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How do the addition and double angle formulae let us expand and simplify trigonometric expressions of combined angles?

Apply the addition formulae for sine, cosine and tangent and the double angle formulae to expand, simplify and evaluate trigonometric expressions

A focused answer to the O-Level A-Maths outcome on compound and double angles. The addition formulae for sine, cosine and tangent, the double angle formulae, and using them to expand, simplify and find exact values.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to use the addition formulae, which expand sin(A±B)\sin(A \pm B), cos(A±B)\cos(A \pm B) and tan(A±B)\tan(A \pm B), and the double angle formulae that follow by setting B=AB = A. With these you can find exact values of unusual angles, expand combined-angle expressions, and prepare equations for solving.

The answer

The addition formulae

For any two angles:

sin(A±B)=sinAcosB±cosAsinB,\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B,

cos(A±B)=cosAcosBsinAsinB,\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B,

tan(A±B)=tanA±tanB1tanAtanB.\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}.

Note the sign reversal in the cosine formula: the right side takes the opposite sign to the left.

The double angle formulae

Setting B=AB = A gives:

sin2A=2sinAcosA,\sin 2A = 2\sin A\cos A,

cos2A=cos2Asin2A=2cos2A1=12sin2A,\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A,

tan2A=2tanA1tan2A.\tan 2A = \frac{2\tan A}{1 - \tan^2 A}.

The three forms of cos2A\cos 2A are interchangeable; choose the one that matches what you know.

Using them for exact values

Split an awkward angle into a sum or difference of special angles, such as 75=45+3075^\circ = 45^\circ + 30^\circ or 15=453015^\circ = 45^\circ - 30^\circ, then apply the addition formula with the known exact values.

Choosing the right cosine form

If a problem gives sinA\sin A, use cos2A=12sin2A\cos 2A = 1 - 2\sin^2 A; if it gives cosA\cos A, use cos2A=2cos2A1\cos 2A = 2\cos^2 A - 1. Matching the form to the data avoids extra work.

Using double angle formulae in reverse for proofs

Read backwards, the double angle formulae let you replace a single-angle expression with a double-angle one, which is the key to many identity proofs and to integrating squared trig functions. Rearranging cos2A=12sin2A\cos 2A = 1 - 2\sin^2 A gives sin2A=1cos2A2\sin^2 A = \tfrac{1 - \cos 2A}{2}, and rearranging cos2A=2cos2A1\cos 2A = 2\cos^2 A - 1 gives cos2A=1+cos2A2\cos^2 A = \tfrac{1 + \cos 2A}{2}. These "power reduction" forms turn a squared ratio into a first-power expression in the double angle, which is exactly what is needed to simplify sin2θ\sin^2\theta in an identity or to integrate cos2θ\cos^2\theta later. Recognising a double angle formula in reverse is a frequently rewarded move in A-Maths proofs.

Combining the formulae to reach a triple angle

The addition and double angle formulae chain together to expand a triple angle such as sin3A\sin 3A. Write 3A=2A+A3A = 2A + A and apply the addition formula: sin3A=sin(2A+A)=sin2AcosA+cos2AsinA\sin 3A = \sin(2A + A) = \sin 2A\cos A + \cos 2A\sin A. Substituting sin2A=2sinAcosA\sin 2A = 2\sin A\cos A and cos2A=12sin2A\cos 2A = 1 - 2\sin^2 A, then simplifying with cos2A=1sin2A\cos^2 A = 1 - \sin^2 A, yields sin3A=3sinA4sin3A\sin 3A = 3\sin A - 4\sin^3 A. The technique of splitting a multiple angle into a double plus a single, then expanding, shows how the basic formulae generate higher-angle identities and is a satisfying way to see them work together.

Examples in context

Example 1. Combining oscillations. Adding two waves of the same frequency but different phase, sin(ωt+ϕ)\sin(\omega t + \phi), expands with the addition formula into a single sine and cosine, the first step in seeing the combined wave, which connects to the R-formula.

Example 2. Halving an angle in optics. A reflection angle that is twice another appears in mirror and lens geometry; the double angle formula relates the two, letting a single measured angle determine the other.

Try this

Q1. Expand sin(θ+30)\sin(\theta + 30^\circ). [2 marks]

  • Cue. sinθcos30+cosθsin30=32sinθ+12cosθ\sin\theta\cos 30^\circ + \cos\theta\sin 30^\circ = \dfrac{\sqrt{3}}{2}\sin\theta + \dfrac{1}{2}\cos\theta.

Q2. Given cosA=35\cos A = \dfrac{3}{5} with AA acute, find cos2A\cos 2A. [2 marks]

  • Cue. cos2A=2cos2A1=29251=725\cos 2A = 2\cos^2 A - 1 = 2\cdot\tfrac{9}{25} - 1 = -\dfrac{7}{25}.

Q3. Write 2sin3θcos3θ2\sin 3\theta\cos 3\theta as a single trigonometric ratio. [2 marks]

  • Cue. This is sin2(3θ)=sin6θ\sin 2(3\theta) = \sin 6\theta.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksWithout a calculator, find the exact value of sin75\sin 75^\circ using the addition formula.
Show worked answer →

Write 75=45+3075^\circ = 45^\circ + 30^\circ and use sin(A+B)=sinAcosB+cosAsinB\sin(A + B) = \sin A\cos B + \cos A\sin B.

sin75=sin45cos30+cos45sin30=1232+1212\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{2}.

=322+122=3+122=6+24= \dfrac{\sqrt{3}}{2\sqrt{2}} + \dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} = \dfrac{\sqrt{6} + \sqrt{2}}{4}.

Markers reward splitting the angle, the correct addition formula, the special values, and a rationalised exact answer.

Original4 marksGiven that sinA=45\sin A = \dfrac{4}{5} and AA is acute, find the exact value of sin2A\sin 2A and cos2A\cos 2A.
Show worked answer →

Since AA is acute, cosA=11625=35\cos A = \sqrt{1 - \tfrac{16}{25}} = \dfrac{3}{5}.

sin2A=2sinAcosA=24535=2425\sin 2A = 2\sin A\cos A = 2\cdot\dfrac{4}{5}\cdot\dfrac{3}{5} = \dfrac{24}{25}.

cos2A=12sin2A=121625=13225=725\cos 2A = 1 - 2\sin^2 A = 1 - 2\cdot\dfrac{16}{25} = 1 - \dfrac{32}{25} = -\dfrac{7}{25}.

Markers reward finding cosA\cos A from the identity, the double angle formula for sin2A\sin 2A, and a correct cos2A\cos 2A.

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