New Zealand
Level 8 - NCEA Level 3

# Evaluate trig expressions using angle sum and difference identities

Lesson

The Angle Sum and Difference identities allow us to expand expressions like $\sin\left(\alpha+\beta\right)$sin(α+β). One way to establish these identities involves first establishing the cosine rule and then using it in an arbitrary triangle with vertices at the centre and circumference of the unit circle. In this chapter, we assume that the cosine rule is already available to us.

We can write two different expressions for the distance $PQ$PQ. First, we use the cosine rule.

The arms of the triangle have unit length because they are radii of the unit circle. Therefore, we have $PQ^2=1^2+1^2-2\times1\times1\times\cos\left(\beta-\alpha\right)$PQ2=12+122×1×1×cos(βα). That is,

$PQ^2=2-2\cos\left(\beta-\alpha\right)$PQ2=22cos(βα)

Next, we write the distance $PQ$PQ using the coordinates of the points and the Distance Formula, which is just an application of Pythagoras's theorem. We have,

$PQ^2=\left(\cos\alpha-\cos\beta\right)^2+\left(\sin\alpha-\sin\beta\right)^2$PQ2=(cosαcosβ)2+(sinαsinβ)2

Equating the two expressions gives

$2-2\cos\left(\beta-\alpha\right)\equiv\left(\cos\alpha-\cos\beta\right)^2+\left(\sin\alpha-\sin\beta\right)^2$22cos(βα)(cosαcosβ)2+(sinαsinβ)2

and this expands first to

$2-2\cos\left(\beta-\alpha\right)\equiv\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta$22cos(βα)cos2α2cosαcosβ+cos2β+sin2α2sinαsinβ+sin2β

and then making use of the Pythagorean identity, $\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ1, we simplify this to

$\cos\left(\beta-\alpha\right)\equiv\cos\beta\cos\alpha+\sin\beta\sin\alpha$cos(βα)cosβcosα+sinβsinα

From this we obtain an expression for  $\cos\left(\beta+\alpha\right)$cos(β+α) by writing this as

$\cos\left(\beta-\left(-\alpha\right)\right)$cos(β(α)) and applying the previous formula. Thus,

$\cos\left(\beta+\alpha\right)\equiv\cos\beta\cos\alpha-\sin\beta\sin\alpha$cos(β+α)cosβcosαsinβsinα

We obtain the corresponding formulas for sine by using the complementary angle fact: $\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$sinθ=cos(π2θ).

If we write $\sin\left(\beta-\alpha\right)$sin(βα) as $\cos\left(\frac{\pi}{2}-\left(\beta-\alpha\right)\right)$cos(π2(βα)), it follows that

$\sin\left(\beta-\alpha\right)=\cos\left(\left(\frac{\pi}{2}-\beta\right)+\alpha\right)=\cos\left(\frac{\pi}{2}-\beta\right)\cos\alpha-\sin\left(\frac{\pi}{2}-\beta\right)\sin\alpha$sin(βα)=cos((π2β)+α)=cos(π2β)cosαsin(π2β)sinα

That is,

$\sin\left(\beta-\alpha\right)\equiv\sin\beta\cos\alpha-\cos\beta\sin\alpha$sin(βα)sinβcosαcosβsinα,

and similarly,

$\sin\left(\beta+\alpha\right)\equiv\sin\beta\cos\alpha+\cos\beta\sin\alpha$sin(β+α)sinβcosα+cosβsinα

The corresponding formulas for the tangent function are obtained by expanding $\tan\left(\alpha+\beta\right)=\frac{\sin\left(\alpha+\beta\right)}{\cos\left(\alpha+\beta\right)}$tan(α+β)=sin(α+β)cos(α+β)

After some simplification, we have

$\tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$tan(α+β)=tanα+tanβ1tanαtanβ

and

$\tan\left(\alpha-\beta\right)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$tan(αβ)=tanαtanβ1+tanαtanβ

#### Example

Find an 'exact value' expression for $\sin\frac{\pi}{12}$sinπ12

We notice that $\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}$π12=π4π6. So, $\sin\frac{\pi}{12}=\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\sin\frac{\pi}{4}\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\cos\frac{\pi}{4}$sinπ12=sin(π4π6)=sinπ4cosπ6sinπ6cosπ4

Then, using known exact values, we have $\sin\frac{\pi}{12}=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$sinπ12=12.3212.12=624

#### Worked Examples

##### Question 1

Given that $\sin A=\frac{24}{25}$sinA=2425 and $\cos B=\frac{20}{29}$cosB=2029, find the exact value of:

1. $\sin\left(A+B\right)$sin(A+B)

2. $\cos\left(A-B\right)$cos(AB)

##### Question 2

Find the exact value of $\cos\frac{11\pi}{12}$cos11π12.

##### Question 3

Suppose $\sin A=\frac{3}{5}$sinA=35 and $\sin B=-\frac{12}{13}$sinB=1213, where $00<A<π2 and$\piπ<B<3π2.

1. Find $\cos A$cosA.

2. Find $\cos B$cosB.

3. Find $\sin\left(A+B\right)$sin(A+B).

4. Find $\sin\left(A-B\right)$sin(AB).

5. Find $\tan\left(A+B\right)$tan(A+B).

6. Find $\tan\left(A-B\right)$tan(AB).

7. Which quadrant does $A+B$A+B lie in?

$III$III

A

$II$II

B

$I$I

C

$IV$IV

D

$III$III

A

$II$II

B

$I$I

C

$IV$IV

D
8. Which quadrant does $A-B$AB lie in?

$III$III

A

$II$II

B

$I$I

C

$IV$IV

D

$III$III

A

$II$II

B

$I$I

C

$IV$IV

D

### Outcomes

#### M8-6

Manipulate trigonometric expressions

#### 91575

Apply trigonometric methods in solving problems