# Trigonometric Functions

# Sum and Difference Identities

First, let's prove the **sine sum and difference identities**.

sin(α + β) = sinα cosβ + cosα sinβ

and

sin(α – β) = sinα co β – cosα sinβ

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 will see that perpendicular lines have been added to form a few right triangles.

Find ∠BOE.

Notice that ∠BOE = α + β.

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

Now, look at ΔOAE.

Since BD = AC, .

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

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

Since

and

and

and

Subbing in all these finally gives

**sin(α + β ) = sinαco β + cosαsinβ**

Victory!

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

sin(α – β ) = sin[α + (-β)] = sinαcos (-β) + cosαsin(-β)

From our negative angle identities, we know that:

cos(-β) = cos β and sin(-β) = -sinβ

Which means we have our **difference identity**:

**sin(α – β ) = sinαcosβ – cosαsinβ**

Now onto the **cosine sum identity**.

**cos(α + β) = cosαcosβ – sinαsinβ **

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

cos(α + β ) = sin[ 90° – (α + β )]

Removing parentheses

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

Next, plug this in to our sin difference identity

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

Using co-function identities again

cos(α + β ) = cosα cosβ – sinαsinβ

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

cos [α + (-β )] = cosαcos(-β) – sinαsin(-β)

and using your negative angle identities

**cos(α – β ) = cosαcosβ + sinαsinβ**

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°).

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

We see that 75° = 30° + 45°.

So,

cos(75°)= cos(30° + 45°)= cos(30°)cos(45°) – sin(30°)sin(45°)

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

And the answer is

### Sample Problem

Use a sum or difference identity to find the exact value of sin165°.

What quadrant is 165° in?

Quadrant 2.

To work this, we look at the 165° to see if it is 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°).

What quadrant is 255° in?

Quadrant 3

To work this, we look at the 255° to see if it is the sum or difference of our angles from our reference triangles.

255° = 300° – 45°

So

The tangent sum and difference identities can be found from the sine and cosine sum and difference identities.

Plug in the sum identities for both sin and cos.

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

Divide the numerator and the denominator by cosαcosβ.

Now, split up the terms.

Replacing each term with "1" or "tan," wherever appropriate, gives us

Now simplifying just a little more:

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 cosαcosβ.

Now, split up the terms.

Replacing each term with "1" or "tan," wherever appropriate, gives us

We're almost there—there *is *a there—simplifying just a little more will do the trick.

### Sample Problem

Use a sum or difference identity to find tan255°.

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:

And our answer is:

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