NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Expand expressions using angle sum and difference identities
Lesson

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+sinπ4cosθ

$=\frac{1}{\sqrt{2}}\left(\sin\theta+\cos\theta\right)$=12(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)cosαcosβsinαsinβ, we have $\cos\left(x+2y\right)=\cos x\cos2y-\sin x\sin2y$cos(x+2y)=cosxcos2ysinxsin2y.

Example 3

Re-write this expression more concisely as a single trigonometric function: $\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x$32sinx12cosx.

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 

$\cos\frac{\pi}{6}\sin x-\sin\frac{\pi}{6}\cos x$cosπ6sinxsinπ6cosx.

This is,  $\sin\left(x-\frac{\pi}{6}\right)$sin(xπ6).

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$32sinx12cosx=sinπ3sinxcosπ3cosx. This is,

$-\left(\cos\frac{\pi}{3}\cos x-\sin\frac{\pi}{3}\sin x\right)=-\cos\left(\frac{\pi}{3}+x\right)=\cos\left(\frac{\pi}{3}+x\right)$(cosπ3cosxsinπ3sinx)=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

Expand $\tan\left(\theta-x\right)$tan(θx).

QUESTION 2

Express $\sin A\cos2B+\cos A\sin2B$sinAcos2B+cosAsin2B using one trigonometric ratio.

QUESTION 3

QUESTION 4

Express $\cos\left(3\theta+x\right)\cos3\theta-\sin\left(3\theta+x\right)\sin3\theta$cos(3θ+x)cos3θsin(3θ+x)sin3θ in simplest form.

Outcomes

M8-6

Manipulate trigonometric expressions

91575

Apply trigonometric methods in solving problems

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