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NSWMaths Extension 1Syllabus dot point

How do we write $a \sin x + b \cos x$ as a single sinusoid, and what is this used for?

Express $a \sin x + b \cos x$ in the form $R \sin(x + \alpha)$ or $R \cos(x - \alpha)$ and use this to solve equations and find extreme values

A focused answer to the HSC Maths Extension 1 dot point on the auxiliary angle technique. Writing $a \sin x + b \cos x$ as a single sinusoid, finding the amplitude and phase, and using the result to solve equations and identify maxima and minima.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to combine $a \sin x + b \cos x$ into a single sinusoidal expression of the form $R \sin(x + \alpha)$ or $R \cos(x \pm \alpha)$ , find the amplitude $R$ and the phase $\alpha$ , and use this to solve equations or find extreme values.

The answer

The form IMATH_13

Expand using the sine sum identity:

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

For this to equal $a \sin x + b \cos x$ , match coefficients:

R \cos \alpha = a, \qquad R \sin \alpha = b.

Squaring and adding,

R^2 = a^2 + b^2 \implies R = \sqrt{a^2 + b^2} \quad (R > 0).

Dividing,

\tan \alpha = \frac{b}{a}.

Choose $\alpha$ in the quadrant determined by the signs of $a$ and $b$ (since $R \cos \alpha = a$ and $R \sin \alpha = b$ have the same signs as $a$ and $b$ ).

Other equivalent forms

Depending on convenience, $a \sin x + b \cos x$ can also be written as:

R \cos(x - \alpha), \quad \text{where } R \cos \alpha = b, R \sin \alpha = a.

R \sin(x - \alpha) = R \cos \alpha \sin x - R \sin \alpha \cos x, \quad \text{requires the } b \text{ term to be negative}.

For $a \sin x - b \cos x$ , use $R \sin(x - \alpha)$ with $R \cos \alpha = a$ , $R \sin \alpha = b$ .

The exam usually specifies which form to use; if not, $R \sin(x + \alpha)$ with $R > 0$ and $0 \le \alpha < 2 \pi$ (or $-\pi < \alpha \le \pi$ ) is the default.

Why this matters

Once $a \sin x + b \cos x$ is written as $R \sin(x + \alpha)$ :

The maximum value is $R$ , achieved when $x + \alpha = \frac{\pi}{2} + 2 n \pi$ .
The minimum value is $-R$ .
Equations like $a \sin x + b \cos x = c$ become $\sin(x + \alpha) = \frac{c}{R}$ , which solves by standard general solution.
Sketching is reduced to a single shifted, amplitude-scaled sinusoid.

Worked example

Convert a sum

Express $3 \sin x + 4 \cos x$ in the form $R \sin(x + \alpha)$ .

$R = \sqrt{9 + 16} = 5$ , $\tan \alpha = \frac{4}{3}$ .

Both $a = 3$ and $b = 4$ are positive, so $\alpha$ is in Q1: $\alpha = \arctan \frac{4}{3} \approx 53.13^\circ$ or $0.927$ rad.

$3 \sin x + 4 \cos x = 5 \sin(x + 0.927)$ (to 3 dp).

Equation

Solve $\sin x + \sqrt{3} \cos x = 1$ for $0 \le x \le 2 \pi$ .

$R = \sqrt{1 + 3} = 2$ , $\tan \alpha = \sqrt{3}$ , $\alpha = \frac{\pi}{3}$ .

Equation: $2 \sin\!\left( x + \frac{\pi}{3} \right) = 1$ , so $\sin\!\left( x + \frac{\pi}{3} \right) = \frac{1}{2}$ .

General: $x + \frac{\pi}{3} = \frac{\pi}{6} + 2 n \pi$ or $\frac{5 \pi}{6} + 2 n \pi$ .

$x = -\frac{\pi}{6} + 2 n \pi$ or $x = \frac{\pi}{2} + 2 n \pi$ .

In $[0, 2 \pi]$ : $x = \frac{\pi}{2}$ or $x = -\frac{\pi}{6} + 2 \pi = \frac{11 \pi}{6}$ .

Maximum and minimum

Find the maximum value of $5 \sin x - 12 \cos x$ and the smallest positive $x$ at which it occurs.

$R = \sqrt{25 + 144} = 13$ . Write as $13 \sin(x - \alpha)$ where $\cos \alpha = \frac{5}{13}$ and $\sin \alpha = \frac{12}{13}$ .

Maximum value is $13$ when $\sin(x - \alpha) = 1$ , that is $x - \alpha = \frac{\pi}{2}$ , $x = \frac{\pi}{2} + \alpha \approx \frac{\pi}{2} + 1.176 \approx 2.747$ rad.

Sketching

Sketch $y = \sin x + \cos x$ .

$R = \sqrt{2}$ , $\alpha = \frac{\pi}{4}$ , so $y = \sqrt{2} \sin\!\left( x + \frac{\pi}{4} \right)$ .

Amplitude $\sqrt{2}$ , period $2 \pi$ , phase shift $\frac{\pi}{4}$ to the left. The curve crosses zero at $x = -\frac{\pi}{4}$ , peaks at $x = \frac{\pi}{4}$ , returns to zero at $x = \frac{3 \pi}{4}$ .

Common traps

Sign of $\alpha$ misplaced: IMATH_1 has $\alpha$ added; $R \sin(x - \alpha)$ has $\alpha$ subtracted. The form determines whether the phase is left-shifted or right-shifted.
Wrong quadrant for $\alpha$: IMATH_6 has two candidate angles per period. Choose the one consistent with the signs of $R \cos \alpha = a$ and $R \sin \alpha = b$ .
Forgetting to halve when solving the equation: IMATH_9 requires dividing by $R$ , then taking $\arcsin$ , then accounting for both branches.
Confusing amplitude with max/min: Maximum value of $R \sin(x + \alpha) + c$ is $R + c$ , minimum is $c - R$ , amplitude is $R$ . Track each.
Negative $R$: Convention is $R > 0$ . If you write $-R \sin(x + \alpha) = R \sin(x + \alpha + \pi)$ , you can keep $R$ positive by shifting $\alpha$ by $\pi$ .

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q123 marksExpress

\sqrt{3} \sin x + \cos x

in the form

R \sin(x + \alpha)

where

R > 0

and

0 < \alpha < \frac{\pi}{2}

, and hence find the maximum value of

\sqrt{3} \sin x + \cos x

Show worked answer →

Expand $R \sin(x + \alpha) = R \cos \alpha \sin x + R \sin \alpha \cos x$ .

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

$R^2 = (\sqrt{3})^2 + 1^2 = 4$ , so $R = 2$ .

$\tan \alpha = \frac{1}{\sqrt{3}}$ , so $\alpha = \frac{\pi}{6}$ .

$\sqrt{3} \sin x + \cos x = 2 \sin\!\left( x + \frac{\pi}{6} \right)$ .

Maximum value is $2$ (achieved when $\sin = 1$ , that is $x + \frac{\pi}{6} = \frac{\pi}{2}$ , so $x = \frac{\pi}{3}$ ).

Markers reward expansion of $R \sin(x + \alpha)$ , matching coefficients, the Pythagorean step for $R$ , and the maximum value of $R$ .

2021 HSC Q134 marksSolve

2 \sin x - 2 \cos x = \sqrt{2}

for

0 \le x \le 2 \pi

Show worked answer →

Write LHS as $R \sin(x - \alpha)$ with $R \cos \alpha = 2$ and $R \sin \alpha = 2$ .

$R = \sqrt{4 + 4} = 2 \sqrt{2}$ , $\tan \alpha = 1$ , $\alpha = \frac{\pi}{4}$ .

Equation: $2 \sqrt{2} \sin\!\left( x - \frac{\pi}{4} \right) = \sqrt{2}$ , so $\sin\!\left( x - \frac{\pi}{4} \right) = \frac{1}{2}$ .

General solution for $\sin = \frac{1}{2}$ : $x - \frac{\pi}{4} = \frac{\pi}{6} + 2 n \pi$ or $\pi - \frac{\pi}{6} + 2 n \pi$ .

So $x = \frac{\pi}{4} + \frac{\pi}{6} = \frac{5 \pi}{12}$ or $x = \frac{\pi}{4} + \frac{5 \pi}{6} = \frac{13 \pi}{12}$ .

Both are in $[0, 2 \pi]$ .

Markers reward the auxiliary-angle conversion, identifying both general-solution branches of $\sin$ , restricting to the given interval, and a clean final list.