What are solving equations?

To solve asinθ + bcosθ = k, rewrite the left as Rsin(θ + α) = k, so sin(θ + α) = kR, an equation in a single ratio that you solve in the usual way.

What is wrong matching for a cosine form?

For Rcos(θ + α) the cosine coefficient matches Rcosα and the sine coefficient matches Rsinα; align with the chosen form carefully.

What is quadrant errors in α?

With the standard positive R and acute α, α = tan^-1(b/a) is acute; do not add a sign.

Express sinθ + cosθ in the form Rsin(θ + α). [3 marks]

State the maximum value of 6sinθ + 8cosθ. [2 marks]

State the minimum value of 5 + 3sinθ + 4cosθ. [2 marks]

SEAB Original-style practice: Express 3sinθ + 4cosθ in the form Rsin(θ + α), where R > 0 and 0^ < α < 90^. Hence state the maximum value of the expression.

Compare Rsin(θ + α) = Rsinθcosα + Rcosθsinα with 3sinθ + 4cosθ. So Rcosα = 3 and Rsinα = 4. Then R = sqrt(3^2 + 4^2) = 5 and tanα = 43, so α = 53.13^. Thus 3sinθ + 4cosθ = 5sin(θ + 53.13^). The maximum value is R = 5, when sin(θ + α) = 1. Markers reward matching coefficients, R = sqrt(a^2 + b^2), the angle from tanα, and the maximum value 5.

SEAB Original-style practice: The expression 5cosθ - 12sinθ is written as Rcos(θ + α). Find R and the minimum value of the expression.

Expand Rcos(θ + α) = Rcosθcosα - Rsinθsinα and compare with 5cosθ - 12sinθ. So Rcosα = 5 and Rsinα = 12, giving R = sqrt(5^2 + 12^2) = 13. The minimum value of Rcos(θ + α) is -R = -13, when cos(θ + α) = -1. Markers reward correct matching with the cosine expansion, R = 13, and the minimum value -13.

SingaporeAdditional MathematicsSyllabus dot point

How does the R-formula combine a sine and a cosine term into one, and what does it tell us about maximum and minimum values?

Express a sine plus cosine as a single R sine or R cosine function and use it to find maximum and minimum values and to solve equations

A focused answer to the O-Level A-Maths outcome on the R-formula. Writing a cosine plus sine as a single R cosine or R sine, finding R and the angle, and using the form for maxima, minima and equations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to combine an expression of the form $a\sin\theta + b\cos\theta$ into a single trigonometric term $R\sin(\theta + \alpha)$ or $R\cos(\theta \pm \alpha)$ . This is powerful because a single sine or cosine has an obvious maximum and minimum and is easy to solve, so the R-formula unlocks both optimisation and equation-solving.

The answer

The idea

A sum of a sine and a cosine of the same angle is itself a single sinusoid of that angle, just shifted. Writing it as $R\sin(\theta + \alpha)$ makes its amplitude $R$ and phase shift $\alpha$ explicit.

Finding R and the angle

Expand the target form using the addition formula and compare coefficients. For $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ :

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

so $R\cos\alpha = a$ and $R\sin\alpha = b$ . Therefore:

R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \frac{b}{a}.

Take $R$ positive and $\alpha$ acute (the standard convention).

Maximum and minimum

Since $-1 \leq \sin(\theta + \alpha) \leq 1$ , the expression ranges between $-R$ and $R$ :

\text{maximum} = R \ \text{(when the bracket} = 1\text{)}, \qquad \text{minimum} = -R \ \text{(when the bracket} = -1\text{)}.

You can also find the value of $\theta$ at which each occurs by solving $\theta + \alpha = 90^\circ$ (maximum of a sine form) or the appropriate angle.

Solving equations

To solve $a\sin\theta + b\cos\theta = k$ , rewrite the left as $R\sin(\theta + \alpha) = k$ , so $\sin(\theta + \alpha) = \dfrac{k}{R}$ , an equation in a single ratio that you solve in the usual way.

Worked example

Express $\sqrt{3}\sin\theta - \cos\theta$ as $R\sin(\theta - \alpha)$ and solve $\sqrt{3}\sin\theta - \cos\theta = 1$ for $0^\circ \leq \theta \leq 360^\circ$ .

Step 1: Match coefficients

$R\sin(\theta - \alpha) = R\cos\alpha\,\sin\theta - R\sin\alpha\,\cos\theta$ , so $R\cos\alpha = \sqrt{3}$ and $R\sin\alpha = 1$ .

Step 2: Find R and the angle

$R = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ , and $\tan\alpha = \dfrac{1}{\sqrt{3}}$ , so $\alpha = 30^\circ$ . Thus the expression is $2\sin(\theta - 30^\circ)$ .

Step 3: Reduce the equation

$2\sin(\theta - 30^\circ) = 1$ , so $\sin(\theta - 30^\circ) = \dfrac{1}{2}$ .

Step 4: Solve over the range

$\theta - 30^\circ = 30^\circ$ or $150^\circ$ , so $\theta = 60^\circ$ or $\theta = 180^\circ$ .

Common traps

Computing R with a difference: $R = \sqrt{a^2 + b^2}$ always uses a sum of squares, regardless of the sign between the terms.
Wrong matching for a cosine form: For $R\cos(\theta + \alpha)$ the cosine coefficient matches $R\cos\alpha$ and the sine coefficient matches $R\sin\alpha$ ; align with the chosen form carefully.
Forgetting the phase shift when finding where the max occurs: The maximum of $R\sin(\theta + \alpha)$ is at $\theta + \alpha = 90^\circ$ , so $\theta = 90^\circ - \alpha$ .
Reading the minimum as zero: The minimum of the combined expression is $-R$ , not $0$ .
Quadrant errors in $\alpha$: With the standard positive $R$ and acute $\alpha$ , $\alpha = \tan^{-1}(b/a)$ is acute; do not add a sign.

Examples in context

Example 1. Amplitude of a combined signal. Two alternating voltages $a\sin\omega t$ and $b\cos\omega t$ of the same frequency combine to a single oscillation of amplitude $\sqrt{a^2 + b^2}$ , which is exactly the value $R$ , telling an engineer the peak voltage.

Example 2. Greatest projection on a slope. A force resolved into components $a\sin\theta + b\cos\theta$ along a direction is maximised when the direction aligns with the force; the R-formula gives that maximum $R$ and the angle at which it happens.

Try this

Q1. Express $\sin\theta + \cos\theta$ in the form $R\sin(\theta + \alpha)$ . [3 marks]

Cue. $R = \sqrt{2}$ , $\tan\alpha = 1$ so $\alpha = 45^\circ$ : $\sqrt{2}\sin(\theta + 45^\circ)$ .

Q2. State the maximum value of $6\sin\theta + 8\cos\theta$ . [2 marks]

Cue. $R = \sqrt{36 + 64} = 10$ , so the maximum is $10$ .

Q3. State the minimum value of $5 + 3\sin\theta + 4\cos\theta$ . [2 marks]

Cue. The sine-cosine part has minimum $-5$ , so the whole expression has minimum $5 - 5 = 0$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksExpress

3\sin\theta + 4\cos\theta

in the form

R\sin(\theta + \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

. Hence state the maximum value of the expression.

Show worked answer →

Compare $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$ with $3\sin\theta + 4\cos\theta$ .

So $R\cos\alpha = 3$ and $R\sin\alpha = 4$ . Then $R = \sqrt{3^2 + 4^2} = 5$ and $\tan\alpha = \dfrac{4}{3}$ , so $\alpha = 53.13^\circ$ .

Thus $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13^\circ)$ . The maximum value is $R = 5$ , when $\sin(\theta + \alpha) = 1$ .

Markers reward matching coefficients, $R = \sqrt{a^2 + b^2}$ , the angle from $\tan\alpha$ , and the maximum value $5$ .

Original4 marksThe expression

5\cos\theta - 12\sin\theta

is written as

R\cos(\theta + \alpha)

. Find

R

and the minimum value of the expression.

Show worked answer →

Expand $R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$ and compare with $5\cos\theta - 12\sin\theta$ .

So $R\cos\alpha = 5$ and $R\sin\alpha = 12$ , giving $R = \sqrt{5^2 + 12^2} = 13$ .

The minimum value of $R\cos(\theta + \alpha)$ is $-R = -13$ , when $\cos(\theta + \alpha) = -1$ .

Markers reward correct matching with the cosine expansion, $R = 13$ , and the minimum value $-13$ .