Evaluate trig expressions using angle sum and difference identities

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+12−2×1×1×cos(β−α). That is,

$PQ^2=2-2\cos\left(\beta-\alpha\right)$PQ2=2−2cos(β−α)

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=(cosα−cosβ)2+(sinα−sinβ)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$2−2cos(β−α)≡(cosα−cosβ)2+(sinα−sinβ)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$2−2cos(β−α)≡cos2α−2cosαcosβ+cos2β+sin2α−2sinαsinβ+sin2β

and then making use of the Pythagorean identity, $\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ≡1, we simplify this to

$\cos\left(\beta-\alpha\right)\equiv\cos\beta\cos\alpha+\sin\beta\sin\alpha$cos(β−α)≡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)≡cosβcosα−sinβsinα

We obtain the corresponding formulas for sine by using the complementary angle fact: $\sin\theta=\cos\left(90^\circ-\theta\right)$sinθ=cos(90°−θ).

If we write $\sin\left(\beta-\alpha\right)$sin(β−α) as $\cos\left(90^\circ-\left(\beta-\alpha\right)\right)$cos(90°−(β−α)), it follows that 

$\sin\left(\beta-\alpha\right)=\cos\left(\left(90^\circ-\beta\right)+\alpha\right)=\cos\left(90^\circ-\beta\right)\cos\alpha-\sin\left(90^\circ-\beta\right)\sin\alpha$sin(β−α)=cos((90°−β)+α)=cos(90°−β)cosα−sin(90°−β)sinα

That is, 

$\sin\left(\beta-\alpha\right)\equiv\sin\beta\cos\alpha-\cos\beta\sin\alpha$sin(β−α)≡sinβcosα−cosβsinα,

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+tanβ1−tanαtanβ​

and

$\tan\left(\alpha-\beta\right)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$tan(α−β)=tanα−tanβ1+tanαtanβ​

Example

Find an 'exact value' expression for $\sin15^\circ$sin15° 

We notice that $15^\circ=45^\circ-30^\circ$15°=45°−30°. So, $\sin15^\circ=\sin\left(45^\circ-30^\circ\right)=\sin45^\circ\cos30^\circ-\sin30^\circ\cos45^\circ$sin15°=sin(45°−30°)=sin45°cos30°−sin30°cos45°

Then, using known exact values, we have $\sin15^\circ=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$sin15°=1√2​.√32​−12​.1√2​=√6−√24​

 

Worked Examples 

Question 1

Question 2

Question 3