A Level Maths: Compound Angle & Double Angle Formula
Introduction
Trigonometry is a huge topic. It is introduced in year 12, where area’s of GCSE content were revisited such as Pythagoras, SOH CAH TOA as well as the sine and cosine rules.
In year 12 you were introduced to trig identities and how to use them to solve trig equations. Depending on your school or college you were also introduced to the CAST diagram or were instructed how to solve trig equations with the aid of a graph.
In year 13, as a part of your A Level Maths Course, trigonometry is taken to another level.
You are introduced to the Compound Angle Formula and also the Double Angle Formula.
Then there are more trig identities as well as looking at combining trig curves together into one.
Compound Angle Formula – Introduction
If you were given the expression x(y+z) then everything that you have learnt in algebra will tell you that if you expand the brackets then you obtain the expression x y+x z
But what if you are given\sin \left(\theta+60^{\circ}\right)?
Well if you follow the rules of algebra then it would suggest that this would give \sin \theta+\sin 60^{\circ} \text {. }
But is this correct?
Take \theta=30^{\circ} \text { then } \sin \left(30^{\circ}+60^{\circ}\right)=\sin 90^{\circ}=1
And \sin 30^{\circ}+\sin 60^{\circ}=\frac{1}{2}+\frac{\sqrt{3}}{2} \text {. }
Clearly the results are not the same.
In order to solve expressions such as \sin \left(\theta+60^{\circ}\right), compound angle formulas need to be used.
Compound Angle Formula
The compound angle formulas are listed below and it is strongly advised that you memorise them.
\begin{gathered} \sin (A+B)=\sin A \cos B+\cos A \sin B \\ \sin (A-B)=\sin A \cos B-\cos A \sin B \\ \cos (A+B)=\cos A \cos B-\sin A \sin B \\ \cos (A-B)=\cos A \cos B+\sin A \sin B \\ \tan \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \end{gathered}Example
Write \cos 75^{\circ} as a surd.
Solution
\begin{gathered} \cos 75^{\circ}-\cos (45+30) \\ =\cos 45 \cos 30-\sin 45 \sin 30 \\ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ =\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \\ =\frac{\sqrt{3}-1}{2 \sqrt{2}} \end{gathered}Example
Solve \sin \left(60^{\circ}-\theta\right)=\cos \theta, 0^{\circ} \leq \theta \leq 360^{\circ}
Solution
Expanding the left hand side using the compound angle formula gives:
\begin{gathered} \sin 60^{\circ} \cos \theta-\cos 60^{\circ} \sin \theta=\cos \theta \\ \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=\cos \theta \\ \frac{\sqrt{3}}{2} \cos \theta-\cos \theta=\frac{1}{2} \sin \theta \\ \cos \theta\left(\frac{\sqrt{3}}{2}-1\right)=\sin \theta \\ 2\left(\frac{\sqrt{3}}{2}-1\right)=\tan \theta \\ \theta=-15^{\circ} \\ \therefore \theta=165^{\circ}, 345^{\circ} \end{gathered}
Double Angle Formula
In the case where A = B, the compound angle formulae become double angle formulae:
\begin{gathered} \sin 2 A-2 \sin A \cos A \\ \cos 2 A=\cos ^2 A-\sin ^2 A \end{gathered}Using \cos ^2 A+\sin ^2 A=1 to give two additional compound angle formulae::
\begin{gathered} \cos 2 A=1-2 \sin ^2 A \\ \cos 2 A=2 \cos ^2 A-1 \\ \tan 2 A=\frac{2 \tan A}{1-\tan ^2 A} \end{gathered}What is important to remember is that Trigonometry at A Level Maths is a very large topic and there are a great deal of skills and knowledge that you need. It is hoped that this article will give you some assistance but should you find that you are needing additional help then you can also consider an online maths tutor for A Level. He or she will guide you through how to answer questions effectively and how to look for clues within a question so that you can determine a starting point as that is often the most challenging part of A Level Maths.
You do get some trigonometric formulas given to you but it is best for you to learn them. If you learn them, you understand them and you have a better chance of doing the question better. Plus you are saving time not having to look up a formula compared to if you already know what it is.
Example
Using double angle formula show that \tan 2 \theta \equiv \frac{2 \tan \theta}{1-\tan ^2 \theta}
Solution
Starting with the left hand side:
\begin{gathered} \tan 2 \theta=\frac{\sin 2 \theta}{\cos 2 \theta} \\ =\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta-\sin ^2 \theta} \end{gathered}Looking at the result that is required, dividing numerator and denominator by \cos ^2 \theta
\begin{gathered} =\frac{\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta}}{\frac{\cos ^2 \theta}{\cos ^2 \theta}-\frac{\sin ^2 \theta}{\cos ^2 \theta}} \\ =\frac{\frac{2 \sin \theta}{\cos \theta}}{1-\frac{\sin ^2 \theta}{\cos ^2 \theta}} \\ =\frac{2 \tan \theta}{1-\tan ^2 \theta} \end{gathered}Example
Solve \cos 2 \theta-2=5 \cos \theta, 0 \leq \theta \leq 2 \pi
Solution
\begin{gathered} \cos ^2 \theta-\sin ^2 \theta-2=5 \cos \theta \\ \cos ^2 \theta-\left(1-\cos ^2 \theta\right)-2=5 \cos \theta \\ 2 \cos ^2 \theta-1-2=5 \cos \theta \\ 2 \cos ^2 \theta-5 \cos \theta-3=0 \\ (\cos \theta-3)(2 \cos \theta+1)=0 \\ \cos \theta=3, \text { has no solution } \\ \cos \theta=-\frac{1}{2} \therefore \theta=\frac{2 \pi}{3}, \frac{4 \pi}{3} \end{gathered}The last example has an instance where there is no solution and sometimes this does happen. It is important to realise that you have not done anything wrong and that you need to be confident with the maths that you have done.
Also, the last example gave a question in terms of π. This is telling you that you are working in terms of radians so your calculator needs to be in radian mode. Where possible do give answers in terms of a fraction but it is ok if this is not possible.
One of the biggest challenges is knowing which compound or double angle formula to use especially when it comes to proof questions. Quite often it is a case of “try it and see”.
To help students gain more confidence with such topic areas we hold various A Level Maths Revision Courses for Year 13 students. Courses take place during half term and there five time frames when these take place:
October, Christmas, February, Easter and May. We also hold pre-exam courses which can be anything from 1 – 2 days and are generally exam paper specific.
Below are some additional questions to help you get to grips with the Compound and Double Angle Formulas.
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