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hscscience Ext 1 · Y12
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Module 7 · L2 of 20 ~40 min ⚡ +100 XP available

The Auxiliary Angle Transformation

Any expression of the form $a\cos x + b\sin x$ can be written as a single sinusoidal wave $R\cos(x - \alpha)$ or $R\sin(x + \alpha)$. This transformation reveals the amplitude, phase shift and maximum/minimum in one elegant step.

Today's hook, What is the maximum value of $3\cos x + 4\sin x$? Before reading on, make your best guess and explain your reasoning. You'll verify it precisely after card 05.
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Recall, your gut answer first

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Recall, your gut answer first
+5 XP warm-up

The expression $3\cos x + 4\sin x$ is a sum of two sinusoidal waves of different amplitudes. Before any calculation what do you think its maximum value is, and at what value of $x$ does it occur? Write your reasoning.

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Orientation and key terms

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The auxiliary angle idea
+5 XP to read

Any expression $a\cos x + b\sin x$ behaves like a single sinusoidal wave. The auxiliary angle method makes this precise: we find an amplitude $R$ and a phase shift $\alpha$ so that the sum collapses into one term.

The key insight is to match coefficients with an expanded compound angle identity:

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

Comparing with $a\cos x + b\sin x$, we need $R\cos\alpha = a$ and $R\sin\alpha = b$, which gives $R = \sqrt{a^2+b^2}$ and $\tan\alpha = \frac{b}{a}$.

A right triangle with legs a and b, hypotenuse R and angle α.
$R = \sqrt{a^2 + b^2}$, $\;\tan\alpha = \dfrac{b}{a}$
Two R-forms
$a\cos x + b\sin x = R\cos(x-\alpha)$ and also $= R\sin(x+\beta)$ where $\tan\beta = \frac{a}{b}$. Choose whichever form suits the context.
Maximum and minimum
Once in R-form, the maximum of $a\cos x + b\sin x$ is $R$ and the minimum is $-R$, since $\cos(x-\alpha)$ ranges over $[-1,1]$.
Quadrant check
$\alpha$ is in $[0°, 360°)$. The signs of $a = R\cos\alpha$ and $b = R\sin\alpha$ determine the quadrant, do not rely on $\tan^{-1}$ alone.
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What you'll master
Know

Key facts

  • $a\cos x + b\sin x = R\cos(x - \alpha)$ where $R = \sqrt{a^2+b^2}$ and $\tan\alpha = \frac{b}{a}$
  • Alternatively: $a\cos x + b\sin x = R\sin(x + \beta)$ where $\tan\beta = \frac{a}{b}$
  • Maximum value is $R$; minimum value is $-R$
Understand

Concepts

  • Why expanding $R\cos(x-\alpha)$ via the compound angle formula generates $a\cos x + b\sin x$
  • Why $\alpha$ must be placed in the correct quadrant using the signs of both $a$ and $b$
  • How the R-form reveals amplitude and phase shift simultaneously
Can do

Skills

  • Convert any $a\cos x + b\sin x$ expression into R-form, with $\alpha$ in correct quadrant
  • State the maximum and minimum values directly from R-form
  • Recognise when to use $R\cos(x-\alpha)$ versus $R\sin(x+\beta)$
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Key terms
Auxiliary angle ($\alpha$)The phase shift angle in R-form: $\alpha$ satisfies $\cos\alpha = \frac{a}{R}$ and $\sin\alpha = \frac{b}{R}$. Its quadrant is determined by the signs of both $a$ and $b$.
Amplitude ($R$)$R = \sqrt{a^2+b^2}$. The maximum value of $a\cos x + b\sin x$ and the amplitude of the equivalent single sinusoid.
R-formThe single-term form $R\cos(x-\alpha)$ or $R\sin(x+\beta)$ that an expression $a\cos x + b\sin x$ is converted into.
Phase shiftThe horizontal translation of the sinusoidal graph; equal to $\alpha$ in $R\cos(x-\alpha)$, meaning the maximum occurs at $x = \alpha$ rather than $x = 0$.
Coefficient matchingThe technique of comparing coefficients of $\cos x$ and $\sin x$ on both sides of an identity to extract the unknowns $R$ and $\alpha$.
$\tan^{-1}$ vs quadrant$\tan^{-1}(\frac{b}{a})$ gives an angle in $(-90°, 90°)$ only. The correct quadrant for $\alpha$ requires checking the signs of $\cos\alpha = \frac{a}{R}$ and $\sin\alpha = \frac{b}{R}$ independently.
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Deriving the auxiliary angle formula

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Deriving the auxiliary angle formula
core concept

We want to write $a\cos x + b\sin x$ as $R\cos(x - \alpha)$. Expand the right side using the compound angle identity:

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

Comparing coefficients of $\cos x$ and $\sin x$:

  • Coefficient of $\cos x$: $R\cos\alpha = a$
  • Coefficient of $\sin x$: $R\sin\alpha = b$

Squaring and adding: $R^2\cos^2\alpha + R^2\sin^2\alpha = a^2 + b^2$, so $R^2 = a^2 + b^2$, giving:

$$R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \frac{b}{a} \quad \text{(with quadrant from signs of }a, b\text{)}$$

Dividing: $\frac{R\sin\alpha}{R\cos\alpha} = \frac{b}{a}$, so $\tan\alpha = \frac{b}{a}$. But always confirm the quadrant from the signs of $a = R\cos\alpha$ and $b = R\sin\alpha$.

Answer to the hook. For $3\cos x + 4\sin x$: $R = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$. So the maximum value is $\mathbf{5}$. It occurs when $\cos(x-\alpha) = 1$, i.e., at $x = \alpha = \tan^{-1}\!\frac{4}{3} \approx 53.1°$.

a x + b x = R(x-) where R = a^2+b^2 and = b{a}; Derivation: expand R(x-), match coefficients, square and add to get R

Pause, copy the auxiliary angle derivation into your book: expand $R\sin(x-\alpha)$, match coefficients to get $R\cos\alpha = a$ and $R\sin\alpha = b$; square and add to give $R = \sqrt{a^2+b^2}$.

Quick check: What is the amplitude $R$ of $5\cos x + 12\sin x$?

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Handling all four sign cases
core concept

We just saw that $R = \sqrt{a^2+b^2}$ comes from squaring and adding the coefficient equations. That raises a question: we have $R\cos\alpha = a$ and $R\sin\alpha = b$, but how do we find the exact angle $\alpha$, and does $\tan\alpha = b/a$ give a unique answer? This card answers it → find the reference angle $\arctan|b/a|$ first, then use the signs of $a$ and $b$ to pin down the correct quadrant.

The signs of $a$ and $b$ determine which quadrant $\alpha$ lies in. This is the step where most errors occur, always check both signs explicitly.

$$R\cos\alpha = a \quad \Rightarrow \quad \cos\alpha \text{ has the same sign as } a$$
$$R\sin\alpha = b \quad \Rightarrow \quad \sin\alpha \text{ has the same sign as } b$$

Using the ASTC rule (All, Sin, Tan, Cos positive by quadrant):

  • $a > 0$, $b > 0$: $\alpha$ in Q1, $0° < \alpha < 90°$
  • $a < 0$, $b > 0$: $\alpha$ in Q2, $90° < \alpha < 180°$
  • $a < 0$, $b < 0$: $\alpha$ in Q3, $180° < \alpha < 270°$
  • $a > 0$, $b < 0$: $\alpha$ in Q4, $270° < \alpha < 360°$ (or equivalently $-90° < \alpha < 0°$)
Example with negative signs. Write $-\sqrt{3}\cos x + \sin x$ in R-form. Here $a = -\sqrt{3}$ and $b = 1$, so $R = \sqrt{3+1} = 2$. Since $\cos\alpha = \frac{-\sqrt{3}}{2} < 0$ and $\sin\alpha = \frac{1}{2} > 0$, $\alpha$ is in Q2. The reference angle is $30°$, so $\alpha = 150°$. Thus $-\sqrt{3}\cos x + \sin x = 2\cos(x - 150°)$.

From R = a and R = b: signs of a and b determine quadrant of; Always find the reference angle ^{-1}\!|b{a}| first, then adjust for quadrant

Pause, copy the sign-case rule into your book: $\cos\alpha = a/R$ and $\sin\alpha = b/R$, negative $a$ means $\alpha$ in Q2 or Q3; negative $b$ means Q3 or Q4; their combination fixes the quadrant uniquely.

Did you get this? True or false: for $-3\cos x - 4\sin x = R\cos(x-\alpha)$, the angle $\alpha$ is in the third quadrant.

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · STANDARD R-FORM

Write $3\cos x + 4\sin x$ in the form $R\cos(x - \alpha)$ where $0° \leq \alpha \leq 360°$.

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Identify $a = 3$, $b = 4$. Calculate $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Always find $R$ first. It is always positive. For 3–4–5, recognise this as a Pythagorean triple.
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PROBLEM 2 · NEGATIVE COEFFICIENTS

Express $\cos x - \sqrt{3}\sin x$ in the form $R\cos(x + \alpha)$ where $0° \leq \alpha \leq 90°$.

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Expand $R\cos(x+\alpha) = R\cos\alpha\cos x - R\sin\alpha\sin x$. Match: $R\cos\alpha = 1$, $R\sin\alpha = \sqrt{3}$.
The target form is $R\cos(x+\alpha)$, so expand that form, not $R\cos(x-\alpha)$. This changes the sign pattern.
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PROBLEM 3 · MAXIMUM VALUE

Find the maximum value of $f(x) = -2\cos x + 2\sqrt{3}\sin x$ and the smallest positive value of $x$ at which it occurs.

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$a = -2$, $b = 2\sqrt{3}$. $R = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.
$R$ is always positive regardless of the signs of $a$ and $b$.
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Fill the gap: For $\sqrt{3}\cos x + \sin x = R\cos(x - \alpha)$, we get $R = \sqrt{3+1} = $ and $\tan\alpha = \frac{1}{\sqrt{3}}$ so $\alpha = 30°$.

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Using $\tan^{-1}$ without quadrant check
$\tan^{-1}\!\frac{b}{a}$ only returns values in $(-90°, 90°)$. If $a < 0$, the angle is in Q2 or Q3, not Q1 or Q4. Always verify using $\cos\alpha = \frac{a}{R}$ and $\sin\alpha = \frac{b}{R}$ separately. Skipping this check is the most penalised auxiliary angle error in HSC marking.
Trap 02
Wrong target form → wrong sign pattern
$R\cos(x-\alpha)$ expands to $R\cos\alpha\cos x + R\sin\alpha\sin x$ (positive $\sin$ term). $R\cos(x+\alpha)$ expands to $R\cos\alpha\cos x - R\sin\alpha\sin x$ (negative $\sin$ term). If the original expression has a negative $\sin$ coefficient, use $R\cos(x+\alpha)$, not $R\cos(x-\alpha)$.
Trap 03
Forgetting $R > 0$ always
$R = \sqrt{a^2+b^2}$ is always the positive square root. Never write $R = \pm\sqrt{a^2+b^2}$. The sign information is entirely captured in $\alpha$. A negative "R" would change the amplitude and is not a valid R-form.

Did you get this? True or false: the maximum value of $5\cos x - 12\sin x$ is $13$.

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Activities

Activities · practice with the ideas

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Write $\sqrt{3}\cos x + \sin x$ in the form $R\cos(x - \alpha)$, giving $\alpha$ exactly.

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Find the maximum and minimum values of $f(x) = 5\cos x + 12\sin x$ and the values of $x$ in $[0°, 360°]$ at which they occur.

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Write $\cos x - \sin x$ in the form $R\sin(x + \beta)$ where $0° \leq \beta \leq 360°$.

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A function is defined by $g(x) = -\cos x - \sin x$. Find the amplitude and the value of $x$ in $[0°, 360°]$ where $g(x)$ is maximum.

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Show that $a\cos x + b\sin x$ can also be written as $R\sin(x + \beta)$ where $R = \sqrt{a^2+b^2}$ and $\tan\beta = \frac{a}{b}$, by expanding $R\sin(x+\beta)$ and matching coefficients.

Odd one out: Three of these are correct statements about $a\cos x + b\sin x = R\cos(x-\alpha)$. Which one is FALSE?

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Revisit your thinking

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Revisit your thinking

In card 01 you estimated the maximum of $3\cos x + 4\sin x$. The exact answer is $R = \sqrt{3^2+4^2} = 5$, occurring at $x = \tan^{-1}\!\frac{4}{3} \approx 53.1°$. Did your intuition converge on $5$? Many students guess $3+4=7$ (adding amplitudes), this overcounts because the two waves are not in phase with each other.

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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

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Short answer
ApplyBand 32 marks

Q1. Write $\sqrt{3}\cos x + \sin x$ in the form $R\cos(x - \alpha)$, giving $\alpha$ exactly. (2 marks)

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ApplyBand 43 marks

Q2. Find the maximum value of $f(x) = -\sqrt{3}\cos x + \sin x$ and the smallest positive $x$ (in degrees) at which it occurs. (3 marks)

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AnalyseBand 53 marks

Q3. Prove that $a\cos x + b\sin x = \sqrt{a^2+b^2}\,\cos(x-\alpha)$ for some angle $\alpha$, by expanding the right side and identifying $\alpha$ in terms of $a$ and $b$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\sqrt{3}\cos x + \sin x$: $R=2$, $\tan\alpha = \frac{1}{\sqrt{3}}$, $\alpha = 30°$ (Q1). Answer: $2\cos(x-30°)$.

2. $5\cos x + 12\sin x$: $R=13$, $\alpha = \tan^{-1}\!\frac{12}{5} \approx 67°23'$ (Q1). Max $=13$ at $x \approx 67°23'$, Min $=-13$ at $x \approx 247°23'$.

3. $\cos x - \sin x = R\sin(x+\beta)$: expand $R\sin(x+\beta)=R\cos\beta\sin x + R\sin\beta\cos x$. Match: $R\cos\beta = -1$, $R\sin\beta = 1$. $R=\sqrt{2}$, $\cos\beta < 0$, $\sin\beta > 0$ → Q2. Ref angle $= 45°$, so $\beta = 135°$. Answer: $\sqrt{2}\sin(x+135°)$.

4. $-\cos x - \sin x$: $a=-1$, $b=-1$, $R=\sqrt{2}$, $\alpha$ in Q3. Ref angle $45°$, so $\alpha=225°$. $g(x) = \sqrt{2}\cos(x-225°)$. Max $=\sqrt{2}$ at $x=225°$.

Q1 (2 marks): $R = \sqrt{(\sqrt{3})^2+1^2} = 2$ [1]; $\tan\alpha = \frac{1}{\sqrt{3}}$, both coefficients positive so $\alpha = 30°$ [1]. $\sqrt{3}\cos x + \sin x = 2\cos(x-30°)$.

Q2 (3 marks): $a=-\sqrt{3}$, $b=1$, $R = \sqrt{3+1} = 2$ [1]. $\cos\alpha = \frac{-\sqrt{3}}{2}<0$, $\sin\alpha=\frac{1}{2}>0$ → Q2; ref angle $= 60°$, so $\alpha = 120°$ [1]. Max $= 2$ when $x - 120° = 0$, i.e., $x = 120°$ [1].

Q3 (3 marks): $\sqrt{a^2+b^2}\cos(x-\alpha) = \sqrt{a^2+b^2}(\cos\alpha\cos x + \sin\alpha\sin x)$ [1]. Setting $\sqrt{a^2+b^2}\cos\alpha = a$ and $\sqrt{a^2+b^2}\sin\alpha = b$, squaring and adding confirms $R^2 = a^2+b^2$ [1]. Dividing gives $\tan\alpha = \frac{b}{a}$ with quadrant determined by signs of $a$ and $b$ [1].

Module overview · Extension 1 · Checkpoint 1