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Learn7357Paper 1 · PureTrigonometryCompound-angle formulae
Trigonometry · Subtopic 4 of 12

Compound-Angle Formulae

sin⁡(A+B)\sin(A+B) is not sin⁡A+sin⁡B\sin A+\sin B. The correct formulae for sin⁡(A±B)\sin(A\pm B), cos⁡(A±B)\cos(A\pm B) and tan⁡(A±B)\tan(A\pm B), how to use them for exact values, and how they turn awkward equations into tan⁡θ=k\tan\theta=k.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • State and use the formulae for sin⁡(A±B)\sin(A\pm B), cos⁡(A±B)\cos(A\pm B) and tan⁡(A±B)\tan(A\pm B).
  • Find exact values such as sin⁡75∘\sin75^\circ and tan⁡15∘\tan15^\circ.
  • Find sin⁡(A±B)\sin(A\pm B) from the ratios of AA and BB, using right-angled triangles.
  • Expand compound angles to solve equations and prove identities.
Pause and think

Is sin⁡(30∘+60∘)\sin(30^\circ+60^\circ) equal to sin⁡30∘+sin⁡60∘\sin30^\circ+\sin60^\circ? Check with numbers.

Section 2 of 14

Core idea

Start hereAdding angles is not adding sines

sin⁡30∘=0.5\sin30^\circ=0.5 and sin⁡60∘≈0.866\sin60^\circ\approx0.866. Is sin⁡90∘=0.5+0.866=1.366\sin90^\circ=0.5+0.866=1.366? No: sin⁡90∘=1\sin90^\circ=1, and no sine can exceed 11. Combining angles needs a proper formula, and proving it is a lovely piece of geometry.

The big idea

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B; cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B; tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}.

1. The tempting mistake

sin(x + 60°) against sin x + sin 60°
50100150200250300350−1.5−1−0.50.511.520x (degrees)ysin(x + 60°)sin x + sin 60° (wrong)
The two graphs are completely different. The correct expansion is sin⁡xcos⁡60∘+cos⁡xsin⁡60∘\sin x\cos60^\circ+\cos x\sin60^\circ.

2. Proof for cos(A − B)

Using the unit circle and the distance formula
  1. P=(cos⁡A,sin⁡A),Q=(cos⁡B,sin⁡B)P=(\cos A,\sin A),\quad Q=(\cos B,\sin B)
    Two points on the unit circle; the angle between OPOP and OQOQ is A−BA-B.
  2. PQ2=(cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2
    Distance formula.
  3. =(cos⁡2A+sin⁡2A)+(cos⁡2B+sin⁡2B)−2(cos⁡Acos⁡B+sin⁡Asin⁡B)=(\cos^2A+\sin^2A)+(\cos^2B+\sin^2B)-2(\cos A\cos B+\sin A\sin B)
    Expand both squares and regroup.
  4. =2−2(cos⁡Acos⁡B+sin⁡Asin⁡B)=2-2(\cos A\cos B+\sin A\sin B)
    Each bracket is 11.
  5. PQ2=12+12−2cos⁡(A−B)=2−2cos⁡(A−B)PQ^2=1^2+1^2-2\cos(A-B)=2-2\cos(A-B)
    Cosine rule in triangle OPQOPQ.
  6. cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B)=\cos A\cos B+\sin A\sin B
    Equate the two expressions.
Replacing BB by −B-B (using cos⁡(−B)=cos⁡B\cos(-B)=\cos B, sin⁡(−B)=−sin⁡B\sin(-B)=-\sin B) gives cos⁡(A+B)\cos(A+B); using sin⁡x=cos⁡(90∘−x)\sin x=\cos(90^\circ-x) gives the sine formulas; dividing gives tan.
Compound-angle formulaesin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\begin{aligned}\sin(A\pm B)&=\sin A\cos B\pm\cos A\sin B\\\cos(A\pm B)&=\cos A\cos B\mp\sin A\sin B\\\tan(A\pm B)&=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}\end{aligned}
Common misconceptionThe cos sign flip cos⁡(A+B)\cos(A+B) has a minus; cos⁡(A−B)\cos(A-B) has a plus.

3. Using them

Worked exampleExact sin⁡75∘\sin75^\circ

Find exactly.

Goal
Surd form.
Strategy
75=45+3075=45+30.
  1. sin⁡45cos⁡30+cos⁡45sin⁡30\sin45\cos30+\cos45\sin30
    Formula.
  2. =22⋅32+22⋅12=6+24=\tfrac{\sqrt2}2\cdot\tfrac{\sqrt3}2+\tfrac{\sqrt2}2\cdot\tfrac12=\frac{\sqrt6+\sqrt2}4
    Exact values.
Answer6+24\dfrac{\sqrt6+\sqrt2}4.
✓ Check2.449+1.4144=0.966=sin⁡75∘\frac{2.449+1.414}{4}=0.966=\sin75^\circ ✓.
Worked examplesin⁡(x+30∘)=2cos⁡x\sin(x+30^\circ)=2\cos x, 0≤x≤360∘0\le x\le360^\circ

Solve.

Goal
All xx.
Strategy
Expand, collect, get tan⁡x\tan x.
  1. 32sin⁡x+12cos⁡x=2cos⁡x\tfrac{\sqrt3}2\sin x+\tfrac12\cos x=2\cos x
    Expand.
  2. 32sin⁡x=32cos⁡x\tfrac{\sqrt3}2\sin x=\tfrac32\cos x
    Subtract 12cos⁡x\frac12\cos x.
  3. tan⁡x=3/23/2=3\tan x=\frac{3/2}{\sqrt3/2}=\sqrt3
    Divide by 32cos⁡x\frac{\sqrt3}2\cos x.
Answerx=60∘,240∘x=60^\circ,240^\circ.
✓ Checkx=60x=60: sin⁡90=1\sin90=1, 2cos⁡60=12\cos60=1 ✓.
Worked examplecos⁡(A−B)\cos(A-B) from given ratios

sin⁡A=35\sin A=\frac35 (AA acute), cos⁡B=−513\cos B=-\frac5{13} (BB obtuse). Find cos⁡(A−B)\cos(A-B).

Goal
Exact value.
Strategy
Find missing ratios with signs; apply formula.
  1. cos⁡A=45,sin⁡B=1213\cos A=\tfrac45,\quad\sin B=\tfrac{12}{13}
    Triangles; BB obtuse so sin⁡B>0\sin B>0.
  2. 45⋅(−513)+35⋅1213=−20+3665\tfrac45\cdot(-\tfrac5{13})+\tfrac35\cdot\tfrac{12}{13}=\frac{-20+36}{65}
    Formula.
Answer1665\frac{16}{65}.
✓ Check∣16/65∣<1|16/65|<1 ✓ (plausible cosine).
Pause and think

Use the formula to simplify cos⁡(90∘−x)\cos(90^\circ-x).

Summary

  • Proof via unit circle + cosine rule.
  • Mind the cos sign flip.
  • Split awkward angles into 30, 45, 60.
Key words
Compound angle
An angle like A+BA+B.
HistoryPtolemy's tables. Around 150 AD, the astronomer Ptolemy built a table of chords (essentially sines) for every half degree. To do it, he used a geometric theorem equivalent to the compound-angle formulae, starting from a few exact values and combining them. His table was used for over a thousand years.
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