The Logic of Decomposition#
In Grade 11, you worked with single angles. In Grade 12, we learn that a single angle can be broken into two parts (Compound) or doubled (Double).
The “Addition Trap” Analogy#
Imagine a blender.
- If you blend an Apple and a Banana, the taste isn’t just “Taste of Apple + Taste of Banana”. It’s a completely new mixture.
- Trigonometry is the same: $\sin(A + B)$ is NOT simply $\sin A + \sin B$.
- We use specific “recipes” (Identities) to find the true mixture.
1. The Complete Formula Sheet#
Compound Angle Identities#
$$ \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta $$$$ \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta $$$$ \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta $$$$ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta $$Pattern to remember:
- Sine formulas: sin·cos ± cos·sin (the trig functions alternate)
- Cosine formulas: cos·cos ∓ sin·sin (the trig functions match)
- The sign flips for cosine: $\cos(\alpha + \beta)$ has a minus in the middle; $\cos(\alpha - \beta)$ has a plus.
Double Angle Identities#
These are the special case where $\beta = \alpha$:
$$ \sin 2\alpha = 2\sin\alpha\cos\alpha $$$$ \cos 2\alpha = \cos^2\alpha - \sin^2\alpha $$$$ \cos 2\alpha = 2\cos^2\alpha - 1 $$$$ \cos 2\alpha = 1 - 2\sin^2\alpha $$The three $\cos 2\alpha$ forms are all equivalent — you choose the one that simplifies your problem. See the Double Angle Deep Dive and Proving Identities pages for guidance on choosing.
2. Calculating Exact Values of Non-Special Angles#
Compound angles let you find exact trig values for angles that aren’t on the special triangle table.
Example: Find $\cos 75°$ without a calculator#
Break it down: $75° = 45° + 30°$
$$ \cos 75° = \cos(45° + 30°) $$$$ = \cos 45°\cos 30° - \sin 45°\sin 30° $$$$ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} $$$$ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} $$$$ = \frac{\sqrt{6} - \sqrt{2}}{4} $$Example: Find $\sin 15°$ without a calculator#
$15° = 45° - 30°$
$$ \sin 15° = \sin(45° - 30°) $$$$ = \sin 45°\cos 30° - \cos 45°\sin 30° $$$$ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} $$$$ = \frac{\sqrt{6} - \sqrt{2}}{4} $$Notice that $\cos 75° = \sin 15°$. This makes sense because $\cos\theta = \sin(90° - \theta)$.
3. The “Reverse Match” — Pattern Recognition#
Exams frequently give you the expanded form and ask you to simplify. You must recognize which compound angle formula was used.
Example: Simplify $\sin 70°\cos 10° - \cos 70°\sin 10°$#
This matches the pattern $\sin\alpha\cos\beta - \cos\alpha\sin\beta = \sin(\alpha - \beta)$:
$$ = \sin(70° - 10°) = \sin 60° = \frac{\sqrt{3}}{2} $$Example: Simplify $\cos 20°\cos 40° - \sin 20°\sin 40°$#
This matches $\cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha + \beta)$:
$$ = \cos(20° + 40°) = \cos 60° = \frac{1}{2} $$Example: Simplify $2\sin 3x\cos 3x$#
This matches $2\sin\alpha\cos\alpha = \sin 2\alpha$:
$$ = \sin 2(3x) = \sin 6x $$4. Using a Given Ratio to Find Compound Values#
This is a classic exam question: you’re given information about $\alpha$ and $\beta$ and must find $\sin(\alpha + \beta)$ or $\cos(\alpha - \beta)$.
Example#
Given: $\sin\alpha = \frac{3}{5}$ with $\alpha \in (90°; 180°)$ and $\cos\beta = -\frac{12}{13}$ with $\beta \in (180°; 270°)$.
Find $\cos(\alpha - \beta)$.
Step 1: Find the missing ratios using Pythagoras and CAST.
For $\alpha$ (Quadrant II — sin positive, cos negative):
$$ \cos\alpha = -\frac{4}{5} $$For $\beta$ (Quadrant III — sin negative, cos negative):
$$ \sin\beta = -\frac{5}{13} $$Step 2: Apply the formula:
$$ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta $$$$ = \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) + \left(\frac{3}{5}\right)\left(-\frac{5}{13}\right) $$$$ = \frac{48}{65} + \left(-\frac{15}{65}\right) $$$$ = \frac{48 - 15}{65} = \frac{33}{65} $$5. The Negative Angle Identities#
These are useful shortcuts:
- $\sin(-\theta) = -\sin\theta$ (sine is an odd function)
- $\cos(-\theta) = \cos\theta$ (cosine is an even function)
- $\tan(-\theta) = -\tan\theta$ (tangent is an odd function)
🚨 Common Mistakes#
- The $\sin(A+B) = \sin A + \sin B$ Error: This is the single most common mistake in Grade 12 Trig. Never distribute a trig function into brackets like it’s a number.
- Sign flip for cosine: $\cos(\alpha + \beta)$ has a MINUS in the middle. $\cos(\alpha - \beta)$ has a PLUS. Many students get this backwards.
- CAST diagram errors: When given $\sin\alpha = \frac{3}{5}$ in Q2, students often forget that $\cos\alpha$ must be negative. Always draw a quick sketch.
- Square Root Signs: When solving $\sin^2 A = \frac{1}{4}$, remember that $\sin A = \pm \frac{1}{2}$. Check your CAST diagram to determine the correct sign.
- Not recognizing the reverse pattern: If you see $\cos x\cos y + \sin x\sin y$ and don’t recognize it as $\cos(x - y)$, you’ll waste time trying to simplify it the hard way.
💡 Pro Tip: The “Reverse” Match#
Exam questions often give you the expanded form (e.g., $\sin 40°\cos 10° - \cos 40°\sin 10°$) and ask you to simplify it. Don’t try to evaluate each term separately. Look for the formula that matches the pattern! This is $\sin(40° - 10°) = \sin 30° = 0.5$.