Pure · Trigonometric Modelling · Year 2

Addition Formulae

Sine and cosine don't distribute over addition — but they do obey their own precise rule for splitting a compound angle into two simpler ones.

9 min read 4 worked examples

The addition formulae

It's tempting to think $\sin(A+B) = \sin A + \sin B$, but this is false in general. The correct identities, called the addition formulae (or compound angle formulae), are:

$\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A\pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

The $\pm$ and $\mp$ symbols mean the formulae compress two identities into one line: read the top sign throughout for the "$+$" version, and the bottom sign throughout for the "$-$" version. Notice the sign flips in the cosine formula (a "$+$" on the left becomes a "$-$" on the right, and vice versa) — a common source of error.

Where these come from

These identities can be established geometrically, by constructing two right-angled triangles at angles $A$ and $B$ inside a unit circle (or a carefully constructed diagram of overlapping triangles) and comparing side lengths and areas. The full geometric argument is intricate, but the core idea is that rotating a unit vector first by angle $B$ then by angle $A$ lands it at angle $A+B$ — and expanding the coordinates of that rotated point using right-angled trigonometry on each stage produces exactly the $\cos A\cos B - \sin A\sin B$ and $\sin A\cos B+\cos A\sin B$ combinations seen above. At this level, the formulae are typically quoted and used directly rather than proved from scratch.

The tangent formula follows from the sine and cosine ones, since $\tan(A\pm B) = \dfrac{\sin(A\pm B)}{\cos(A\pm B)}$; dividing numerator and denominator through by $\cos A\cos B$ turns the sines and cosines into tangents.

Using the formulae for exact values

Angles that aren't among the standard exact-value angles ($30°,45°,60°,\dots$) can sometimes be built by adding or subtracting two that are — e.g. $75° = 45°+30°$, or $15° = 45°-30°$. Substituting into the addition formulae then produces an exact (often surd-heavy) value.

Worked example 1

Exact value via addition formula
Find the exact value of $\sin 75°$, giving your answer as a single simplified surd expression.
MethodWrite 75° as 45° + 30° and apply the sine addition formula
1
$\sin 75° = \sin(45°+30°) = \sin45°\cos30° + \cos45°\sin30°$.
Apply sin(A+B) = sin A cos B + cos A sin B with A = 45°, B = 30°.
2
$= \dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2} + \dfrac{1}{\sqrt2}\cdot\dfrac{1}{2}$.
Substitute the exact values of sin, cos at 45° and 30°.
3
$= \dfrac{\sqrt3}{2\sqrt2} + \dfrac{1}{2\sqrt2} = \dfrac{\sqrt3+1}{2\sqrt2}$.
Combine over a common denominator.
4
Rationalise: $\dfrac{\sqrt3+1}{2\sqrt2}\times\dfrac{\sqrt2}{\sqrt2} = \dfrac{\sqrt6+\sqrt2}{4}$.
Multiply top and bottom by √2 to remove the surd from the denominator.
▪ solved
Quick check

$\sin 75° \approx 0.966$; evaluating $\dfrac{\sqrt6+\sqrt2}{4} \approx \dfrac{2.449+1.414}{4} \approx 0.966$. Matches.

Worked example 2

Exact value using cosine and given ratios
Given that $\sin A = \dfrac{3}{5}$ with $A$ acute, and $\cos B = \dfrac{12}{13}$ with $B$ acute, find the exact value of $\cos(A+B)$.
MethodFind the missing ratios using right-angled triangles, then apply the formula
1
For $A$: a 3-4-5 right triangle gives $\cos A = \dfrac{4}{5}$ (both A acute, so cosine is positive).
sin A = 3/5 defines a right triangle with opposite 3, hypotenuse 5, so adjacent = 4 by Pythagoras.
2
For $B$: a 5-12-13 right triangle gives $\sin B = \dfrac{5}{13}$ (B acute, so sine is positive).
cos B = 12/13 defines adjacent 12, hypotenuse 13, so opposite = 5 by Pythagoras.
3
$\cos(A+B) = \cos A\cos B - \sin A\sin B = \dfrac45\cdot\dfrac{12}{13} - \dfrac35\cdot\dfrac{5}{13}$.
Apply the cosine addition formula (top sign: "+" case uses "−").
4
$= \dfrac{48}{65} - \dfrac{15}{65} = \dfrac{33}{65}$.
Combine over the common denominator 65.
▪ solved

Worked example 3

Solving an equation using the addition formula
Solve $\sin(x+60°) = \cos x$ for $0 \le x < 360°$.
MethodExpand the compound angle, then collect and solve for tan x
1
Expand: $\sin x\cos60° + \cos x\sin60° = \cos x$.
Apply sin(A+B) = sin A cos B + cos A sin B with A = x, B = 60°.
2
$\sin x\left(\dfrac12\right) + \cos x\left(\dfrac{\sqrt3}{2}\right) = \cos x$.
Substitute exact values for cos 60° and sin 60°.
3
$\dfrac12\sin x = \cos x - \dfrac{\sqrt3}{2}\cos x = \left(1-\dfrac{\sqrt3}{2}\right)\cos x$.
Collect all cos x terms on one side.
4
Divide both sides by $\cos x$: $\dfrac12\tan x = 1-\dfrac{\sqrt3}{2} \Rightarrow \tan x = 2-\sqrt3$.
sin x / cos x = tan x; multiply both sides by 2.
5
$\tan x = 2-\sqrt3 \approx 0.268 \Rightarrow x = 15°$ (principal value); tan has period $180°$, so $x = 15°$ or $x=195°$.
Tangent repeats every 180°, giving two solutions in [0°, 360°).
▪ solved
Examiner's eye

Dividing by $\cos x$ is valid here since $\cos x=0$ would force $\sin x=0$ too (from the rearranged equation), which is impossible simultaneously — so no solutions are lost by dividing.

Worked example 4

Proving a further identity using the addition formulae · Challenge
Show that $\dfrac{\sin(A+B)}{\cos A\cos B} = \tan A + \tan B$.
MethodExpand the numerator, then split the fraction into two simpler ones
1
Start from the left-hand side: $\dfrac{\sin(A+B)}{\cos A\cos B} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B}$.
Expand the numerator using the sine addition formula.
2
Split into two fractions: $= \dfrac{\sin A\cos B}{\cos A\cos B} + \dfrac{\cos A\sin B}{\cos A\cos B}$.
A sum in the numerator over a shared denominator splits into a sum of fractions.
3
Cancel common factors: $= \dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}$.
cos B cancels in the first fraction, cos A cancels in the second.
4
$= \tan A + \tan B$, which is the right-hand side.
sin/cos = tan by definition, for each angle.
▪ proven
Look deeper

This kind of derived identity is exactly the pattern used to build the tangent addition formula itself — dividing the sine and cosine addition formulae through by $\cos A\cos B$ and simplifying gives $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ directly.

Common mistake

Writing $\sin(A+B) = \sin A + \sin B$ or $\cos(A+B)=\cos A+\cos B$ — trigonometric functions do not distribute over addition. Always expand a compound angle using the full addition formula, never by splitting the function across the sum.

Check yourself

Find the exact value of $\cos 15°$, giving your answer as a single simplified surd expression.
(Hint: write 15° = 45° − 30° and use the cosine subtraction formula.)

Method
1
$\cos15° = \cos(45°-30°) = \cos45°\cos30° + \sin45°\sin30°$.
Apply cos(A−B) = cos A cos B + sin A sin B (bottom sign case).
2
$= \dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2} + \dfrac{1}{\sqrt2}\cdot\dfrac12 = \dfrac{\sqrt3+1}{2\sqrt2}$.
Substitute exact values and combine.
3
Rationalise: $\dfrac{\sqrt3+1}{2\sqrt2}\times\dfrac{\sqrt2}{\sqrt2} = \dfrac{\sqrt6+\sqrt2}{4}$.
Remove the surd from the denominator.
▪ solved