Compound Angle Formulas
A navigator calculates a bearing split into two components. A physicist decomposes a wave into two frequencies. Behind both lies the same elegant machinery: the compound angle formulas. These six identities let you expand $\sin(A \pm B)$, $\cos(A \pm B)$, and $\tan(A \pm B)$ into products of simpler functions — and once you see why sine doesn't distribute over addition, you'll never make that mistake again.
Is $\sin(60° + 30°) = \sin 60° + \sin 30°$? Without using the compound angle formula, test this by calculating both sides numerically. What do you notice? Write your prediction and working before reading on.
Every compound angle question requires exactly two moves — and missing either one costs marks.
Move 1 — Recognise the split. Identify which compound angle formula applies. Write $75° = 45° + 30°$, or spot that $\sin(A - B)$ has a minus sign in front of the $\cos A \sin B$ term. The key is to rewrite the angle as a sum or difference of standard angles before reaching for any formula.
Move 2 — Apply and simplify. Substitute known exact values for $\sin$ and $\cos$ of the component angles, multiply out, and simplify surds. Resist the urge to use a calculator — the whole point of compound angle formulas is to find exact answers from exact inputs.
Key facts
- The six compound angle formulas for $\sin$, $\cos$, and $\tan$ of $A \pm B$
- $\sin(A+B) \neq \sin A + \sin B$ — sine does not distribute over addition
- Standard exact values: $\sin/\cos$ of $30°, 45°, 60°$
Concepts
- Why the compound angle formulas involve cross-products of sine and cosine
- The sign pattern: sine keeps the sign, cosine flips the sign
- How to split non-standard angles (e.g., 75°, 15°, 105°) into standard components
Skills
- Find exact values of angles like $\sin 75°$, $\cos 15°$, $\tan 105°$
- Evaluate $\sin(A \pm B)$ given individual trig ratios without a calculator
- Simplify expressions such as $\sin(\theta + \frac{\pi}{2})$ using the formulas
These six formulas are the foundation of this lesson. They hold for any angles $A$ and $B$ — whether degrees or radians.
Memory pattern:
- Sine: mix the functions — $\sin\cos \pm \cos\sin$. The $\pm$ sign matches the original sign in $A \pm B$.
- Cosine: same functions — $\cos\cos \mp \sin\sin$. The sign in front of the second term is opposite to the original.
- Tangent: fraction with $\pm$ on top; $\mp$ on the denominator (always opposite).
(A B) = A B A B — mix the functions, keep the sign; (A B) = A B A B — same functions, flip the sign
Pause — copy all six compound angle formulas into your book: $\sin(A\pm B)$, $\cos(A\pm B)$, and $\tan(A\pm B)$, noting sine keeps the sign, cosine flips it, and tangent has opposite signs top and bottom.
Quick check: Which of the following correctly expands $\cos(A - B)$?
We just saw that $\sin(A+B) = \sin A\cos B + \cos A\sin B$ — not simply $\sin A + \sin B$. That raises a question: how do we use this formula to produce exact values for non-standard angles like $75°$ or $\frac{7\pi}{12}$? This card answers it → by splitting the angle into two standard angles (e.g. $75° = 45°+30°$) and substituting the exact values from the standard table.
Many non-standard angles can be written as the sum or difference of two standard angles. This is how exact values for angles like 75°, 15°, and 105° are computed.
Useful splits:
- $75° = 45° + 30°$
- $15° = 45° - 30°$
- $105° = 60° + 45°$
- $\dfrac{7\pi}{12} = \dfrac{\pi}{4} + \dfrac{\pi}{3}$
- $\dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}$
Reference exact values table:
To find exact trig values: split the angle into two standard angles, apply the formula, substitute exact values; Common splits: 75° = 45°+30°; 15° = 45°-30°; 105° = 60°+45°
Pause — copy the common angle splits into your book: $75°=45°+30°$, $15°=45°-30°$, $105°=60°+45°$, $\frac{7\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$, with the exact value table for $30°$, $45°$, $60°$.
Did you get this? True or false: $\sin(A - B) = \sin A \cos B - \cos A \sin B$.
Worked examples · 3 in a row, reveal as you go
Find the exact value of $\sin(75°)$.
If $\sin A = \dfrac{3}{5}$ and $\cos B = \dfrac{5}{13}$, where $A$ and $B$ are acute, find $\sin(A+B)$.
Simplify $\sin\!\left(\theta + \dfrac{\pi}{2}\right)$.
Fill the gap: To find $\sin(75°)$ exactly, write $75° = 45° +$ $°$. After applying the compound angle formula and substituting, $\sin(75°) = \dfrac{\sqrt{6} +$ $}{4}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\cos(A+B) = \cos A \cos B + \sin A \sin B$.
Odd one out: Three of these are correct compound angle formulas. Which one is the odd one out (i.e., incorrect)?
Activities · practice with the ideas
Find the exact value of $\cos(15°)$ by writing $15° = 45° - 30°$ and applying the compound angle formula.
If $\sin\alpha = \dfrac{4}{5}$ and $\cos\beta = \dfrac{12}{13}$, with both angles acute, find $\cos(\alpha + \beta)$.
Simplify $\cos(\theta - \pi)$ using the compound angle formula for cosine.
Find the exact value of $\tan(75°)$ by writing $75° = 45° + 30°$ and applying the tangent compound angle formula.
Explain in your own words why $\sin(A+B) \neq \sin A + \sin B$, and give a numerical example to support your answer.
Earlier you were asked: Is $\sin(60° + 30°) = \sin 60° + \sin 30°$?
$\sin(60° + 30°) = \sin 90° = 1$. But $\sin 60° + \sin 30° = \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} = \dfrac{\sqrt{3}+1}{2} \approx 1.366$. These are not equal — so sine does not distribute over addition. The compound angle formula gives the correct result: $\sin(A+B) = \sin A \cos B + \cos A \sin B$, which accounts for the cross-product interaction between the two angles.
Why is the sign pattern for cosine the opposite of sine? In $\cos(A+B) = \cos A \cos B - \sin A \sin B$, the minus sign comes directly from the geometry of how cosine projects onto the horizontal axis when two rotations are compounded.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the exact value of $\cos(15°)$. (2 marks)
Q2. If $\sin\alpha = \dfrac{4}{5}$ and $\cos\beta = \dfrac{12}{13}$, with both angles acute, find $\cos(\alpha + \beta)$. (3 marks)
Q3. Simplify $\sin\!\left(\theta + \dfrac{\pi}{2}\right)$ and explain what this result tells you about the relationship between the sine and cosine functions. (2 marks)
Comprehensive answers (click to reveal)
Activity 1: 1. $\cos(15°) = \cos(45°-30°) = \cos45°\cos30° + \sin45°\sin30° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$ · 2. $\cos\alpha = \frac{3}{5}$, $\sin\beta = \frac{5}{13}$; $\cos(\alpha+\beta) = \frac{3}{5}\cdot\frac{12}{13}-\frac{4}{5}\cdot\frac{5}{13} = \frac{36}{65}-\frac{20}{65} = \frac{16}{65}$ · 3. $\cos(\theta-\pi) = \cos\theta\cos\pi+\sin\theta\sin\pi = \cos\theta\cdot(-1)+\sin\theta\cdot0 = -\cos\theta$ · 4. $\tan(75°) = \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{2} = 2+\sqrt{3}$ · 5. Sine measures the vertical height of a point on the unit circle after rotation; adding two rotation angles produces a combined rotation whose height is not the sum of individual heights. E.g. $A=B=90°$: $\sin(180°)=0$ but $\sin90°+\sin90°=2$.
Q1 (2 marks): $\cos(15°) = \cos(45°-30°) = \cos45°\cos30° + \sin45°\sin30°$ [1] $= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$ [1].
Q2 (3 marks): $\cos\alpha = \frac{3}{5}$ [0.5]; $\sin\beta = \frac{5}{13}$ [0.5]; $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{3}{5}\cdot\frac{12}{13} - \frac{4}{5}\cdot\frac{5}{13}$ [1] $= \frac{36-20}{65} = \frac{16}{65}$ [1].
Q3 (2 marks): $\sin(\theta+\frac{\pi}{2}) = \sin\theta\cos\frac{\pi}{2}+\cos\theta\sin\frac{\pi}{2} = 0+\cos\theta = \cos\theta$ [1]. This shows that the cosine function is the sine function shifted left by $\frac{\pi}{2}$ — they are the same wave, just out of phase [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arena- $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ — mixed products, same sign as the original $\pm$.
- $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ — same-function products, sign is opposite.
- $\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ — numerator keeps, denominator flips.
- Use known splits (e.g., $75° = 45°+30°$) to find exact values of non-standard angles.
- When given trig ratios, always find missing values using $\sin^2 + \cos^2 = 1$ before substituting.
- Next lesson: Double Angle Formulas — a special case where $B = A$.
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