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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 formulawhat 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

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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.

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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