# Trigonometric Functions

Sum and Difference Identities

# Sum and Difference Identities

First, allow us to introduce you to the sine sum and difference identities.

sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

and

sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

Pretty wild, right? Let's prove that they actually work.

Take a look at the figure below, which shows two angles α and β.

Imagine Line OB is swinging around O and sweeps out the angle α and then sweeps out the angle β. (Things are definitely moving here in trig land.)

Look carefully, and you'll see that perpendicular lines have been added to form a few right triangles.

First off, notice that ∠BOE = α + β.

This means that sin(∠BOE) = sin(α + β).

Now, look at ΔOAE and bust out our trig ratio for the sine.

Since BD = AC, we can rewrite that as .

Stay with us now…it's time to do some rearranging.

Let's shuffle the denominators. (Poker, anyone?)

Hold onto that equation for a sec while we take a quick side trip. We need to show that ∠CED is the same as angle α for the rest of the proof to work.

Okay, see how the line OD crosses both the parallel lines CD and OB? That means ∠ODC has the same measure as angle α because of the alternate interior angles theorem from back in Geometry. We can also see from the drawing that ∠ODC and ∠CDE form a right angle, so m∠CDE = 90° – m∠ODC. And since ∠ODC = α, we can rewrite that as m∠CDE = 90° – α. Right? Right.

Since the three angles in triangle CDE are gonna add up to 180° (like any triangle), that means:

m∠DCE + m∠CDE + m∠CED = 180°

90° + (90° – α) + m∠CED = 180°

180° – α + m∠CED = 180°

-α + m∠CED = 0

m∠CED = α

Whew. Good deal; that angle way up at the top of the figure has the same measure as angle α. Trust us, that's important.

Now we'll look back at the diagram one last time and write down a bunch of trig ratios for the angles α and β.

(This one works because of all that work we just did to show that m∠CED = α.)

How do those guys help us? Welp, we can now substitute all those gnarly ratios we came up with earlier:

sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

Victory! That's our sine sum identity.

Now, we can find the difference identity by plugging in -β for β like this:

sin(α – β) = sin[α + (-β)] = sin(α)cos(-β) + cos(α)sin(-β)

From our negative angle identities, we know that:

cos(-β) = cos(β)

and

sin(-β) = -sin(β)

Which means we have our sine difference identity:

sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

## Cosine Gets a Turn

Now onto the cosine sum identity.

cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

We can prove the cosine sum identity by applying our co-function identity:

cos(α + β) = sin[90° – (α + β)]

Remove those pesky parentheses:

cos(α + β) = sin(90° – α – β) = sin[(90° – α) – β]

Next, plug this into our sine difference identity:

cos(α + β) = sin[(90° – α) – β] = sin(90° – α)cos(β) – cos(90° – α)sin(β)

Using co-function identities again, we can swap out sin(90° – α) for cos(α), and swap cos(90° – α) for sin(α) to finish up.

cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

Bam. Nailed it.

To find the cosine difference identity, plug in -β for β, just like you did for the sine difference identity.

cos(α – β) = cos[α + (-β)] = cos(α)cos(-β) – sin(α)sin(-β)

Sub in those negative angle identities to get the cosine difference identity:

cos(α – β) = cos(α)cos(β) + sin(α)sin(β)

Now let's take our hard-earned sum and difference identities, and use them to solve problems.

### Sample Problem

Use a sum or difference identity to find the exact value of cos(75°) without a calculator.

To work this, we look at the 75° to see if it's the sum or difference of any angles from our reference triangles.

We see that 75° = 30° + 45°.

So:

cos(75°) = cos(30° + 45°)

We can use the cosine sum identity.

cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

cos(30° + 45°) = cos(30°)cos(45°) – sin(30°)sin(45°)

Now plug in values from your reference triangles. (Oh, it's them again.)

### Sample Problem

Use a sum or difference identity to find the exact value of sin(165°).

It's bigger than 90° but smaller than 180°, so it's in Quadrant II.

To work this, we look at the 165° to see if it's the sum or difference of our angles from our reference triangles.

165° = 120° + 45°

So:

### Sample Problem

Use a sum or difference identity to find the exact value of cos(255°).

Yup, it's in Quadrant III. Let's look at the 255° to see if it is the sum or difference of any special angles.

255° = 300° – 45°

So:

## Going Off on a Quick Tangent

The tangent sum and difference identities can be found from the sine and cosine sum and difference identities. Lucky for us, the tangent of an angle is the same thing as sine over cosine.

Plug in the sum identities for both sine and cosine.

Next, a little division gets us on our way (fractions never hurt).

Divide the numerator and the denominator by cos(α)cos(β).

Now, split up the terms.

Replace each term with "1" or "tan," wherever appropriate. Remember, tangent is sine over cosine.

Now simplifying just a little more gives us the tangent sum identity:

Next, we can find tan(α – β) the same way.

Once again, with a little help from our friend division, divide the numerator and the denominator by cos(α)cos(β).

Now, split up the terms and replace each term with "1" or "tan," wherever appropriate.

We're almost there—there is a there, we promise. Simplifying just a little more will do the trick. Here's our tangent difference identity:

### Sample Problem

Use a sum or difference identity to find tan(255°).

First, rewrite 255° as a sum or difference.

255° = 300° – 45°

Which means

We have to clean up our denominator (so far, it's looking u-u-u-ugly).

So multiply by the conjugate over the conjugate:

We've made out it of trig land—alive (or barely alive). See, the world of angles, triangulation and all, isn't so bad.