The Double Angle Formulas#
A Double Angle is a compound angle where $\beta = \alpha$ (i.e., $\alpha + \alpha = 2\alpha$).
Sine Double Angle#
$$ \sin 2\theta = 2\sin\theta\cos\theta $$Cosine Double Angle — The “Three Faces”#
$$ \cos 2\theta = \cos^2\theta - \sin^2\theta \tag{Form 1} $$$$ \cos 2\theta = 2\cos^2\theta - 1 \tag{Form 2} $$$$ \cos 2\theta = 1 - 2\sin^2\theta \tag{Form 3} $$All three are equivalent. The power is in choosing the right one for each situation.
1. The Selection Guide#
| Situation | Best Form | Why |
|---|---|---|
| Expression has $1 + \cos 2\theta$ | Form 2: $2\cos^2\theta - 1$ | $1 + (2\cos^2\theta - 1) = 2\cos^2\theta$ |
| Expression has $1 - \cos 2\theta$ | Form 3: $1 - 2\sin^2\theta$ | $1 - (1 - 2\sin^2\theta) = 2\sin^2\theta$ |
| Only $\sin$ terms in the expression | Form 3 | Keeps everything in $\sin$ |
| Only $\cos$ terms in the expression | Form 2 | Keeps everything in $\cos$ |
| Both $\sin$ and $\cos$ present | Form 1 | Factorises as $(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)$ |
| Need to create a quadratic in $\sin$ | Form 3 | Replaces $\cos 2\theta$ with $\sin$ terms |
2. The Half-Angle Trick (Reversed Double Angles)#
Sometimes you need to go backwards — expressing $\sin^2\theta$ or $\cos^2\theta$ in terms of $\cos 2\theta$:
From Form 2: $\cos 2\theta = 2\cos^2\theta - 1$
$$ \cos^2\theta = \frac{1 + \cos 2\theta}{2} $$From Form 3: $\cos 2\theta = 1 - 2\sin^2\theta$
$$ \sin^2\theta = \frac{1 - \cos 2\theta}{2} $$These are called power-reducing (or half-angle) formulas. You’ll use them when an expression has $\sin^2$ or $\cos^2$ and you need to simplify.
3. Worked Examples#
Example 1: Basic Simplification#
Simplify $\frac{\sin 2x}{\cos x}$
$$ = \frac{2\sin x\cos x}{\cos x} = 2\sin x $$Example 2: Using $1 + \cos 2\theta$#
Simplify $\frac{\sin 2\theta}{1 + \cos 2\theta}$
Use Form 2 for $\cos 2\theta$:
$$ = \frac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta $$Example 3: Using $1 - \cos 2\theta$#
Simplify $\frac{1 - \cos 2A}{\sin 2A}$
Use Form 3 for $\cos 2A$:
$$ = \frac{1 - (1 - 2\sin^2 A)}{2\sin A\cos A} = \frac{2\sin^2 A}{2\sin A\cos A} = \frac{\sin A}{\cos A} = \tan A $$Example 4: Given a Ratio, Find Double Angle Value#
Given $\sin\theta = \frac{3}{5}$ with $\theta \in (90°; 180°)$. Find $\sin 2\theta$ and $\cos 2\theta$.
First find $\cos\theta$: In Q2, $\cos\theta$ is negative. $\cos\theta = -\frac{4}{5}$
$$ \sin 2\theta = 2\sin\theta\cos\theta = 2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) = -\frac{24}{25} $$$$ \cos 2\theta = 1 - 2\sin^2\theta = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25} $$Example 5: Creating a Quadratic Equation#
Solve $3\cos 2x - 5\sin x + 1 = 0$ for $x \in [0°; 360°]$.
Replace $\cos 2x$ with Form 3 (because the equation also has $\sin x$):
$$ 3(1 - 2\sin^2 x) - 5\sin x + 1 = 0 $$$$ 3 - 6\sin^2 x - 5\sin x + 1 = 0 $$$$ -6\sin^2 x - 5\sin x + 4 = 0 $$$$ 6\sin^2 x + 5\sin x - 4 = 0 $$Factor: $(3\sin x + 4)(2\sin x - 1) = 0$
$\sin x = -\frac{4}{3}$ (impossible — reject) or $\sin x = \frac{1}{2}$
$x = 30° + n \cdot 360°$ or $x = 150° + n \cdot 360°$
In range: $x = 30°$ or $x = 150°$
🚨 Common Mistakes#
- Wrong form of $\cos 2\theta$: If your calculation gets messier after expanding, you chose the wrong face. Go back and try another.
- Forgetting to double-check the quadrant: When given $\sin\theta$ in a specific quadrant, you must determine $\cos\theta$’s sign before calculating $\sin 2\theta$.
- Not recognising double angles in disguise: $4\sin x\cos x = 2(2\sin x\cos x) = 2\sin 2x$. Train yourself to spot these patterns.
💡 Pro Tip: The “Double Angle Detection” Skill#
Whenever you see $2\sin\theta\cos\theta$ anywhere in an expression, immediately recognise it as $\sin 2\theta$. And whenever you see $\cos^2\theta - \sin^2\theta$ or $2\cos^2\theta - 1$ or $1 - 2\sin^2\theta$, recognise it as $\cos 2\theta$. This “reverse detection” skill saves huge amounts of time in exams.
⏮️ Compound Angles | 🏠 Back to Trigonometry | ⏭️ Proving Identities