Expand expressions using angle sum and difference identities (deg)

In the chapter on Angle Sum and Difference Identities, we showed how identities involving a trigonometric function of a sum or difference have been derived. 

Here we show some examples of their use. The identities can be used to write an expression in expanded form. That is, avoiding functions of a sum or difference. It is also sometimes useful to recognise when it may be possible to perform this process in reverse.

Example 1

Expand $\sin\left(\theta+\frac{\pi}{4}\right)$sin(θ+π4​).

We use the identity: $\sin\left(\alpha+\beta\right)\equiv\sin\alpha\cos\beta+\sin\beta\cos\alpha$sin(α+β)≡sinαcosβ+sinβcosα. It follows that $\sin\left(\theta+\frac{\pi}{4}\right)=\sin\theta\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos\theta$sin(θ+π4​)=sinθcosπ4​+sinπ4​cosθ

$=\frac{1}{\sqrt{2}}\left(\sin\theta+\cos\theta\right)$=1√2​(sinθ+cosθ)

Example 2

Rewrite $\cos\left(x+2y\right)$cos(x+2y) using functions of $x$x and $y$y only.

Using the identity $\cos\left(\alpha+\beta\right)\equiv\cos\alpha\cos\beta-\sin\alpha\sin\beta$cos(α+β)≡cosαcosβ−sinαsinβ, we have $\cos\left(x+2y\right)=\cos x\cos2y-\sin x\sin2y$cos(x+2y)=cosxcos2y−sinxsin2y.

Example 3

Re-write this expression more concisely as a single trigonometric function: $\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x$√32​sinx−12​cosx.

We recognise $\frac{\sqrt{3}}{2}$√32​ as either $\sin\frac{\pi}{3}$sinπ3​ or as $\cos\frac{\pi}{6}$cosπ6​; and $\frac{1}{2}$12​ as either $\sin\frac{\pi}{6}$sinπ6​ or $\cos\frac{\pi}{3}$cosπ3​. So, the original expression could be written as 

$\cos\frac{\pi}{6}\sin x-\sin\frac{\pi}{6}\cos x$cosπ6​sinx−sinπ6​cosx.

This is,  $\sin\left(x-\frac{\pi}{6}\right)$sin(x−π6​).

Alternatively we could put 

$\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x=\sin\frac{\pi}{3}\sin x-\cos\frac{\pi}{3}\cos x$√32​sinx−12​cosx=sinπ3​sinx−cosπ3​cosx. This is,

$-\left(\cos\frac{\pi}{3}\cos x-\sin\frac{\pi}{3}\sin x\right)=-\cos\left(\frac{\pi}{3}+x\right)=\cos\left(\frac{\pi}{3}+x\right)$−(cosπ3​cosx−sinπ3​sinx)=−cos(π3​+x)=cos(π3​+x).

We can show that these two apparently different results are the same, by calling on facts about complementary functions.

More Worked Examples

Question 1

QUESTION 2

QUESTION 3

QUESTION 4