# 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 a variety of ways.

Here is a relatively simple proof using the unit circle:

Proof 1

### Proof 1 - Using the Unit Circle

We start with a unit circle (which means it has radius 1), with center O.

We construct angles `BOA = alpha` and `AOP = beta` as shown.

Next, we drop a perpendicular from P to the *x*-axis at T. Point C is the intersection of OA and PT.

We then construct line PR perpendicular to OA.

Finally, we drop a perpendicular from R to the *x*-axis at S, and another from R to PT at Q, as shown.

We note the following:

(1) `/_TPR = alpha` since triangles OTC and PRC are similar. (`/_OTC = /_PRC = 90°`, and `/_OCT = /_PCR = 90°- alpha`.)

(2) Length |QT| = |RS|

(3) sin (Î± + Î²) = |PT| = |PQ| + |QT| = |PQ| + |RS|

(4) |PR| = sin (Î²)

(5) In triangle PQR, |PQ| = |PR| cos (Î±)

(6) So from (4) and (5), |PQ| = sin (Î²) cos (Î±).

(7) |OR| = cos (Î²)

(8) In triangle OSR, |RS| = |OR| sin (Î±)

(9) So from (7) and (8), |RS| = cos (Î²) sin (Î±)

(10) Thus from (3), (6) and (9), we have proved:

sin (Î± + Î²) = sin (Î²) cos (Î±) + cos (Î²) sin (Î±)

Rearranging gives:

sin (Î± + Î²) = sin (Î±) cos (Î²) + cos (Î±) sin (Î²)

(11) From Even and Odd Functions, we have: cos (−*β*) = cos( β) and sin (−*β*) = −sin(β)

(12) So replacing *β* with (−*β*), the identity in (10) becomes

sin (

α−β) = sinαcosβ− cosαsinβ

[Thank you to David McIntosh for providing the outline of the above proof.]

#### The Cosine Proofs

We now need to prove

cos (

α+β) = cosαcosβ− sinαsinβ

(13) |OT| = cos (*α* + * β*)

(14) In triangle ORS, we have: `cos alpha = |OS|/|OR|`.

(15) cos (Î²) = |OR|, from (7) above.

(16) From (14) and (15), we obtain `cos alpha cos beta= |OS|/|OR|xx|OR| = |OS|`.

(17) In triangle QPR, we have `sin alpha = |QR|/|PR|`.

(18) sin (Î²) = |PR|, from (4) above.

(19) From (17) and (18), we obtain `sin alpha sin beta= |QR|/|PR|xx|PR| = |QR|`.

(20) Now |OS| − |QR| = |OT|.

(21) So `cos alpha cos beta - sin alpha sin beta` ` = cos(alpha+beta)`.

(22) Rearranging, we have:

cos (

α+β) = cosαcosβ− sinαsinβ

(23) Once again, we replace *β* with (−*β*), and the identity in (22) becomes:

cos (

α−β) = cosαcosβ+ sinαsinβ

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.

Proof 2

### Proof 2 - Using the Unit Circle

We will prove the cosine of the sum of two angles identity first, and then show that this result can be extended to all the other identities given.

cos (α+β) = cos α cos β − sin α sin β

We draw a circle with radius 1 unit, with point *P* on the circumference at (1, 0).

We draw an angle α from the centre with terminal point *Q* at (cos α, sin α), as shown. [*Q* is (cos α, sin α) because the hypotenuse is 1 unit.]

We extend this idea by drawing:

a. The angle β with terminal points at *Q* (cos α, sin α) and *R* (cos (α + β), sin (α + β))

b. The angle −β with terminal point at *S *(cos (−β), sin (−β))

c. The lines *PR* and *QS*, which are equivalent in length.

Now, using the distance formula from Analytical Geometry, we have:

*PR*^{2} = (cos (α + β) − 1)^{2} + sin^{2}(α + β)

= cos^{2}(α + β) − 2 cos (α + β) + 1 + sin^{2}(α + β)

= 2 − 2cos (α + β)

[since sin^{2}(α + β) + cos^{2}(α + β) = 1]

Now using the distance formula on distance *QS*:

*QS*^{2} = (cos α − cos (−β))^{2} + (sin α − sin (−β))^{2}

= cos^{2} α − 2 cos α cos(−β) + cos^{2}(−β) + sin ^{2}α − 2sin α sin(−β) + sin^{2}(−β)

= 2 − 2cos α cos(−β) − 2sin α sin(−β)

[since

sin^{2}α + cos^{2}α = 1 and

sin^{2}(−β) + cos^{2}(−β) = 1]

= 2 − 2cos α cos β + 2sin α sin β

[since

cos(−β) = cos β (cosine is an even function) and

sin(−β) = −sinβ (sine is an odd function − see Even and Odd Functions)]

Since *PR* = *QS*, we can equate the 2 distances we just found:

2 − 2cos (α + β) = 2 − 2cos α cos β + 2sin α sin β

Subtracting 2 from both sides and dividing throughout by −2, we obtain:

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

If we replace β with (−β), this identity becomes:

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

[since cos(−β) = cos β and sin(−β) = −sinβ]

### Sine of the Sum of Two Angles

We aim to prove that

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

Recall that (see phase shift)

sin (

θ) = cos (π/2−θ)

If *θ* = α + β, then we have:

sin (α + β)

= cos [π/2 − (α + β)]

= cos [π/2 − α − β)]

Next, we re-group the angles inside the cosine term, since we need this for the rest of the proof:

cos [π/2 − α − β)] = cos [(π/2 − α) − β]

Using the cosine of the difference of 2 angles identity that we just found above [which said

cos (α − β) = cos α cos β + sin α sin β],

we have:

cos [(π/2 − α) − β]

= cos (π/2 − α) cos (β) + sin (π/2 − α) sin (β)

= sin α cos β + cos α sin β

[Since cos (π/2 − α) = sin α; and sin (π/2 − α) = cos α]

Therefore:

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

Replacing β with (−β), this identity becomes (because of Even and Odd Functions):

sin (α − β) = sin α cos β − cos α sin β

We have proved the 4 identities involving sine and cosine of the sum and difference of two angles.

#### Summary:

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

sin (α − β) = sin α cos β − cos α sin β

cos (α+β) = cos α cos β − sin α sin β

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

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

Proof 3

### Proof 3 - Using Complex Numbers

The exponential and polar forms of a complex number provide an easy way to prove the fundamental trigonometric identities.

Assume we have 2 complex numbers which we write as:

r_{1}e^{jα}=r_{1}(cos α +jsin α)

and

r_{2}e^{jβ}=r_{2}(cos β +jsin β)

We multiply these complex numbers together.

Multiplying the left hand sides:

r_{1}e^{jα}×r_{2}e^{jβ}=r_{1}r_{2}e^{j(α+β)}

We can write this answer as:

*r*_{1}*r*_{2}*e*^{j(α+β)} = *r*_{1}*r*_{2}(cos (α+β) + *j* sin (α+β)) ... (1)

Multiplying the right hand sides:

*r*_{1}(cos α + *j* sin α) × *r*_{2}(cos β + *j* sin β)

*= r*_{1} *r*_{2}(cos α cos β +* **j* cos α sin β + *j* sin α cos β − sin α sin β)

= *r*_{1} *r*_{2}(cos α cos β − sin α sin β +* **j* (cos α sin β +* *sin α cos β)) .... (2)

[since *j*^{2} = −1]

Now, equating (1) and (2) and dividing both parts by *r*_{1} *r*_{2}:

cos (α+β) + *j* sin (α+β) = cos α cos β − sin α sin β + *j* (cos α sin β +* *sin α cos β)

Equating the real parts gives:

cos (α+β) = cos α cos β − sin α sin β

Equating the imaginary parts gives:

sin (α+β) = sin α cos β + cos α sin β

We would then proceed to replace β with (−β) as before, to obtain the identities for sin (α − β) and cos (α − β).

## 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}.

### 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^{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)`

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.