The $R\cos(\theta \pm \alpha)$ Form

Builds on: aqa-alevel-maths-pure-2 / further-trigonometry — this lesson is the A2-level deep dive.

This lesson covers a powerful technique for rewriting expressions of the form $a\sin\theta + b\cos\theta$ as a single trigonometric function $R\sin(\theta + \alpha)$ or $R\cos(\theta + \alpha)$ . This technique — often called the "harmonic form" or " $R$ - $\alpha$ method" — is essential for solving equations, finding maximum and minimum values, and modelling periodic phenomena.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the harmonic form $R\sin(\theta \pm \alpha)$ and $R\cos(\theta \pm \alpha)$ for linear combinations of $\sin$ and $\cos$ content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

The General Result

Any expression of the form $a\sin\theta + b\cos\theta$ can be written as:

a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)

where $R > 0$ and $0 < \alpha < 2\pi$ (or equivalently $0° < \alpha < 360°$ ).

Similarly, it can be written in the forms:

\begin{aligned} a\sin\theta + b\cos\theta &\equiv R\cos(\theta - \beta) \quad\text{(alternative form)} \\ a\cos\theta + b\sin\theta &\equiv R\cos(\theta - \alpha) \\ a\cos\theta - b\sin\theta &\equiv R\cos(\theta + \alpha) \end{aligned}

The choice of form depends on the question.

Finding R and α

Method for $a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$

Expand the right-hand side using the addition formula:

R\sin(\theta + \alpha) = R\sin\theta \cos\alpha + R\cos\theta \sin\alpha

Compare coefficients with $a\sin\theta + b\cos\theta$ :

\begin{aligned} R \cos \alpha = a ... (1) \\ R \sin \alpha = b ... (2) \end{aligned}

From these two equations:

Finding R: Square and add equations (1) and (2):

\begin{aligned} R^{2}\cos^{2}\alpha + R^{2}\sin^{2}\alpha &= a^{2} + b^{2} \\ R^{2}(\cos^{2}\alpha + \sin^{2}\alpha) &= a^{2} + b^{2} \\ R^{2} &= a^{2} + b^{2} \\ R &= \sqrt{a^{2} + b^{2}} \end{aligned}

Finding α: Divide equation (2) by equation (1):

\tan \alpha = b/a

Use the signs of $R\cos\alpha = a$ and $R\sin\alpha = b$ to determine the correct quadrant for $\alpha$ .

Method for $a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)$

Expand:

R\cos(\theta - \alpha) = R\cos\theta \cos\alpha + R\sin\theta \sin\alpha

Compare coefficients:

\begin{aligned} R \cos \alpha = a ... (1) \\ R \sin \alpha = b ... (2) \end{aligned}

Same process: $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = b/a$ .

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examples

Example 1: Express $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ , where $R > 0$ and $0° < \alpha < 90°$ .

R\sin(\theta + \alpha) = R\sin\theta \cos\alpha + R\cos\theta \sin\alpha

Comparing: $R\cos\alpha = 3$ and $R\sin\alpha = 4$ .

\begin{aligned} R &= \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 \\ \tan\alpha &= \frac{4}{3} \\ \alpha &= \arctan\!\left(\frac{4}{3}\right) = 53.13^\circ \ (\text{2 d.p.}) \end{aligned}

Therefore: $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13°)$

Example 2: Express $\sqrt{3}\cos\theta + \sin\theta$ in the form $R\cos(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ .

R\cos(\theta - \alpha) = R\cos\theta \cos\alpha + R\sin\theta \sin\alpha

Comparing: $R\cos\alpha = \sqrt{3}$ and $R\sin\alpha = 1$ .

\begin{aligned} R &= \sqrt{3 + 1} = \sqrt{4} = 2 \\ \tan\alpha &= \frac{1}{\sqrt{3}} \\ \alpha &= \frac{\pi}{6} \end{aligned}

Therefore: $\sqrt{3}\cos\theta + \sin\theta = 2\cos(\theta - \tfrac{\pi}{6})$

Example 3: Express $5\cos x - 12\sin x$ in the form $R\cos(x + \alpha)$ , where $R > 0$ and $0° < \alpha < 90°$ .

R\cos(x + \alpha) = R\cos x \cos\alpha - R\sin x \sin\alpha

Comparing: $R\cos\alpha = 5$ and $R\sin\alpha = 12$ .

\begin{aligned} R &= \sqrt{25 + 144} = \sqrt{169} = 13 \\ \tan\alpha &= \frac{12}{5} \\ \alpha &= \arctan\!\left(\frac{12}{5}\right) = 67.38^\circ \ (\text{2 d.p.}) \end{aligned}

Therefore: $5\cos x - 12\sin x = 13\cos(x + 67.38°)$

Finding Maximum and Minimum Values

Since $-1 \leq \sin(\theta + \alpha) \leq 1$ and $-1 \leq \cos(\theta + \alpha) \leq 1$ , the expressions have:

Maximum value of $R$ (when the $\sin$ or $\cos$ term equals 1)
Minimum value of $-R$ (when the $\sin$ or $\cos$ term equals $-1$ )

Example 4: Find the maximum and minimum values of $3\sin\theta + 4\cos\theta$ , and the values of $\theta$ in $[0°, 360°]$ at which they occur.

From Example 1: $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13°)$

Maximum value = 5, when $\sin(\theta + 53.13°) = 1$ , i.e., $\theta + 53.13° = 90°$ , so $\theta = 36.87° \approx 36.9°$ .

Minimum value = $-5$ , when $\sin(\theta + 53.13°) = -1$ , i.e., $\theta + 53.13° = 270°$ , so $\theta = 216.87° \approx 216.9°$ .

Example 5: Find the range of $f(x) = 7 - 2\sin x + 3\cos x$ .

First express $-2\sin x + 3\cos x$ in $R$ form:

R = \sqrt{4 + 9} = \sqrt{13}

So $-2\sin x + 3\cos x$ ranges from $-\sqrt{13}$ to $\sqrt{13}$ .

Therefore $f(x)$ ranges from $7 - \sqrt{13}$ to $7 + \sqrt{13}$ .

That is: $7 - \sqrt{13} \leq f(x) \leq 7 + \sqrt{13}$ (approximately 3.39 to 10.61).

Solving Equations

Example 6: Solve $3\sin\theta + 4\cos\theta = 2$ for $0° \leq \theta \leq 360°$ .

From Example 1: $5\sin(\theta + 53.13°) = 2$

\begin{aligned} \sin(\theta + 53.13^\circ) &= \frac{2}{5} = 0.4 \\ \theta + 53.13^\circ &= \arcsin(0.4) = 23.58^\circ \ \text{or} \ 180^\circ - 23.58^\circ = 156.42^\circ \end{aligned}

$\theta + 53.13° = 23.58°$ : $\theta = -29.55°$ (not in range) $\theta + 53.13° = 156.42°$ : $\theta = 103.29°$

For the next cycle, add $360°$ : $\theta + 53.13° = 23.58° + 360° = 383.58°$ : $\theta = 330.45°$

Solutions: $\theta = 103.3°$ and $\theta = 330.4°$ (1 d.p.)

Example 7: Solve $\sqrt{3}\sin x - \cos x = 1$ for $0 \leq x \leq 2\pi$ .

Express $\sqrt{3}\sin x - \cos x$ as $R\sin(x - \alpha)$ :

R \sin(x - \alpha ) = R \sin x \cos \alpha - R \cos x \sin \alpha

Comparing: $R\cos\alpha = \sqrt{3}$ and $R\sin\alpha = 1$ .

\begin{aligned} R &= \sqrt{3 + 1} = 2 \\ \tan\alpha &= \frac{1}{\sqrt{3}}, \ \text{so} \ \alpha = \frac{\pi}{6} \end{aligned}

So: $2\sin(x - \tfrac{\pi}{6}) = 1$ , meaning $\sin(x - \tfrac{\pi}{6}) = \tfrac{1}{2}$ .

\begin{aligned} x - \pi /6 = \pi /6 or x - \pi /6 = 5 \pi /6 \\ x = \pi /3 or x = \pi \end{aligned}

Solutions: $x = \tfrac{\pi}{3}$ and $x = \pi$

Modelling Applications

The $R\cos(\theta \pm \alpha)$ form is particularly useful in modelling situations where two oscillations combine, such as:

Tidal heights: $h = a + b\sin(ct) + d\cos(ct)$
Temperature models: $T = p + q\cos(rt) + s\sin(rt)$
Sound waves: superposition of sinusoidal signals

By converting to a single trigonometric function, you can easily identify the amplitude ( $R$ ), phase shift ( $\alpha$ ), and find maximum/minimum values.

Synoptic Links

This topic connects to:

Simple harmonic motion (aqa-alevel-maths-mechanics / simple-harmonic-motion) — superposed sinusoidal displacements combine into $R\sin(\omega t + \alpha)$ form.
Stationary points of trig functions (aqa-alevel-maths-calculus-applications / stationary-points) — converting to harmonic form makes max/min trivial: extrema are $\pm R$ .
Solving trig equations (aqa-alevel-maths-trigonometry-depth / solving-trigonometric-equations) — $a\sin\theta + b\cos\theta = k$ becomes $R\sin(\theta + \alpha) = k$ , a single-argument equation.

Summary

$a\sin\theta + b\cos\theta$ can be written as $R\sin(\theta + \alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = b/a$ .
$a\cos\theta + b\sin\theta$ can be written as $R\cos(\theta - \alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = b/a$ .
The maximum value is $R$ , the minimum value is $-R$ .
To solve equations, rewrite in $R$ form, then solve the resulting single trig equation.
Always check the quadrant for $\alpha$ using the signs of $R\cos\alpha$ and $R\sin\alpha$ .

Exam Tip: The choice between $R\sin(\theta + \alpha)$ and $R\cos(\theta - \alpha)$ depends on the form of the question — use whichever the question specifies. When finding $\alpha$ , always use the equations $R\cos\alpha = \ldots$ and $R\sin\alpha = \ldots$ to check which quadrant $\alpha$ is in, rather than blindly using $\tan^{-1}$ . Show the comparison of coefficients clearly and state the values of $R$ and $\alpha$ before writing the final answer.

A-Level Deep Dive: R cos(θ ± α) and the Harmonic Form

Spec mapping

AQA A-Level Mathematics (7357) specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), Year 2 sub-strand covers express $a \cos\theta + b \sin\theta$ in the equivalent forms $R \sin(\theta \pm \alpha)$ or $R \cos(\theta \pm \alpha)$ (refer to the official specification document for exact wording). This is one of the highest-value Year 2 trigonometric techniques: it is examined directly on Paper 2 and surfaces synoptically in Section J (Differentiation) when stationary points of $f(\theta) = a \cos\theta + b \sin\theta$ are sought, in Section L (Numerical methods) when iterative solvers run on harmonic functions, and in Mechanics Paper 3, Section O (Kinematics) under simple harmonic motion. The AQA formula booklet does list the compound-angle identities that underpin the derivation, but the harmonic form itself must be derived in every solution — it is not given.

Worked example with full mark scheme

Question (8 marks):

(a) Express $3 \sin\theta - 4 \cos\theta$ in the form $R \sin(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ , giving the exact value of $R$ and $\alpha$ to 3 decimal places. (4)

(b) Hence solve $3 \sin\theta - 4 \cos\theta = 2.5$ for $0 \leq \theta \leq 2\pi$ , giving answers to 3 decimal places. (4)

Solution with mark scheme:

(a) Step 1 — expand the target form.

$R \sin(\theta - \alpha) = R \sin\theta \cos\alpha - R \cos\theta \sin\alpha$

Comparing coefficients with $3 \sin\theta - 4 \cos\theta$ :

$R \cos\alpha = 3, \qquad R \sin\alpha = 4$

M1 — correct expansion of $R \sin(\theta - \alpha)$ using the compound-angle identity, and correct comparison of coefficients. The most common error is matching $R \sin\alpha = -4$ (carrying the sign incorrectly); the minus sign in $R \sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$ is already absorbed by writing $R \sin\alpha = 4$ to match the coefficient $-4$ of $\cos\theta$ .

Step 2 — find $R$ .

Square and add: $(R\cos\alpha)^2 + (R\sin\alpha)^2 = R^2(\cos^2\alpha + \sin^2\alpha) = R^2$ . So:

$R^2 = 3^2 + 4^2 = 9 + 16 = 25 \implies R = 5$

A1 — $R = 5$ , taking the positive root since the question states $R > 0$ .

Step 3 — find $\alpha$ .

Divide: $\dfrac{R \sin\alpha}{R \cos\alpha} = \tan\alpha = \dfrac{4}{3}$ .

So $\alpha = \arctan(4/3) = 0.927$ rad (3 d.p.).

M1 — using $\tan\alpha = (R\sin\alpha)/(R\cos\alpha)$ to obtain $\alpha$ . Both $R\cos\alpha$ and $R\sin\alpha$ are positive here, so $\alpha$ is in the first quadrant — consistent with the stated range $0 < \alpha < \pi/2$ .

A1 — $3 \sin\theta - 4 \cos\theta = 5 \sin(\theta - 0.927)$ .

(b) Step 1 — substitute the harmonic form.

$5 \sin(\theta - 0.927) = 2.5 \implies \sin(\theta - 0.927) = 0.5$

M1 — substituting and isolating $\sin(\theta - \alpha)$ .

Step 2 — solve over the shifted interval.

Let $\phi = \theta - 0.927$ . The original interval $0 \leq \theta \leq 2\pi$ becomes $-0.927 \leq \phi \leq 5.356$ . Solve $\sin\phi = 0.5$ on this interval.

Principal value: $\phi = \arcsin(0.5) = \pi/6 \approx 0.524$ . The supplementary solution: $\phi = \pi - \pi/6 = 5\pi/6 \approx 2.618$ . Adding $2\pi$ to the first: $\phi \approx 6.807$ (out of range). Subtracting $2\pi$ from the first: $\phi \approx -5.759$ (out of range).

M1 — generating the principal and supplementary solutions, and checking which lie in the shifted interval.

Step 3 — back-substitute.

$\theta = \phi + 0.927$ :

$\theta_1 = 0.524 + 0.927 = 1.451$ rad
$\theta_2 = 2.618 + 0.927 = 3.545$ rad

A1 — both solutions found.

A1 — answers given to 3 decimal places as requested: $\theta = 1.451$ rad, $3.545$ rad.

Total: 8 marks (M3 A4 split as shown, plus 1 method/accuracy mark for the systematic shifted-interval treatment).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): $f(\theta) = 7 \cos\theta + 24 \sin\theta$ for $0 \leq \theta \leq 2\pi$ .

(a) Express $f(\theta)$ in the form $R \cos(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ . (3)

(b) Hence find the maximum value of $f(\theta)$ and the smallest non-negative value of $\theta$ at which it occurs. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — expanding $R \cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$ and matching coefficients: $R\cos\alpha = 7$ , $R\sin\alpha = 24$ .
A1 (AO1.1b) — $R = \sqrt{49 + 576} = \sqrt{625} = 25$ .
A1 (AO1.1b) — $\alpha = \arctan(24/7) \approx 1.287$ rad.

(b)

B1 (AO2.2a) — recognising that $\cos$ is bounded above by $1$ , so the maximum of $25 \cos(\theta - \alpha)$ is $25$ .
M1 (AO2.5) — maximum occurs when $\cos(\theta - \alpha) = 1$ , i.e. $\theta - \alpha = 0$ (smallest non-negative case).
A1 (AO1.1b) — $\theta = \alpha \approx 1.287$ rad.

Total: 6 marks split AO1 = 4, AO2 = 2. Harmonic-form questions reward AO2 reasoning at the moment the candidate connects "max of $\cos$ is 1" to the value of $\theta$ — a small piece of mathematical interpretation that earns the bulk of the AO2 credit.

Synoptic links

Connects to:

Compound-angle identities (Section E earlier in Year 2): the harmonic form is derived from $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ and $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ . Without fluency in these, the harmonic conversion cannot be reconstructed in an exam.
Trigonometric equations (Section E): equations of the form $a \cos\theta + b \sin\theta = c$ are unsolvable directly but trivial after harmonic conversion. The technique transforms the equation into $R \sin(\theta - \alpha) = c$ or $R \cos(\theta - \alpha) = c$ , which is solved by the standard $\arcsin$ / $\arccos$ + symmetry routine.
Maximum / minimum problems without calculus (Section J adjacent): because $|\sin| \leq 1$ and $|\cos| \leq 1$ , the function $a\cos\theta + b\sin\theta$ has maximum $R = \sqrt{a^2 + b^2}$ and minimum $-R$ . This is faster than differentiating and solving $f'(\theta) = 0$ — and AQA examiners reward the elegance.
Mechanics Paper 3 — Simple Harmonic Motion (Section O): the displacement of a particle in SHM is $x(t) = a\cos\omega t + b \sin\omega t$ , which converts to $x(t) = R \cos(\omega t - \phi)$ with amplitude $R$ and phase $\phi$ . The harmonic form is the natural physical description of an oscillator.
Mathematical modelling (Section A AO3 — problem-solving): any superposition of two sinusoids of the same frequency — tides, AC voltage, sound waves — collapses to a single sinusoid via the harmonic form. Modelling questions on Paper 2 routinely use this in coastal-tide, traffic-noise, or pendulum contexts.

Mark-scheme literacy

Harmonic-form questions on AQA 7357 Paper 2 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Expanding the target form, comparing coefficients, computing $R$ via $\sqrt{a^2 + b^2}$ , computing $\alpha$ via $\arctan$
AO2 (reasoning / interpretation)	25–35%	Choosing whether the question demands $R\sin$ or $R\cos$ , identifying the correct quadrant for $\alpha$ , connecting $
AO3 (problem-solving)	0–15%	Modelling-style "tide reaches its maximum at …" or "noise level exceeds 80 dB when …" prompts

The R cos(θ ± α) Form

The $R\cos(\theta \pm \alpha)$ Form

The General Result

Finding R and α

Method for $a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$

Method for $a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)$

Specimen Question (Modelled on the AQA Paper Format)

Worked Examples

Finding Maximum and Minimum Values

Solving Equations

Modelling Applications

Synoptic Links

Summary

A-Level Deep Dive: R cos(θ ± α) and the Harmonic Form

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

More in Mathematics

The R cos(θ ± α) Form

The Rcos⁡(θ±α)R\cos(\theta \pm \alpha)Rcos(θ±α) Form

The General Result

Finding R and α

Method for asin⁡θ+bcos⁡θ≡Rsin⁡(θ+α)a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)asinθ+bcosθ≡Rsin(θ+α)

Method for acos⁡θ+bsin⁡θ≡Rcos⁡(θ−α)a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)acosθ+bsinθ≡Rcos(θ−α)

Specimen Question (Modelled on the AQA Paper Format)

Worked Examples

Finding Maximum and Minimum Values

Solving Equations

Modelling Applications

Synoptic Links

Summary

A-Level Deep Dive: R cos(θ ± α) and the Harmonic Form

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

More in Mathematics

The $R\cos(\theta \pm \alpha)$ Form

Method for $a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$

Method for $a\cos\theta + b\sin\theta \equiv R\cos(\theta - \alpha)$