# 2. Sin, Cos and Tan of Sum and Difference of Two Angles

by M. Bourne

The **sine** of the sum and difference of two angles is as follows:

### On this page...

sin(

+αβ) = sinαcosβ+ cosαsinβ

sin(

α−β) = sinαcosβ− cosαsinβ

The **cosine** of the sum and difference of two angles is as follows:

cos(

α+β) = cosαcosβ− sinαsinβcos(

α−β) = cosαcosβ+ sinαsinβ

### Proofs of the Sine and Cosine of the Sums and Differences of Two Angles

We can prove these identities in several different ways.

Here is a proof using the unit circle:

(This is quite long, but we can learn a lot from proofs, especially how trigonometric identities work.)

Here is an easier proof, using complex numbers:

## Tangent of the Sum and Difference of Two Angles

We have the following identities for the tangent of the sum and difference of two angles:

`tan(alpha+beta)=(tan\ alpha+tan\ beta)/(1-tan\ alpha\ tan\ beta)`

and

`tan(alpha-beta)=(tan\ alpha-tan\ beta)/(1+tan\ alpha\ tan\ beta)`

### Proof of the Tangent of the Sum and Difference of Two Angles

Our proof for these uses the trigonometric identity for tan that we met before.

### Example 1

Find the **exact **value of cos 75° by using 75° = 30° + 45°.

Answer

**Recall** the 30-60 and 45-45 triangles from Values of Trigonometric Functions:

We use the exact sine and cosine ratios from the triangles to answer the question as follows:

`cos 75^"o"=cos(30^("o")+45^("o"))`

`=cos 30^("o")\ cos 45^("o")-sin 30^("o")\ sin 45^("o")`

`=sqrt3/2(1)/sqrt2-1/2(1)/sqrt2`

`=(sqrt3-1)/(2sqrt2)`

This is the **exact** value for cos 75°.

### Example 2

If `sin\ α = 4/5` (in Quadrant I) and `cos\ β = -12/13` (in Quadrant II) evaluate `sin(α − β).`

Answer

We use

sin(

α−β) = sinαcosβ− cosαsinβ

We firstly need to find `cos α` and `sin β`.

If `sin α = 4/5`, then we can draw a triangle and find the value of the unknown side using Pythagoras' Theorem (in this case, 3):

We do the same thing for `cos β = 12/13`, and we obtain the following triangle.

**Note 1:** We are using the positive value `12/13` to calculate the required reference angle relating to `beta`.

**Note 2:** The sine ratio is positive in both Quadrant I and Quadrant II.

**Note 3:** We have used Pythagoras' Theorem to find the unknown side, 5.

Now for the unknown ratios in the question:

`cos\ α = 3/5 `

(positive because in quadrant I)

`sin\ β = 5/13`(positive because in quadrant II)

We are now ready to find the required value, sin(*α* − *β*):

`sin(alpha-beta)=sin\ alpha\ cos\ beta-cos\ alpha\ sin\ beta`

`=4/5(-12/13)-3/5(5/13)`

`=(-48-15)/65`

`=(-63)/65`

This is the **exact** value for sin(*α* − *β*).

### Exercises

1. Find the **exact** value of cos 15° by using 15° = 60° − 45°

2. If `sin\ α = 4/5` (in Quadrant I) and `cos\ β = -12/13` (in Quadrant II) evaluate `cos(β − α).`

[This is not the same as Example 2 above. This time we need to find the **cosine** of the difference.]

3. Reduce the following to a single term. Do not expand.

cos(

x+y)cosy+ sin(x+y)siny

4. Prove that

`cos(30^"o"+x)=(sqrt3 cos x-sin x)/2`

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