya utopian society, you're going to end up with sine of a cosine of b plus you're going to get cosine of a sine of b. So you'll notice that sines and cosines live together in harmony, and there's no preference to one product over the other. Sine, of course, lives as a dystopian despot of some kind, right? Cosine of a plus b becomes cosine of a cosine of b minus sine of a sine of b, which you'll notice what's dystopian about cosine here. The cosines only live together and the sines only live together, so the trig functions are segregated, right? And then there is preference here placed, right? There's a negative sign on the sine functions. Oh, no, that's horrible. So this is the expanded form for sine of a plus b and sine of cosine of a plus b. What we're going to do is we're going to times the top and the bottom of this fraction by cosine of a cosine of b, and then we have to do that to the denominator as well, one over cosine of a cosine of b. And so we distribute these onto the appropriate parts like so. So in bread here is just a strategic number one. So look at the first product, sine of a times cosine of b. What would happen would be when you distribute that fraction is the cosine of b's would cancel out and you're left with a sine of a over a cosine of a. And which case that gives us sine of a over cosine of a. Let's do the next one. On the next group, you see that the cosines of a would cancel out. You have a sine of a on top, you have a cosine of b on the bottom. And so we end up with sine of b over cosine of b. That is then the numerator. In the denominator, we'll get something similar when you distribute the one over cosine a and cosine b with the cosine a and b here, the cosines of a will cancel. The cosines of b will cancel and you're left with just a single number one. And then you're going to subtract from that. On the next one, there's no cancellation, sine a, sine b, cosine a, cosine b, no cancellation there. So let me just write it in the following manner. You're going to get sine of a over cosine of a and then you're going to get sine of b over cosine of b. And perhaps you can see how we organize things. We have a bunch of sine of, we have some sines a over cosines a and then sines b over cosines b. This will then simplify to b tangent of a plus tangent of b over one minus tangent of a tangent of b, which then gives us the right hand side, which establishes the identity we wanted. This gives us the angle sum identity. So the advantage of this angle sum identity is that we can relate the tangent of the sum of angles to other tangents. So you can relate the tangent of a plus b to the tangent of a and the tangent of b in some type of combination. I should mention that the proof for tangent of a minus b is very similar. Suppose what happens is that the signs change. So when you have plus versus minus, in the original you have a plus on top, a minus on the bottom. In the minus version you're going to have a minus on the top and a plus on the bottom. You just switch the signs. This plus became a minus and this minus became a plus. This is something we saw a lot when we talked about the angle sum identities for sine and cosine as we transition to the angle difference identities. And that's going to because the reason you get that is that as you change the signs, this plus will become a minus and this minus becomes a plus. So I won't do the details of the angle difference. I'll leave it up to the viewer to finish that. So could we do tangent of 75 degrees? The answer is yep, we could do it because tangent of 75 degrees, we're going to treat it as tangent of 45 degrees plus 30 degrees. And using the angle sum identity from the previous, from the previous screen, we're going to get tangent of 45 degrees. We're then going to add to that tangent of 30 degrees. This will sit above one minus tangent, tangent of 45 degrees and then tangent of 30 degrees like so. In which case let's compute this. Well tangent is sine over cosine. So at 45 degrees sine and cosine are both they're both equal to root two over two. So tangent of 45 degrees is going to equal one. So I'm going to put that into my formula here one minus one times tangent of 30 degrees. Well at tangent of 30 degrees tangent at 30, excuse me, sine of 30 degrees is one half. So let's put this as an aside, you have tangent of 30 degrees. This is equal to sine of 30 degrees over cosine of 30 degrees. Sine of 30 degrees is one half, cosine is root three over two, if you multiply by the reciprocal, you get one half times two over root three, the two's cancel and you're going to end up with just the one over the square root of three. You could rationalize that if you want to and maybe I will but I'll do that. I'll worry about that later one over square root of three like so. Well I don't have a problem with a square root being in the denominator. I do have a problem with fractions being set of fractions. So I am actually going to rationalize this but mostly by clearing the denominators. So if you distribute the square root of three through, you're going to end up with the square root of three times one square root of three. One over root three times root three is just going to be one. So you get the square root of three plus one in the denominator. You're going to get the square root of three because that's one times square root of three. Then you're going to get the fraction that cancels out the square root of three so you get a minus one. So it turns out that the tangent to 75 degrees is gonna be the square root of three plus one over the square root of three minus one, all right? I wanna compare, how could we have done this alternatively, right? It's also true that tangent of 75 degrees, this is equal to sine of 75 degrees over cosine of 75 degrees, which we've computed these previously. So sine of 75 degrees, we got as the square root of six plus the square root of two over four. Cosine of 75 degrees, we got as the square root of six minus the square root of two over four, which if you multiply by the reciprocal here, you get root six plus root two over four times four over the root six minus root two. The square root of fours will cancel, or excuse me, just the fours will cancel, and you end up with the value square root of six plus the square root of two over the square root of six minus square root of two. Now you compare those, it's like, those don't look like the same answer. Turns out they are equivalent to each other. On the top, you can actually factor out a square root of three, excuse me, the square root of two, there you go. That'll leave behind the square root of three plus one on top. You can factor out a square root of two on the bottom, that'll leave behind the square root of three minus one, in which case then you see the square root of two cancel out, and then it becomes much more decided that the two values are in fact one of the same thing. So it looks a little bit different, but the answers do turn out to be the same. So how do we do this? Basically, in this second approach, what we did is we recognized that tangent was sine over cosine, and then we used the angle sum identities to compute sine of 75 and angle sum identities to do the cosine of 75 degrees. So we just used the angle sum identities for sine and cosine, and combined that with tangent. In the first approach, we actually used the angle sum identity for tangent, which was derived from the angle sum identities first from sine and cosine. So when you wanna compute the angle sum of tangent or the angle difference, you don't actually need the tangent formula for sums of differences, because you can't just get it from sine and cosine recognizing the tangent to sine over cosine. So whatever approach you want is appropriate. It just kind of depends on what's the easiest thing to do. Since we already knew sine and cosine for 75 degrees, it wasn't so hard to do it directly, but if you didn't know sine or cosine for the angle, but you didn't know tangent, then the angle sum identity for tangent is very, very useful. Let me give you another example to illustrate this. Suppose we have two angles, A and B, which terminate in the first and third quadrant, respectively. And suppose we know that sine of A is equal to three-fifths, and suppose we know that cosine of B is equal to negative 513ths. Let's use this information to find sine of A plus B and cosine of A plus B and tangent of A plus B, okay? How are we gonna do that? Well, we know sine of A, what's cosine of A in that situation? Well, if we think about our right triangle diagrams, let's think of a diagram associated to angle A right here. Well, if sine is three over five, that's opposite over a hypotenuse, what's the adjacent side? By the Pythagorean equation, we can find out very quickly that the other side's four, right, this is a three, four, five triangle. Since we're in the first quadrant, the X coordinates should be positive, so this would be three, this is three, four, five. So we get cosine of A would be four-fifths, like so. We could also very easily see that tangent of A would be equal to three-fourths in the situation, all right? Well, what about, well, since we have cosine of B, what about sine of B? Could we do that one? Sine of B, well, we need to think of the triangle in that situation, right? So we get something like this. Now notice I'm using the reference angle of B and I drew the triangle upside down to indicate that we're in the third quadrant right now, but using this reference angle B in the third quadrant, we know that the cosine is gonna be negative five over 13, right? The hypotenuse is always positive, since we're in the third quadrant, the X coordinates should be negative, that's where the negative five came from. By the Pythagorean equation, we see that the other side should be 12, but we're in the third quadrant, so the Y coordinate would be negative 12. So this tells us that sine of B would be negative 12-13. We also know that tangent of B would equal 12 over five. It'll be positive in that situation. So with this, we can then compute all these values, right? So let's consider sine of A plus B. By the angle identity we saw earlier, we would get sine of A, cosine of B, plus cosine of A, sine of B. And now as we've computed all these values, we don't actually know the angle measurements of A at B, but we do know these ratios. So using the information that was provided and the information we've computed, sine of A was three-fifths, cosine of B was negative five over 13. Then for the second one, we have to do cosine of A, which we found out was four-fifths. And then we have to do sine of B, which we found out was negative 12-13, like so. So notice with the denominator, you get five times 13, which is 65. You're gonna get five times, excuse me, three times negative five, which is negative 15. And then for the next one, you get four times negative 12, which is negative 48. Those will combine together to give us a negative 63 over 65. One good thing to check with your answer is that sine is always bounded between negative one, sine of theta, and one. So if you look at the absolute value of your fraction here, it should always be less than one. 63 is less than 65. So that is a feasible calculation for sine. So sine of A plus B is gonna be negative 63 over 65. Let's do now cosine, cosine of A plus B. By its formula, we get cosine of A, cosine of B minus sine of A, sine of B, for which, as we saw earlier, cosine of A is four-fifths. Cosine of B, it's off the screen now, but recall that was negative five-thirteenths. Then for the next one, we're gonna get minus sine of A, that's off the screen, but that was three-fifths. And then for sine of B, we computed that one to be negative 12-thirteenths. So same thing as before, five times 13 is gonna be 65. That's our denominator. Four times negative five is negative 20. And we're gonna get negative three times negative 12. That's a positive 36, right? And then so you put that together, you're going to get 16 over 65, like so. So let's put a little indicator here. We had negative 63 over 65 here. Notice that the sine, which represents the Y coordinate here is negative, but cosine, which represents the X coordinate is positive. So this actually tells us that our angle A plus B lives inside of the fourth quadrant, right, where Y is negative and X is positive. Now let's do tangent. Honestly, the easiest way to do tangent of A plus B would just be to take the sine ratio, which we had negative 63 over 65 and divide it by 16 over 65, which is just gonna give us negative 63 over 16. If you already have sine and cosine in hand, then tangent's super easy. It's just gonna be their ratio, like usual. Now, if you wanted to check this with the angle identities we did before, that would be appropriate, right? So this should look like tangent of A plus tangent of B over one minus tangent of A times tangent of B. Tangent of B. For which tangent of A is now off the screen. What did we have from before? Let's just double check. Tangent of A was three fourths. Tangent of B was 12 fifths. So you're gonna get three fourths plus 12 fifths right there over one minus three fourths times 12 fifths. Like so. For which case we wanna add these things together. Let's just make life a little bit easier and clear out the denominators. Rune times top and bottom by 20. Distribute those right here. Distribute this on the bottom. And so we end up with three times five plus we're gonna get 12 times four over and the denominator gonna get 20 minus three times 12. For which case that'll simplify. You're gonna get 15 plus 48 over 20 minus 36. And so you end up with 63 over negative 12, excuse me, negative 16, which then agrees with the answer we had previously. So the angle, some identity for tangent definitely works. And it's good to use if you have the tangent values, but if you've already done cosine and sine, then the easiest way to get tangent is just to use the ratio identities. And that's gonna end our discussion then of the angle sum and angle difference identities. You'll notice that as we conclude this lecture, we never talked about secant, cosecant or cotangent for angle identities. While they exist, for the most part, like we saw with tangent, if you can do it for sine and cosine, that's probably good enough. We don't necessarily need angle sum identities for all six trigonometric functions. We can just do reciprocity to do secant, cosecant and cotangent from sine, cosine and tangent as well.