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.
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.
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 $\tan\alpha = \frac{b}{a}$.
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 $-R$
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
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 $R\sin(x+\beta)$
We want to write $a\cos x + b\sin x$ as $R\cos(x - \alpha)$. Expand the right side using the compound angle identity:
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:
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 $b = R\sin\alpha$.
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 $5\cos x + 12\sin x$?
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.
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 $-90° < \alpha < 0°$)
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.
Worked examples · 3 in a row, reveal as you go
Write $3\cos x + 4\sin x$ in the form $R\cos(x - \alpha)$ where $0° \leq \alpha \leq 360°$.
Express $\cos x - \sqrt{3}\sin x$ in the form $R\cos(x + \alpha)$ where $0° \leq \alpha \leq 90°$.
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.
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 $\alpha = 30°$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the maximum value of $5\cos x - 12\sin x$ is $13$.
Activities · practice with the ideas
Write $\sqrt{3}\cos x + \sin x$ in the form $R\cos(x - \alpha)$, giving $\alpha$ exactly.
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.
Write $\cos x - \sin x$ in the form $R\sin(x + \beta)$ where $0° \leq \beta \leq 360°$.
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.
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?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write $\sqrt{3}\cos x + \sin x$ in the form $R\cos(x - \alpha)$, giving $\alpha$ exactly. (2 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)
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)
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].
Five timed questions on the auxiliary angle transformation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering auxiliary angle questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.