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Module 3 · L12 of 15 ~35 min ⚡ +95 XP available

Solving Trigonometric Equations, Multiple Angles

$\sin\theta = \tfrac{1}{2}$ has 2 solutions in $[0, 2\pi)$. But $\sin(2\theta) = \tfrac{1}{2}$ has 4, and $\sin(3\theta) = \tfrac{1}{2}$ has 6. More cycles fit into the same interval, so you need more solutions. The trick is deceptively simple: expand the interval before you solve, then divide back at the end.

Today's hook, A student solves $\cos(2\theta) = \tfrac{1}{2}$ and writes $\theta = \tfrac{\pi}{6}$. They scored 0 marks. Another student writes $\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$ and gets full marks. The difference? One student forgot to expand the interval and found only half the solutions. By the end of this lesson, that will never happen to you.

0/5QUESTS

Recall, your gut answer first

+5 XP warm-up

How many solutions does $\sin(2\theta) = \tfrac{1}{2}$ have in $0 \le \theta < 2\pi$? Is it the same number as $\sin\theta = \tfrac{1}{2}$? Write your gut answer before reading oneven if you're not sure.

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The four-step method

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Multiple-angle equations seem harder but follow a rigid four-step process. Master the steps and you'll never miss a solution:

Substitutelet $u = n\theta$ (the multiple angle), where $n$ is the multiplier.
Expand the intervalif the original interval is $[0, 2\pi)$, the interval for $u$ is $[0, 2n\pi)$. Multiply both bounds by $n$.
Solvefind all values of $u$ in the expanded interval satisfying $\sin u = k$, $\cos u = k$, or $\tan u = k$.
Dividecompute $\theta = u/n$ to get all solutions. Check each falls in the original interval.

$\cos(2\theta) = \tfrac{1}{2}$ in $[0, 2\pi)$

Why expand the interval?

$u = 2\theta$ means $u$ ranges twice as fast as $\theta$. In the same $\theta$ interval, $u$ covers twice as many complete cycles, so there are twice as many solutions.

Non-zero lower bounds

If the interval is $-\pi < \theta \le \pi$, multiply both bounds: $-2\pi < u \le 2\pi$. Many students forget to expand the lower bound, this loses solutions in the negative region.

Number of solutions

In general, $\sin(n\theta) = k$ has $2n$ solutions in $[0, 2\pi)$ (for $|k| < 1$). $\cos(n\theta) = k$ also gives $2n$ solutions. $\tan(n\theta) = k$ gives $n$ solutions per period, so $n$ solutions in $[0, \pi)$ and $2n$ in $[0, 2\pi)$.

What you'll master

Know

Key facts

For $\sin(n\theta) = k$ in $[0, 2\pi)$, solve for $n\theta$ in $[0, 2n\pi)$ then divide by $n$
Both bounds of the interval are multiplied by $n$, including a non-zero lower bound
Multiple-angle equations generally have $n$ times as many solutions as single-angle equations

Understand

Concepts

Why the trig function completes $n$ cycles in the original interval when the angle is $n\theta$
How to count the expected number of solutions as a check on your working
Why the substitution $u = n\theta$ is structurally identical to the method for single-angle equations

Can do

Skills

Solve $\sin(n\theta) = k$, $\cos(n\theta) = k$, and $\tan(n\theta) = k$ in any specified interval
Handle non-standard intervals (e.g. $-\pi < \theta \le \pi$, $0 \le \theta \le \pi$) by expanding both bounds
Verify solutions by substituting back into the original equation

Key terms

Multiple angleAn expression of the form $n\theta$ where $n$ is an integer greater than 1, e.g. $2\theta$, $3\theta$.

Adjusted (expanded) intervalWhen solving for $u = n\theta$, multiply both bounds of the original interval by $n$ to get the interval for $u$.

Period multiplierReplacing $\theta$ with $n\theta$ compresses the period by a factor of $n$, fitting $n$ cycles into the original range.

Back-substitutionDividing each solution $u$ by $n$ at the end to recover the values of $\theta$.

Worked Example 1, cosine, double angle

core concept

Let's work through the example from the hero section, fully and carefully, to establish the method.

Worked Example 1, cosine, double angle: the key definition, formula, and conditions, write this into your book.

Pause, copy the four-step double-angle equation method into your book as shown in Worked Example 1: choose the $\cos 2\theta$ form that matches remaining terms → substitute → solve quadratic → apply domain check.

PROBLEM 1 · DOUBLE ANGLE, $\cos$

Solve $\cos(2\theta) = \dfrac{1}{2}$ for $0 \le \theta < 2\pi$.

Let $u = 2\theta$.
Interval for $u$: since $0 \le \theta < 2\pi$, multiply by 2: $0 \le u < 4\pi$.

Expanding the interval is the key step. The function $\cos u$ goes through two full cycles in $[0, 4\pi)$, so we expect four solutions.

Quick check: When solving $\sin(2\theta) = \tfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$, what is the correct interval to use for $u = 2\theta$?

PROBLEM 2 · TRIPLE ANGLE, $\tan$

Solve $\tan(3\theta) = 1$ for $0 \le \theta < \pi$.

Let $u = 3\theta$.
Interval for $u$: $0 \le \theta < \pi$ multiplied by 3 gives $0 \le u < 3\pi$.

The interval for $\theta$ is $[0, \pi)$, not $[0, 2\pi)$. Still multiply by 3.

Did you get this? True or false: $\sin(2\theta) = \tfrac{1}{2}$ has the same solutions as $\sin\theta = \tfrac{1}{2}$ in $[0, 2\pi)$.

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

PROBLEM 3 · DOUBLE ANGLE, NEGATIVE INTERVAL

Solve $\cos(2\theta) = -\dfrac{1}{2}$ for $-\pi < \theta \le \pi$.

Let $u = 2\theta$.
Expand: since $-\pi < \theta \le \pi$, multiply by 2: $-2\pi < u \le 2\pi$.

The lower bound is $-\pi$, not 0. It must also be multiplied by 2. Many students forget this and find only half the solutions.

Fill the gap: To solve $\sin(2\theta) = \tfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$, the correct interval for $u = 2\theta$ is $0 \le u < $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Not expanding the interval

This is the single most common error. A student solves $\cos(2\theta) = \tfrac{1}{2}$ by finding $2\theta = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$ (both in $[0, 2\pi)$) and divides to get $\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$. They found only 2 of the 4 solutions because they didn't expand the interval. The rule: if the angle is $n\theta$, the interval for $u$ must be $n$ times as long.

Trap 02

Forgetting to expand the lower bound

For the interval $-\pi < \theta \le \pi$ with $n=2$: many students write $0 < u \le 2\pi$ (forgetting to multiply $-\pi$ by 2). The correct expanded interval is $-2\pi < u \le 2\pi$. A non-zero lower bound must be multiplied just like the upper bound.

Trap 03

Dividing the interval instead of expanding it

A confused student sees $\cos(2\theta)$ and thinks "the angle is halved, so I should halve the interval to $[0, \pi)$." This is backwards: we are solving for $u = 2\theta$, which is larger than $\theta$, so the interval for $u$ is larger. Always multiply.

Did you get this? True or false: to solve $\tan(3\theta) = \sqrt{3}$ for $0 \le \theta < \pi$, the expanded interval for $u = 3\theta$ is $0 \le u < 3\pi$.

Work mode · how are you completing this lesson?

Activities · practice the four steps

Solve $\sin(2\theta) = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$. How many solutions should you expect?

Solve $\tan(3\theta) = 1$ for $0 \le \theta < \pi$. Show the expanded interval clearly.

Solve $\cos(2\theta) = -\dfrac{1}{2}$ for $-\pi < \theta \le \pi$. What is the expanded interval?

A student claims $\cos(2\theta) = \tfrac{1}{2}$ has only 2 solutions in $[0, 2\pi)$. What mistake did they make? Correct their work.

Explain in your own words why $\sin(n\theta) = k$ has $n$ times as many solutions as $\sin\theta = k$ in $[0, 2\pi)$.

Odd one out: Three of these statements about solving $\sin(2\theta) = \tfrac{1}{2}$ in $[0, 2\pi)$ are correct. Which is WRONG?

Revisit your thinking

At the start you were asked how many solutions $\sin(2\theta) = \tfrac{1}{2}$ has in $[0, 2\pi)$. The answer is 4 not 2.

Letting $u = 2\theta$ and working in $[0, 4\pi)$: $\sin u = \tfrac{1}{2}$ gives reference angle $\tfrac{\pi}{6}$, and $\sin > 0$ in Q1 and Q2. In cycle 1: $u = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$. In cycle 2: $u = \tfrac{13\pi}{6}, \tfrac{17\pi}{6}$. Dividing by 2: $\theta = \tfrac{\pi}{12}, \tfrac{5\pi}{12}, \tfrac{13\pi}{12}, \tfrac{17\pi}{12}$.

Why does replacing $\theta$ with $n\theta$ give $n$ times as many solutions in the same interval?

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Multiple choice

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

Short answer

ApplyBand 43 marks

Q1. Solve $\sin(2\theta) = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$. (3 marks)

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ApplyBand 43 marks

Q2. Solve $\tan(3\theta) = 1$ for $0 \le \theta < \pi$. (3 marks)

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AnalyseBand 53 marks

Q3. Solve $\cos(2\theta) = -\dfrac{1}{2}$ for $-\pi < \theta \le \pi$. (3 marks)

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Comprehensive answers (click to reveal)

Activity 1:

1. $u = 2\theta \in [0,4\pi)$. $\sin u = \frac{\sqrt{3}}{2}$, ref angle $\frac{\pi}{3}$. Cycle 1: $u = \frac{\pi}{3}, \frac{2\pi}{3}$. Cycle 2: $u = \frac{\pi}{3}+2\pi = \frac{7\pi}{3}$, $u = \frac{2\pi}{3}+2\pi = \frac{8\pi}{3}$. $\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$.

2. $u = 3\theta \in [0,3\pi)$. $\tan u = 1$, ref angle $\frac{\pi}{4}$. $u = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}$. $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}$.

3. $u = 2\theta \in (-2\pi,2\pi]$. $\cos u = -\frac{1}{2}$, ref angle $\frac{\pi}{3}$. $u = \frac{2\pi}{3}, \frac{4\pi}{3}, -\frac{2\pi}{3}, -\frac{4\pi}{3}$. $\theta = \frac{\pi}{3}, \frac{2\pi}{3}, -\frac{\pi}{3}, -\frac{2\pi}{3}$.

4. Mistake: only used $[0, 2\pi)$ for $u = 2\theta$ instead of $[0, 4\pi)$. Correct: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.

Q1 (3 marks): Correct expanded interval $[0,4\pi)$ stated [1]. All four values of $u$ found [1]. Correct $\theta$ values: $\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$ [1].

Q2 (3 marks): Correct expanded interval $[0,3\pi)$ [1]. Three $u$ values identified [1]. Correct $\theta$ values: $\frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}$ [1].

Q3 (3 marks): Correct expanded interval $(-2\pi, 2\pi]$ [1]. All four $u$ values (positive and negative) found [1]. Correct $\theta$ values: $\pm\frac{\pi}{3}, \pm\frac{2\pi}{3}$ [1].

Boss battle · The Cycle Counter

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Five timed questions on multiple-angle trig equations. 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 multiple-angle trig questions. Lighter alternative to the boss.

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Module overview · Extension 1 subject page · Checkpoint 3 · Module quiz