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Module 7 · L3 of 20 ~40 min ⚡ +100 XP available

Auxiliary Angle — Worked Examples

You know the formula $R\cos(x - \alpha)$. Now you need to use it without making sign errors. In this lesson you'll work through three complete examples — easy, medium, and tricky — and build the quadrant instinct that separates students who get the right $\alpha$ every time from those who guess and lose marks.

Today's hook — For $f(x) = 3\cos x + 4\sin x$, what do you think the maximum value is, and at what angle does it occur? Jot a guess before card 05.

0/5QUESTS

Recall — your gut answer first

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Consider $f(x) = 3\cos x + 4\sin x$. Without using a formula — what do you think the maximum value of $f$ is, and roughly where does it occur? Write your reasoning.

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The conversion recipe — two moves only

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Every auxiliary angle conversion reduces to two calculations: find $R$, then find $\alpha$. From Lesson 2, expanding $R\cos(x-\alpha) = R\cos\alpha\cos x + R\sin\alpha\sin x$ gives the matching conditions.

Given $a\cos x + b\sin x$, match coefficients with $R\cos(x-\alpha)$:

$R\cos\alpha = a$ (coefficient of $\cos x$)
$R\sin\alpha = b$ (coefficient of $\sin x$)

Square and add: $R^2 = a^2 + b^2$, so $R = \sqrt{a^2+b^2}$.

Divide: $\tan\alpha = \dfrac{b}{a}$ (then choose the quadrant matching the signs of $a$ and $b$).

$R = \sqrt{a^2+b^2}$ $\tan\alpha = \dfrac{b}{a}$

$R$ is always positive

Take the positive square root. The sign information lives entirely in $\alpha$.

Quadrant of $\alpha$

$\alpha$ lives in Q1 when $a>0, b>0$; Q2 when $a<0, b>0$; Q3 when both negative; Q4 when $a>0, b<0$.

$R\sin(x+\alpha)$ variant

If you prefer $R\sin(x+\alpha)$, match $R\cos\alpha = b$ and $R\sin\alpha = a$. Same $R$, different $\alpha$.

What you'll master

Know

Key facts

$R = \sqrt{a^2+b^2}$; $\tan\alpha = b/a$ (with quadrant check)
$a\cos x + b\sin x \equiv R\cos(x-\alpha)$ for unique $R>0$, $\alpha\in(-\pi,\pi]$
The maximum of $a\cos x + b\sin x$ is $R$ and the minimum is $-R$

Understand

Concepts

Why matching coefficients forces $R$ and $\alpha$ to be unique
How the quadrant of $\alpha$ is determined by the signs of $a$ and $b$, not just $\tan\alpha$
The connection between $R$ and the amplitude of the combined wave

Can do

Skills

Convert any $a\cos x + b\sin x$ to $R\cos(x-\alpha)$ in full working
State the maximum and minimum values and where they occur
Verify a conversion by expanding and collecting

Key terms

Auxiliary angle $\alpha$The phase-shift angle in the converted form. Its quadrant is fixed by the signs of the original coefficients $a$ and $b$.

Amplitude $R$The positive constant $R = \sqrt{a^2+b^2}$ that scales the sinusoid. It equals the maximum value of the expression.

Coefficient matchingExpanding the target form and equating coefficients of $\cos x$ and $\sin x$ to determine $R$ and $\alpha$.

Principal value of $\alpha$The value of $\alpha$ in $(-\pi,\pi]$ that satisfies both $\cos\alpha = a/R$ and $\sin\alpha = b/R$ simultaneously.

Pythagorean identity checkAfter finding $\alpha$, verify: $(R\cos\alpha)^2 + (R\sin\alpha)^2 = R^2$. This catches arithmetic errors.

Verification by expansionExpand $R\cos(x-\alpha) = R\cos\alpha\cos x + R\sin\alpha\sin x$ and check you recover the original $a$ and $b$.

The full conversion method

core concept

To convert $a\cos x + b\sin x$ into $R\cos(x-\alpha)$, follow three steps every time:

Step 1 — Find $R$

$$R = \sqrt{a^2 + b^2}$$

Step 2 — Find $\alpha$

$$\cos\alpha = \frac{a}{R}, \quad \sin\alpha = \frac{b}{R} \quad\Rightarrow\quad \tan\alpha = \frac{b}{a}$$

Use both the cosine and sine conditions to pin down the correct quadrant.

Step 3 — Write the answer

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

Why does $\tan\alpha = b/a$ alone not determine $\alpha$? Because $\tan$ has period $\pi$, so $\tan\alpha = b/a$ gives two values of $\alpha$ in $[0,2\pi)$. You need to know the quadrant — determined by the signs of $\cos\alpha = a/R > 0$ iff $a>0$ and $\sin\alpha = b/R > 0$ iff $b > 0$.

Maximum and minimum. Because $-1 \leq \cos\theta \leq 1$ for any $\theta$, we immediately have $-R \leq R\cos(x-\alpha) \leq R$. So the maximum of $a\cos x + b\sin x$ is $\boldsymbol{R = \sqrt{a^2+b^2}}$, achieved when $x = \alpha$.

Answer to today's hook: $3\cos x + 4\sin x$ — here $a=3$, $b=4$, so $R = \sqrt{9+16} = \sqrt{25} = 5$. The maximum is 5. How close was your estimate?

Recipe: R = a^2+b^2; = a/R; = b/R; then a x + b x = R(x-); Always find from both and — alone gives two candidates

Pause — copy the full conversion recipe into your book: $R = \sqrt{a^2+b^2}$; $\cos\alpha = a/R$; $\sin\alpha = b/R$; always use both equations to fix the quadrant — $\tan\alpha = b/a$ alone is ambiguous.

Quick check: For $5\cos x + 12\sin x$, what is the amplitude $R$?

Dealing with negative coefficients

core concept

We just saw the recipe: $R = \sqrt{a^2+b^2}$, then find $\alpha$ from $\cos\alpha = a/R$ and $\sin\alpha = b/R$. That raises a question: when $a$ or $b$ is negative, the quadrant of $\alpha$ changes — how do you read the sign combination correctly? This card answers it → negative $a$ means $\cos\alpha < 0$ (Q2 or Q3); negative $b$ means $\sin\alpha < 0$ (Q3 or Q4); find the quadrant where both hold.

When $a$ or $b$ is negative, $\alpha$ moves out of the first quadrant. The key is to still compute $\cos\alpha = a/R$ and $\sin\alpha = b/R$ separately, then identify the quadrant.

$a > 0,\; b > 0$: $\cos\alpha > 0$, $\sin\alpha > 0$ → $\alpha \in$ Q1, so $0 < \alpha < \tfrac{\pi}{2}$
$a < 0,\; b > 0$: $\cos\alpha < 0$, $\sin\alpha > 0$ → $\alpha \in$ Q2, so $\tfrac{\pi}{2} < \alpha < \pi$
$a < 0,\; b < 0$: $\cos\alpha < 0$, $\sin\alpha < 0$ → $\alpha \in$ Q3, so $-\pi < \alpha < -\tfrac{\pi}{2}$
$a > 0,\; b < 0$: $\cos\alpha > 0$, $\sin\alpha < 0$ → $\alpha \in$ Q4, so $-\tfrac{\pi}{2} < \alpha < 0$

Exact angle trick. If $|\tan\alpha|$ is a standard ratio ($1, \tfrac{1}{\sqrt{3}}, \sqrt{3}$), $\alpha$ is an exact angle ($\tfrac{\pi}{4}, \tfrac{\pi}{6}, \tfrac{\pi}{3}$ or their supplements/negatives). Always check for exact angles before reaching for a calculator.

Negative a: < 0, so Q2 or Q3; Negative b: < 0, so Q3 or Q4

Pause — copy the negative-coefficient rule into your book: negative $a$ means $\cos\alpha < 0$ (Q2 or Q3); negative $b$ means $\sin\alpha < 0$ (Q3 or Q4); find the quadrant where both conditions hold.

Did you get this? True or false: for $\sqrt{3}\cos x - \sin x$, the auxiliary angle $\alpha$ lies in Q4 (i.e. $\alpha = -\pi/6$).

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

PROBLEM 1 · STANDARD CONVERSION (Q1 ANGLE)

Express $3\cos x + 4\sin x$ in the form $R\cos(x-\alpha)$, where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$. State the maximum value and the value of $x$ at which it occurs for $x \in [0, 2\pi]$.

$R = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$

Apply $R = \sqrt{a^2+b^2}$ with $a=3$, $b=4$. This is a 3-4-5 Pythagorean triple.

PROBLEM 2 · NEGATIVE COEFFICIENT (Q2 ANGLE)

Write $-\cos x + \sqrt{3}\sin x$ in the form $R\cos(x-\alpha)$ where $R > 0$ and $\alpha \in (0, \pi)$.

$R = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$

$a = -1$, $b = \sqrt{3}$. Squaring removes the sign: $R = 2$.

PROBLEM 3 · EXACT ANGLE IN Q4

Express $\cos x - \sin x$ in the form $R\cos(x-\alpha)$ with $R>0$ and $\alpha \in (-\tfrac{\pi}{2}, 0)$. Hence find the minimum value of $f(x) = \cos x - \sin x$ and where it occurs.

$R = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

$a=1$, $b=-1$. $R = \sqrt{1+1} = \sqrt{2}$.

Fill the gap: $\cos x - \sin x = \sqrt{2}\cos(x +$ $)$ . (Enter the exact angle.)

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using $\tan\alpha$ alone to find $\alpha$

$\tan\alpha = b/a$ has two solutions in $[0, 2\pi)$: one in Q1/Q3 and one in Q2/Q4. Without checking the signs of $a$ and $b$ separately, you will pick the wrong quadrant roughly half the time. Always confirm with $\cos\alpha = a/R$ and $\sin\alpha = b/R$.

Trap 02

Negative $R$

$R$ is defined as the positive square root. Writing $R = -\sqrt{a^2+b^2}$ produces a different function entirely. If you absorb a minus sign into $R$, your answer is wrong. All sign information must go into the angle $\alpha$.

Trap 03

Forgetting to verify

Under exam pressure, students skip the check step. Expanding $R\cos(x-\alpha) = R\cos\alpha\cos x + R\sin\alpha\sin x$ takes 20 seconds and catches every sign error. Always do it.

Did you get this? True or false: the maximum value of $a\cos x + b\sin x$ is always $\sqrt{a^2+b^2}$, regardless of the signs of $a$ and $b$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Express $\cos x + \sin x$ in the form $R\cos(x-\alpha)$, $R>0$, $0<\alpha<\pi/2$. State the maximum value.

Express $5\cos x + 12\sin x$ in the form $R\cos(x-\alpha)$. Hence find the minimum value of the expression.

Express $-\sqrt{3}\cos x - \sin x$ in the form $R\cos(x-\alpha)$ with $\alpha \in (-\pi, -\pi/2)$. Identify the quadrant of $\alpha$.

Verify that $3\cos x + 4\sin x = 5\cos(x - \tan^{-1}\tfrac{4}{3})$ by expanding and collecting.

For $f(x) = 2\cos x + 2\sqrt{3}\sin x$, find $R$ and the exact value of $\alpha$, then state when $f(x)$ is maximum on $[0, 2\pi]$.

Odd one out: Three of the following auxiliary angle conversions are correct. Which one is WRONG?

Revisit your thinking

Earlier you estimated the maximum of $3\cos x + 4\sin x$.

The exact answer is $R = \sqrt{9+16} = \mathbf{5}$, occurring at $x = \tan^{-1}\tfrac{4}{3} \approx 53.1°$. The key insight is that combining two sinusoids of the same frequency always produces a sinusoid with amplitude $\sqrt{a^2+b^2}$ — a direct consequence of the Pythagorean theorem on the coefficient triangle.

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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.

Short answer

ApplyBand 32 marks

Q1. Express $\cos x + \sqrt{3}\sin x$ in the form $R\cos(x-\alpha)$ where $R > 0$ and $0 < \alpha < \pi/2$. (2 marks)

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ApplyBand 43 marks

Q2. Express $-\cos x + \sqrt{3}\sin x$ in the form $R\cos(x-\alpha)$, $R>0$, $\alpha\in(0,\pi)$. Hence state the minimum value and the value of $x \in [0,2\pi]$ at which it occurs. (3 marks)

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AnalyseBand 53 marks

Q3. Show that $\sin x - \cos x$ can be written as $\sqrt{2}\sin\!\left(x - \dfrac{\pi}{4}\right)$. Hence find all solutions of $\sin x - \cos x = 1$ in $[0, 2\pi]$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\cos x + \sin x$: $R = \sqrt{2}$, $\cos\alpha = \sin\alpha = 1/\sqrt{2}$, so $\alpha = \pi/4$. Result: $\sqrt{2}\cos(x-\pi/4)$. Maximum $= \sqrt{2}$.

2. $5\cos x + 12\sin x$: $R = 13$, $\alpha = \tan^{-1}(12/5)$. Minimum $= -13$.

3. $-\sqrt{3}\cos x - \sin x$: $R = 2$, $\cos\alpha = -\sqrt{3}/2 < 0$, $\sin\alpha = -1/2 < 0$ — Q3. $\alpha = -5\pi/6$. Result: $2\cos(x+5\pi/6)$.

4. Expand: $5\cos(x-\alpha) = 5\cos\alpha\cos x + 5\sin\alpha\sin x = 3\cos x + 4\sin x$ ✓ (since $5\cos\alpha = 3$ and $5\sin\alpha = 4$).

5. $2\cos x + 2\sqrt{3}\sin x$: $R = 4$, $\tan\alpha = \sqrt{3}$, Q1 $\Rightarrow \alpha = \pi/3$. Maximum at $x = \pi/3$.

Q1 (2 marks): $R = \sqrt{1+3} = 2$ [1]. $\cos\alpha = 1/2$, $\sin\alpha = \sqrt{3}/2 \Rightarrow \alpha = \pi/3$ [1]. $\cos x + \sqrt{3}\sin x = 2\cos(x-\pi/3)$.

Q2 (3 marks): $R = 2$, $\alpha = 2\pi/3$ (Q2, since $\cos\alpha = -1/2$ and $\sin\alpha = \sqrt{3}/2$) [2]. $-\cos x + \sqrt{3}\sin x = 2\cos(x-2\pi/3)$. Minimum $= -2$, occurs when $x - 2\pi/3 = \pi$, i.e. $x = 5\pi/3$ [1].

Q3 (3 marks): Expand $\sqrt{2}\sin(x-\pi/4) = \sqrt{2}(\sin x\cos\tfrac{\pi}{4} - \cos x\sin\tfrac{\pi}{4}) = \sqrt{2}\cdot\tfrac{1}{\sqrt{2}}\sin x - \sqrt{2}\cdot\tfrac{1}{\sqrt{2}}\cos x = \sin x - \cos x$ ✓ [1]. Equation: $\sqrt{2}\sin(x-\pi/4) = 1 \Rightarrow \sin(x-\pi/4) = 1/\sqrt{2}$. Let $u = x - \pi/4$: $\sin u = 1/\sqrt{2} \Rightarrow u = \pi/4$ or $u = 3\pi/4$. So $x = \pi/2$ or $x = \pi$ [2].

Boss battle · The Auxiliary Angle Master

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering auxiliary angle questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 2 · The Auxiliary Angle Transformation Lesson 4 · Solving a cos x + b sin x = c →

Module overview · Extension 1 · Checkpoint 1