Lesson Video: Angle Sum and Difference Identities Mathematics

Video Transcript

In this video, we will learn how to derive the angle sum and difference identities graphically and use them to find trigonometric values. The trigonometric angle sum and difference identities have been used in mathematics for centuries to solve real-world problems. The ancient Greeks used these formulas to solve astronomy problems, like distances from Earth to the Sun. Before we try and derive some of these identities or use them to solve problems, let’s remind ourselves what the sum and difference identities are. For the angle sum identities, we have sin of 𝐴 plus 𝐵 is equal to sin of 𝐴 times cos of 𝐵 plus cos of 𝐴 times sin of 𝐵. The cos of 𝐴 plus 𝐵 is equal to the cos of 𝐴 times the cos of 𝐵 minus the sin of 𝐴 times the sin of 𝐵. And finally, the tan of 𝐴 plus 𝐵 will be equal to the tan of 𝐴 plus the tan of 𝐵 all over one minus the tan of 𝐴 times the tan of 𝐵.

Note here that the variables 𝐴 and 𝐵 can represent any angle measure, and this will work when we’re operating in radians or degrees. In some places, you might see these angle measures represented with the variables 𝑥 and 𝑦 or even the Greek letters 𝛼 and 𝛽. We use the angle sum identity when we want to take a larger angle and break it up into two smaller angles. Generally, we already know the sine and cosine values for the two smaller angles, which help us calculate the sine, cosine, or tangent of a larger angle measure. So we know sin of 𝐴 minus 𝐵 is equal to sin of 𝐴 times cos of 𝐵 minus cos of 𝐴 times sin of 𝐵. cos of 𝐴 minus 𝐵 is equal to cos of 𝐴 times cos of 𝐵 plus sin of 𝐴 times sin of 𝐵. And tan of 𝐴 minus 𝐵 is equal to the tan of 𝐴 minus the tan of 𝐵 all over one plus the tan of 𝐴 times the tan of 𝐵.

Now, we’re gonna take this first angle sum identity and form a geometric proof to show that it’s true. To prove the sin of 𝐴 plus 𝐵 is equal to the sin of 𝐴 times the cos of 𝐵 plus the cos of 𝐴 times the sin of 𝐵, we can use what we know about right-angled triangles. Suppose we have a right-angled triangle and we know that one of these angles measures 𝐵 degrees. Stacked on top of that is the right triangle 𝐷𝐶𝑋, which has an angle measure of 𝐴 degrees. By drawing a perpendicular line from 𝐷 to the line 𝑋𝑌, we’ve created a third right triangle, and this third right triangle has an angle of 𝐴 plus 𝐵.

We can then use this figure to form a geometric proof. Let’s let the distance from 𝑋 to 𝐷 be equal to one. In our figure, what is the sin of 𝐴 plus 𝐵? That would be inside triangle 𝐷𝐹𝑋. And we know that in a right-angled triangle, the sin of an angle is equal to the opposite side length over the hypotenuse. So the sin of 𝐴 plus 𝐵 will be equal to the side length 𝐷𝐹 over one. In our figure, 𝐷𝐹 can be broken up into two smaller pieces. If we label this point 𝐸, then we can say that 𝐷𝐹 is equal to 𝐷𝐸 plus 𝐸𝐹.

Now we want to see if we can find the sin or cos of angle 𝐴 and the sin or cos of angle 𝐵. If we look at the right triangle that includes angle 𝐴, we see that it has a hypotenuse of one, which means the sin of angle 𝐴 will be the opposite side length 𝐷𝐶 over one, which simplifies to 𝐷𝐶. And we can label that side length the sin of 𝐴. Similarly, for cos of 𝐴, we know that the cos of any angle is equal to the adjacent side length over the hypotenuse. Here, that’s going to be the side length 𝑋𝐶 over one, which simplifies to 𝑋𝐶. And so we can label that side length cos of 𝐴.

Now we’ll consider the right-angled triangle containing angle 𝐵. To find the sin of angle 𝐵, we see the opposite side length, 𝐶𝑌, over the hypotenuse, which here is the cos of angle 𝐴. At this point, we recognize something interesting. We have an equation in which we have two terms, the cos of 𝐴 and the sin of 𝐵. We can rearrange this equation by multiplying both sides by cos of 𝐴. And then we see that the line segment 𝐶𝑌 is equal to the cos of 𝐴 times the sin of 𝐵. We already know that the sin of 𝐴 plus 𝐵 will be equal to line segment 𝐷𝐸 plus line segment 𝐸𝐹. And now we’re saying that the cos of 𝐴 times the sin of 𝐵 is equal to the line segment 𝐶𝑌. But we want to ask the question, can we relate line segment 𝐸𝐹 to line segment 𝐶𝑌?

If we look at our figure, line segment 𝐶𝑌 and line segment 𝐸𝐹 fall inside a rectangle, and they are opposite sides from each other. This means line segment 𝐶𝑌 is equal in length to line segment 𝐸𝐹. And we can substitute that value in for the cos of 𝐴 times the sin of 𝐵. Now we have line segment 𝐸𝐹 on both sides of this equation. And that means we want to ask the question, can we relate the sin of 𝐴 times the cos of 𝐵 to the line segment 𝐷𝐸? Line segment 𝐷𝐸 is here. And if we notice that line segment 𝐸𝐶 and line segment 𝑋𝑌 are parallel, they’re cut by the transversal 𝑋𝐶, which means we can say that this angle is equal to the angle 𝐵. And then we can say that this angle is 90 minus 𝐵. Therefore, this angle is also angle 𝐵.

Using this angle as our angle 𝐵, we can say that the cos of angle 𝐵 equals the adjacent side length 𝐷𝐸 over its hypotenuse, which is the sin of 𝐴. So we have cos of 𝐵 is equal to 𝐷𝐸 over sin of 𝐴. And if we multiply both sides of this equation by the sin of 𝐴, we get that line segment 𝐷𝐸 is equal to the sin of 𝐴 times the cos of 𝐵. If we substitute line segment 𝐷𝐸 in for the sin of 𝐴 times the cos of 𝐵, we’ve shown that, in this figure, the sin of 𝐴 plus 𝐵 is equal to line segment 𝐷𝐸 plus line segment 𝐸𝐹. The sin of 𝐴 times the cos of 𝐵 is equal to line segment 𝐷𝐸. The cos of 𝐴 times the sin of 𝐵 is equal to line segment 𝐸𝐹. And therefore, line segment 𝐷𝐸 plus line segment 𝐸𝐹 is equal to line segment 𝐷𝐸 plus line segment 𝐸𝐹.

It would also be possible to use this figure to derive the cos of 𝐴 plus 𝐵 and then to find the tan of 𝐴 plus 𝐵. But you can do that another day. For now, let’s look at some examples of using the angle sum and difference identities. To solve problems with the angle sum and difference identities, we’ll often have to recognize patterns and/or rearrange equations. So let’s look at our first example of that.

Simplify cos of two 𝑋 times cos of 22𝑋 minus sin of two 𝑋 times sin of 22𝑋.

When we look at this expression, we see that we have the angle two 𝑋 and the angle 22𝑋. We have the cos and sin of the angle two 𝑋, and we have the cos and sin of the angle 22𝑋, which means we have an expression in the form cos of 𝐴 times cos of 𝐵 minus sin of 𝐴 times sin of 𝐵. And we can simplify this expression using the angle sum identities. We know that the cos of 𝐴 times the cos of 𝐵 minus the sin of 𝐴 times the sin of 𝐵 is equal to the cos of 𝐴 plus 𝐵. In our case, the angle 𝐴 is two 𝑋 and the angle 𝐵 is 22𝑋, making this expression equal to cos of two 𝑋 plus 22𝑋, cos of 24𝑋. And so the simplified form of this expression is cos of 24𝑋.

Now we’re ready to consider another example.

Given that sin of 60 degrees times cos of 30 degrees minus cos of 60 degrees times sin of 30 degrees equals sin of 𝜃 degrees, find the value of 𝜃 in degrees.

When we look at this equation, we notice that we’re dealing with angles of 60 degrees and two angles of 30 degrees. And this should remind us of an angle sum or difference identity. That is sin of 𝐴 times cos of 𝐵 minus cos of 𝐴 times sin of 𝐵 is equal to sin of 𝐴 minus 𝐵. If we let 𝐴 be equal to 60 degrees and 𝐵 be equal to 30 degrees, we can write the angle 𝜃 in terms of 𝐴 and 𝐵. 𝜃 is going to be equal to 𝐴 minus 𝐵. 𝜃 will then be equal to 60 degrees minus 30 degrees, which equals 30 degrees. Using the angle difference identities, we’re able to show that 𝜃 here is equal to 30 degrees.

In our next example, we’re given the cosine values of two smaller angles and asked to find their cosine value combined.

Find the cos of 𝐴 plus 𝐵 given the cos of 𝐴 equals 15 over 17 and the cos of 𝐵 equals five over 13, where 𝐴 and 𝐵 are acute angles.

When we see this cos of 𝐴 plus 𝐵, it should remind us of our angle sum identities. We know that the cos of 𝐴 plus 𝐵 is equal to the cos of 𝐴 times the cos of 𝐵 minus the sin of 𝐴 times the sin of 𝐵. We have enough information for the first term since we’ve already been given the value of cos of 𝐴 and cos of 𝐵. So how should we go about finding the sin of 𝐴 and the sin of 𝐵 if we know the cosine ratios? They are acute angles. And therefore, one strategy to find sin 𝐴 and sin 𝐵 would be to make right-angled triangles with these proportions.

First, we can draw a right triangle with angle 𝐴. We know that the cosine relationship will be equal to the adjacent side length over the hypotenuse. And so we would label the adjacent side to 𝐴 15 and the hypotenuse 17. To find the sine relationship, we’ll need to know this opposite side length, and that means we’ll need to use the Pythagorean theorem. We’ll let our unknown side be lowercase 𝑎. And then we’ll have 17 squared equals 𝑎 squared plus 15 squared, which will give us 289 is equal to 𝑎 squared plus 225. To isolate 𝑎, we subtract 225 from both sides, and then we get 64 is equal to 𝑎 squared. Taking the square root of both sides, we get 𝑎 equal to eight.

We’re only interested in the positive square root since we’re dealing with distance. If 𝑎 equals eight, then the sin of angle 𝐴 will be equal to eight over 17. If we consider a second right-angled triangle with angle 𝐵, its adjacent side length is five and its hypotenuse is 13. We should recognize that this is a Pythagorean triple. It’s a set of positive integers that occur in the ratio 𝑎 squared plus 𝑏 squared equals 𝑐 squared. And because we know the hypotenuse is 13 and one of the sides is five, this is a five, 12, 13 triangle. And therefore, the adjacent side length will be 12 and the sin of angle 𝐵 is 12 over 13. And we can plug that value in. 15 over 17 times five over 13 is 75 over 221. Eight over 17 times 12 over 13 is 96 over 221, which makes the cos of 𝐴 plus 𝐵 equal to negative 21 over 221.

Before we finish, we’ll look at one final example involving right triangles and combining two different angles.

The diagram shows triangle 𝐴𝐵𝐶. Given that 𝐴𝐷 is perpendicular to 𝐵𝐶, 𝐴𝐷 equals 15 centimeters, 𝐵𝐷 equals 10 centimeters, and 𝐶𝐷 equals seven centimeters, find the value of the tan of 𝑥 plus 𝑦.

The first thing we want to do is label our diagram. 𝐴𝐷 equals 15 centimeters, 𝐵𝐷 equals 10 centimeters, and 𝐶𝐷 equals seven centimeters. We’re interested in the tangent of this angle, the angle 𝑥 plus 𝑦. However, this angle is in a triangle that’s not a right triangle, so we’ll need a different strategy to solve for this angle. If we look, we see that angle 𝑥 and angle 𝑦 are both located inside right triangles. This means it’s possible to find the tangent value of 𝑥 and the tangent value of 𝑦. And by our angle sum identity, we know that the tan of 𝐴 plus 𝐵 will be equal to the tan of 𝐴 plus the tan of 𝐵 all over one minus the tan of 𝐴 times the tan of 𝐵.

The tangent of any angle in a right triangle will be equal to the opposite side length over the adjacent side length. For angle 𝑥, the opposite side length is 10 and the adjacent side length is 15. The tan of 𝑥 equals 10 over 15. To find the tan of 𝑦, we have the opposite side length of seven over the adjacent side length of 15. So the tan of 𝑦 equals seven fifteenths. So the tan of 𝑥 plus 𝑦 is equal to 10 over 15 plus seven over 15 all over one minus 10 over 15 times seven over 15. Our numerator becomes 17 over 15.

Before we do this multiplying in the denominator, we can do some simplifying. 10 over 15 simplifies to two-thirds, and two-thirds times seven fifteenths equals 14 over 45. In our denominator, since we have one minus 14 over 45, we can rewrite the one as 45 over 45. 45 minus 14 is 31. This means we’re dividing seventeen fifteenths by 31 over 45. And to divide by a fraction, we multiply by the reciprocal. We have seventeen fifteenths times 45 over 31. 15 goes into 45 three times. 17 times three equals 51, making the tan of 𝑥 plus 𝑦 51 over 31.

Before we finish, let’s review the key points from this video. Angle sum and difference identities can be used to simplify expressions involving two angles. There are three angle sum identities for sin of 𝐴 plus 𝐵, cos of 𝐴 plus 𝐵, and tan of 𝐴 plus 𝐵 and three angle difference identities for sin of 𝐴 minus 𝐵, cos of 𝐴 minus 𝐵, and tan of 𝐴 minus 𝐵, all of which are listed here.