# 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:

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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 a variety of ways.

Here is a relatively simple proof using the unit circle:

The next proof is the standard one that you see in most text books. It also uses the unit circle, but is not as straightforward as the first proof. However, we can still learn a lot from this next proof, especially about the way trigonometric identities work.

Finally, here is an easier proof of the identities, using **complex numbers**:

Continues below ⇩

## 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.

Proof

### Case: `tan(alpha+beta)`

Recall that

`tan theta=(sin theta)/(cos theta)`

So, letting *θ* = α + β, and expanding using our new sine and cosine identities, we have:

`tan(alpha+beta)` `=(sin(alpha+beta))/(cos(alpha+beta))` `=(sin alpha cos beta+cos alpha sin beta)/(cos alpha cos beta-sin alpha sin beta)`

Dividing numerator and denominator by cos α cos β:

`=(sin alpha cos beta+cos alpha sin beta)/(cos alpha cos beta-sin alpha sin beta)` `-:(cos alpha cos beta)/(cos alpha cos beta`

Simplifying gives us:

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

### Case: `tan(alpha-beta)`

Replacing β with (−β) gives us

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

[The tangent function is odd, so tan(−β) = − tan β]

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

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

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

### Example 1

Find the **exact **value of cos 75^{o} by using 75^{o} = 30^{o} + 45^{o}.

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^{o}.

Easy to understand math videos:

MathTutorDVD.com

### 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*).

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### Exercises

1. Find the **exact** value of cos 15^{o} by using 15^{o} = 60^{o} − 45^{o}

Answer

Once again, we use the 30^{o}-60^{o} and 45^{o}-45^{o} triangles to find the exact value.

`cos 15^("o")=cos(60^("o")-45^("o"))`

`=cos 60^("o")\ cos 45^("o")+sin 60^("o")\ sin 45^("o")`

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

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

Easy to understand math videos:

MathTutorDVD.com

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.]

Answer

In this case, for the cosine of the difference of two angles, we have:

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

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

`=(-36+20)/65`

`=(-16)/65`

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

cos(

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

Answer

We recognise this expression as the right hand side of:

cos(

α−β) = cosαcosβ+ cosαcosβ,

with *α* = *x *+* y* and *β* = *y.*

We can now write this in terms of cos(*α* − *β*) as follows:

cos(*x* + *y*)cos *y* + sin(*x* + *y*)sin *y*

= cos[(

x+y) − (y)]= cos

x

We have reduced the expression to a single term.

4. Prove that

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

Answer

We recall the 30-60 triangle from before (in Values of Trigonometric Functions):

Using

cos(

α+β) = cosαcosβ− sinαsinβ

and our 30-60 triangle, we start with the left hand side (LHS) and obtain:

`"LHS"=cos(30^"o"+x)`

`=cos 30^"o"\ cos x - sin 30^"o"\ sin x`

`=sqrt3/2 cos x-1/2sin x`

`=(sqrt3\ cos x-sin x)/2`

`="RHS"`

Since the LHS = RHS, we have proved the identity.

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