The R cos(θ ± α) Form

This lesson covers a powerful technique for rewriting expressions of the form a sin θ + b cos θ as a single trigonometric function R sin(θ + α) or R cos(θ + α). 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.

The General Result

Any expression of the form a sin θ + b cos θ can be written as:

a sin θ + b cos θ ≡ R sin(θ + α)

where R > 0 and 0 < α < 2π (or equivalently 0° < α < 360°).

Similarly, it can be written in the forms:

a sin θ + b cos θ ≡ R cos(θ − β)    (alternative form)
a cos θ + b sin θ ≡ R cos(θ − α)
a cos θ − b sin θ ≡ R cos(θ + α)

The choice of form depends on the question.

Finding R and α

Method for a sin θ + b cos θ ≡ R sin(θ + α)

Expand the right-hand side using the addition formula:

R sin(θ + α) = R sin θ cos α + R cos θ sin α

Compare coefficients with a sin θ + b cos θ:

R cos α = a    ... (1)
R sin α = b    ... (2)

From these two equations:

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

R²cos²α + R²sin²α = a² + b²
R²(cos²α + sin²α) = a² + b²
R² = a² + b²
R = √(a² + b²)

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

tan α = b/a

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

Method for a cos θ + b sin θ ≡ R cos(θ − α)

Expand:

R cos(θ − α) = R cos θ cos α + R sin θ sin α

Compare coefficients:

R cos α = a    ... (1)
R sin α = b    ... (2)

Same process: R = √(a² + b²) and tan α = b/a.

Worked Examples

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

R sin(θ + α) = R sin θ cos α + R cos θ sin α

Comparing: R cos α = 3 and R sin α = 4.

R = √(3² + 4²) = √(9 + 16) = √25 = 5
tan α = 4/3
α = arctan(4/3) = 53.13° (2 d.p.)

Therefore: 3sin θ + 4cos θ = 5sin(θ + 53.13°)

Example 2: Express √3 cos θ + sin θ in the form R cos(θ − α), where R > 0 and 0 < α < π/2.

R cos(θ − α) = R cos θ cos α + R sin θ sin α

Comparing: R cos α = √3 and R sin α = 1.

R = √(3 + 1) = √4 = 2
tan α = 1/√3
α = π/6

Therefore: √3 cos θ + sin θ = 2cos(θ − π/6)

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

R cos(x + α) = R cos x cos α − R sin x sin α

Comparing: R cos α = 5 and R sin α = 12.

R = √(25 + 144) = √169 = 13
tan α = 12/5
α = arctan(12/5) = 67.38° (2 d.p.)

Therefore: 5cos x − 12sin x = 13cos(x + 67.38°)

Finding Maximum and Minimum Values

Since −1 ≤ sin(θ + α) ≤ 1 and −1 ≤ cos(θ + α) ≤ 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 3sin θ + 4cos θ, and the values of θ in [0°, 360°] at which they occur.

From Example 1: 3sin θ + 4cos θ = 5sin(θ + 53.13°)

Maximum value = 5, when sin(θ + 53.13°) = 1, i.e., θ + 53.13° = 90°, so θ = 36.87° ≈ 36.9°.

Minimum value = −5, when sin(θ + 53.13°) = −1, i.e., θ + 53.13° = 270°, so θ = 216.87° ≈ 216.9°.

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

First express −2sin x + 3cos x in R form:

R = √(4 + 9) = √13

So −2sin x + 3cos x ranges from −√13 to √13.

Therefore f(x) ranges from 7 − √13 to 7 + √13.

That is: 7 − √13 ≤ f(x) ≤ 7 + √13 (approximately 3.39 to 10.61).

Solving Equations

Example 6: Solve 3sin θ + 4cos θ = 2 for 0° ≤ θ ≤ 360°.

From Example 1: 5sin(θ + 53.13°) = 2

sin(θ + 53.13°) = 2/5 = 0.4
θ + 53.13° = arcsin(0.4) = 23.58° or 180° − 23.58° = 156.42°

θ + 53.13° = 23.58°: θ = −29.55° (not in range) θ + 53.13° = 156.42°: θ = 103.29°

For the next cycle, add 360°: θ + 53.13° = 23.58° + 360° = 383.58°: θ = 330.45°

Solutions: θ = 103.3° and θ = 330.4° (1 d.p.)

Example 7: Solve √3 sin x − cos x = 1 for 0 ≤ x ≤ 2π.

Express √3 sin x − cos x as R sin(x − α):

R sin(x − α) = R sin x cos α − R cos x sin α

Comparing: R cos α = √3 and R sin α = 1.

R = √(3 + 1) = 2
tan α = 1/√3, so α = π/6

So: 2sin(x − π/6) = 1, meaning sin(x − π/6) = 1/2.

x − π/6 = π/6 or x − π/6 = 5π/6
x = π/3 or x = π

Solutions: x = π/3 and x = π

Modelling Applications

The R cos(θ ± α) 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 (α), and find maximum/minimum values.

Summary

a sin θ + b cos θ can be written as R sin(θ + α) where R = √(a² + b²) and tan α = b/a.
a cos θ + b sin θ can be written as R cos(θ − α) where R = √(a² + b²) and tan α = 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 α using the signs of R cos α and R sin α.

Exam Tip: The choice between Rsin(θ + α) and Rcos(θ − α) depends on the form of the question — use whichever the question specifies. When finding α, always use the equations R cos α = ... and R sin α = ... to check which quadrant α is in, rather than blindly using tan⁻¹. Show the comparison of coefficients clearly and state the values of R and α 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

Examiner-rewarded phrasing: "expanding $R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$ and comparing coefficients"; "since both $R\cos\alpha > 0$ and $R\sin\alpha > 0$ , $\alpha$ lies in the first quadrant"; "the maximum value of $R\cos(\theta - \alpha)$ is $R$ , occurring when $\theta - \alpha = 2k\pi$ for integer $k$ ".

The R cos(θ ± α) Form

The R cos(θ ± α) Form

The General Result

Finding R and α

Method for a sin θ + b cos θ ≡ R sin(θ + α)

Method for a cos θ + b sin θ ≡ R cos(θ − α)

Worked Examples

Finding Maximum and Minimum Values

Solving Equations

Modelling Applications

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

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