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 calculationwhat 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 confidencethat 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
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