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hscscience Ext 1 · Y11
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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.
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Recall, your gut answer first

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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 on even if you're not sure.

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Orientation and key terms

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The four-step method
+5 XP to read

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

  1. Substitute let $u = n\theta$ (the multiple angle), where $n$ is the multiplier.
  2. Expand the interval if the original interval is $[0, 2\pi)$, the interval for $u$ is $[0, 2n\pi)$. Multiply both bounds by $n$.
  3. Solve find all values of $u$ in the expanded interval satisfying $\sin u = k$, $\cos u = k$, or $\tan u = k$.
  4. Divide compute $\theta = u/n$ to get all solutions. Check each falls in the original interval.
Four-step method for solving trig equations with multiple angles
$\cos(2\theta) = \tfrac{1}{2}$ 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 $[0, 2\pi)$.
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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
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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 $\theta$.
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Worked Example 1, cosine, double angle

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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 $0 \le \theta < 2\pi$.

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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.
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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 $0 \le \theta < \pi$.

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

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Worked examples

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

PROBLEM 3 · DOUBLE ANGLE, NEGATIVE INTERVAL

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

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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.
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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 < $ .

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Misconceptions to fix

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 $0 \le u < 3\pi$.

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Activities

Activities · practice the four steps

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

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

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

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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.

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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?

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Revisit your thinking

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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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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.

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Short answer

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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].

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