# Compound Angles and Double Angles

> Expand sin(A + B) and cos(A + B), use double angle formulas, write a sin x + b cos x as one wave, and solve harder trig equations.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/further-trig-identities/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Two speakers, one sound
Two speakers play the same note, but one is a little behind the other. Your ear does not hear two separate waves. It hears **one** wave, a bit louder or softer, shifted in time.

Adding $\sin x$ and $\cos x$ works the same way: the result is a single sine wave. The tools that show this are the compound angle formulas. They are also the key to integrating $\sin^2 x$ and $\cos^2 x$ in calculus.

**In short:** Compound angle formulas expand the trig of a sum or difference, like $\sin(A + B) = \sin A\cos B + \cos A\sin B$. Double angle formulas follow from them: $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 2\cos^2 A - 1$, and $a\sin x + b\cos x$ can be written as $R\sin(x + \alpha)$.

## Starting point: what you already know

- Exact values for $\dfrac{\pi}{6}$, $\dfrac{\pi}{4}$ and $\dfrac{\pi}{3}$, and the unit circle.
- $\sin^2 x + \cos^2 x = 1$ and $\tan x = \dfrac{\sin x}{\cos x}$.
- Solving equations like $\sin x = \dfrac{1}{2}$ for $0 \le x \le 2\pi$.
- $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$.

### Watch out: The big trap
$\sin(A + B)$ is **not** $\sin A + \sin B$. Try $A = B = \dfrac{\pi}{2}$: the left side is $\sin\pi = 0$, but the right side is 2.

## Compound angle formulas

### Key idea: Learn these six
$$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$$
$$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$$
$$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$$

Notice that the cosine formula **swaps** the sign: $\cos(A + B)$ has a minus in the middle.

### Worked example: An exact value
$\sin 75^\circ = \sin(45^\circ + 30^\circ)$
$$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$

## Double angle formulas

Put $B = A$ in the compound formulas.

### Key idea: Double angles
$$\sin 2A = 2\sin A\cos A$$
$$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$
$$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$

Rearranging the cosine versions gives two formulas you will use in calculus:

$$\sin^2 A = \frac{1}{2}(1 - \cos 2A) \qquad \cos^2 A = \frac{1}{2}(1 + \cos 2A)$$

### Worked example: Solve sin 2x = sin x for 0 ≤ x ≤ 2π
$2\sin x\cos x - \sin x = 0$, so $\sin x(2\cos x - 1) = 0$.

- $\sin x = 0$: $x = 0, \pi, 2\pi$
- $\cos x = \dfrac{1}{2}$: $x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$

Never divide both sides by $\sin x$, or you lose the first set of answers.

## One wave: the auxiliary angle

Any $a\sin x + b\cos x$ can be written as a single wave $R\sin(x + \alpha)$, where

$$R = \sqrt{a^2 + b^2}, \qquad \cos\alpha = \frac{a}{R}, \qquad \sin\alpha = \frac{b}{R}$$

This works because $R\sin(x + \alpha) = R\cos\alpha\sin x + R\sin\alpha\cos x$.

### Worked example: Write √3 sin x + cos x as one wave
$R = \sqrt{3 + 1} = 2$. Then $\cos\alpha = \dfrac{\sqrt{3}}{2}$ and $\sin\alpha = \dfrac{1}{2}$, so $\alpha = \dfrac{\pi}{6}$.
$$\sqrt{3}\sin x + \cos x = 2\sin\left(x + \frac{\pi}{6}\right)$$
Its biggest value is 2 and smallest is $-2$. The two speakers make one wave of height 2.

### Calculator tip: R and α in one step
For $a\sin x + b\cos x = R\sin(x + \alpha)$ with $a$ and $b$ positive, $R = \sqrt{a^2 + b^2}$ and $\alpha = \tan^{-1}\dfrac{b}{a}$. Many calculators find both at once with `Pol(`: `Pol(a, b)` gives $r = R$ and $\theta = \alpha$, in the current angle mode. Use radian mode if the answer must be in radians.

**Mind map**

```text
Further trig identities
  Compound angles
    sin(A ± B)
    cos(A ± B) swaps the sign
    tan(A ± B)
  Double angles
    sin 2A = 2 sin A cos A
    cos 2A has three forms
    sin² A and cos² A in terms of cos 2A
  Auxiliary angle
    a sin x + b cos x = R sin(x + α)
    R = √(a² + b²)
  Uses
    Exact values
    Solving equations
    Integrating sin² and cos²
```

### Year 11: Proving identities
Prove $\dfrac{\sin 2A}{1 + \cos 2A} = \tan A$.

Left side $= \dfrac{2\sin A\cos A}{1 + 2\cos^2 A - 1} = \dfrac{2\sin A\cos A}{2\cos^2 A} = \dfrac{\sin A}{\cos A} = \tan A$.

**Products to sums.** Adding and subtracting the compound angle formulas gives
$$2\sin A\cos B = \sin(A + B) + \sin(A - B) \qquad 2\cos A\cos B = \cos(A - B) + \cos(A + B)$$
$$2\sin A\sin B = \cos(A - B) - \cos(A + B)$$
So $2\sin 2x\cos 5x = \sin 7x + \sin(-3x) = \sin 7x - \sin 3x$.

**Sums to products** go the other way: $\sin P + \sin Q = 2\sin\dfrac{P + Q}{2}\cos\dfrac{P - Q}{2}$ and $\cos P + \cos Q = 2\cos\dfrac{P + Q}{2}\cos\dfrac{P - Q}{2}$. For example $\dfrac{\sin x + \sin 5x}{\cos x + \cos 5x} = \dfrac{2\sin 3x\cos 2x}{2\cos 3x\cos 2x} = \tan 3x$.

**Triple angles.** $\cos 3A = \cos(2A + A)$ expands to $4\cos^3 A - 3\cos A$, and $\sin 3A = 3\sin A - 4\sin^3 A$.

**Maximum of a quadratic in $\sin x$.** In $2\cos^2 x + \sin x$, replace $\cos^2 x$ with $1 - \sin^2 x$ and let $s = \sin x$: $-2s^2 + s + 2$ for $-1 \leq s \leq 1$. Its vertex is at $s = \frac{1}{4}$, giving the maximum $\frac{17}{8}$.

### Year 12: Solving with the auxiliary angle
Solve $\sqrt{3}\sin x + \cos x = 1$ for $0 \le x \le 2\pi$.

$2\sin\left(x + \dfrac{\pi}{6}\right) = 1$, so $\sin\left(x + \dfrac{\pi}{6}\right) = \dfrac{1}{2}$. With $\dfrac{\pi}{6} \le x + \dfrac{\pi}{6} \le \dfrac{13\pi}{6}$:
$$x + \frac{\pi}{6} = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6} \quad\Rightarrow\quad x = 0, \frac{2\pi}{3}, 2\pi$$

**Maximum and minimum.** $R\sin(x + \alpha)$ runs from $-R$ to $R$, so $5\sin x - 12\cos x$ has maximum 13 and minimum $-13$. When $b$ is negative, use $R\sin(x - \alpha)$ with $\tan\alpha = \frac{12}{5}$.

**Inequalities.** To solve $\sqrt{3}\sin x + \cos x \geq 1$, write it as $\sin\left(x + \frac{\pi}{6}\right) \geq \frac{1}{2}$, then read off where the sine is at least $\frac{1}{2}$ in the shifted interval.

### Year 12: The t formulas
With $t = \tan\dfrac{x}{2}$:
$$\sin x = \frac{2t}{1 + t^2} \qquad \cos x = \frac{1 - t^2}{1 + t^2} \qquad \tan x = \frac{2t}{1 - t^2}$$
They turn any equation in $\sin x$ and $\cos x$ into an equation in $t$. Always check $x = \pi$ separately, since $\tan\dfrac{\pi}{2}$ does not exist.

**Simplifying with $t$.** $\dfrac{1 - \cos x}{\sin x} = \dfrac{\frac{2t^2}{1 + t^2}}{\frac{2t}{1 + t^2}} = t$, because $1 - \cos x = \dfrac{2t^2}{1 + t^2}$.

**Solving with $t$.** For $\sin x + \cos x = 1$: $\dfrac{2t + 1 - t^2}{1 + t^2} = 1$ gives $2t + 1 - t^2 = 1 + t^2$, so $2t^2 - 2t = 0$ and $t = 0$ or $t = 1$. Then $\frac{x}{2} = 0$ or $\frac{\pi}{4}$, so $x = 0$ or $\frac{\pi}{2}$ (and check $x = \pi$ separately: $0 - 1 \neq 1$, so it is not a solution).

## Quick summary

- $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$.
- $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$.
- $\sin^2 A = \dfrac{1}{2}(1 - \cos 2A)$ and $\cos^2 A = \dfrac{1}{2}(1 + \cos 2A)$.
- $a\sin x + b\cos x = R\sin(x + \alpha)$ with $R = \sqrt{a^2 + b^2}$.
- Factorise, do not divide, when solving trig equations.

## Quiz

Try these questions from the chapter quiz.

1. $\sin(A + B)$ equals:
   - A. $\sin A + \sin B$
   - B. $\sin A\cos B + \cos A\sin B$
   - C. $\sin A\sin B + \cos A\cos B$
   - D. $\cos A\cos B - \sin A\sin B$
   - Answer: B. $\sin A\cos B + \cos A\sin B$. Sine mixes: sin cos plus cos sin.

2. $\cos(A + B)$ equals:
   - A. $\cos A\cos B + \sin A\sin B$
   - B. $\cos A\cos B - \sin A\sin B$
   - C. $\cos A + \cos B$
   - D. $\sin A\cos B - \cos A\sin B$
   - Answer: B. $\cos A\cos B - \sin A\sin B$. Cosine keeps cos cos and sin sin, and swaps the sign.

3. $\sin 2A$ equals:
   - A. $2\sin A$
   - B. $2\sin A\cos A$
   - C. $\sin^2 A - \cos^2 A$
   - D. $\sin A\cos A$
   - Answer: B. $2\sin A\cos A$. Put $B = A$ in $\sin(A + B)$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/further-trig-identities/#quiz