Mathematics Advanced • Year 11 • Module 2 • Lesson 9

Domains and Ranges of Trigonometric Functions

Build fluency in stating the domain and range of all six trig functions, finding undefined points, and writing general solutions with $n \in \mathbb{Z}$.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the table.

FunctionDomainRange
$\sin x$
$\tan x$
$\sec x$

Q1.2 Why is $\tan x$ undefined where $\cos x = 0$? Answer in one sentence.

Q1.3 Define "bounded" and give one bounded and one unbounded trig function.

Stuck? Revisit lesson § Domain and range at a glance and § Systematic analysis of all six functions.

2. Worked example — domain of $f(x) = \frac{\sin x}{1 + \cos x}$

Every line annotated. Use as the template for the faded version.

Problem. State the domain of $f(x) = \frac{\sin x}{1 + \cos x}$.

Step 1 — Identify what can break the function.

Division by zero is the only risk — $\sin x$ itself is defined for all real $x$.

Reason: numerator (sin) has domain $\mathbb{R}$; the denominator is the only constraint.

Step 2 — Set the denominator equal to zero.

1 + cos x = 0 ⇒ cos x = −1

Reason: forbidden inputs are those that make the denominator vanish.

Step 3 — Solve $\cos x = -1$ over all reals.

cos x = −1 at x = π, then every $2\pi$ apart: x = π + 2nπ, $n \in \mathbb{Z}$.

Reason: cosine has period $2\pi$ and equals $-1$ once per cycle (Trap 3: must use general solution).

Step 4 — Write the domain.

Domain: all real x except x = π + 2nπ, $n \in \mathbb{Z}$.

Conclusion. Domain = $\mathbb{R} \setminus \{\pi + 2n\pi : n \in \mathbb{Z}\}$.

3. Faded example — fill in the missing steps

Find the range of $y = 3 \sin x + 1$. Fill in each blank line. 4 marks

Step 1 — Start with the range of $\sin x$.

________ ≤ sin x ≤ ________

Step 2 — Multiply through by 3 (positive, so inequality direction unchanged).

________ ≤ 3 sin x ≤ ________

Step 3 — Add 1 to every part.

________ ≤ 3 sin x + 1 ≤ ________

Step 4 — Write the range as an interval.

Range = [________, ________]

Conclusion. Range of $3 \sin x + 1$ is ____________.

Stuck? Revisit lesson § Worked Example — Range of a modified function.

4. Graduated practice

State the domain and range. Use general-solution notation ($n \in \mathbb{Z}$) where infinitely many points are excluded.

Foundation — basic six (4 questions)

QFunctionDomainRange
4.1 1$y = \cos x$
4.2 1$y = \sec x$
4.3 1$y = \cot x$
4.4 1$y = \csc x$

Standard — transformed and combined (6 questions)

4.5 State the range of $y = 2 \sin x - 3$. Show the inequality manipulation.    2 marks

4.6 State the range of $y = 4 - 2\cos x$. (Careful with the sign on the cos coefficient.)    2 marks

4.7 Find all $x \in [0, 2\pi]$ where $\csc x$ is undefined.    2 marks

4.8 State the domain of $y = \tan 2x$ in general-solution form (use $n \in \mathbb{Z}$). Then list every undefined value on $[0, \pi]$.    2 marks

4.9 State the domain of $g(x) = \frac{1}{\sin x - 1}$ in general-solution form.    2 marks

4.10 A student writes "the range of $\csc x$ is $[-1, 1]$". State the trap (referencing the lesson by number) and give the correct range.    2 marks

Extension — combine concepts (2 questions)

4.11 Find the domain and range of $y = \sec\left(x - \frac{\pi}{4}\right) + 2$. Use general-solution form for the domain.    3 marks

4.12 State the domain of $h(x) = \frac{\cos x}{\sin^2 x - \sin x}$. Hint: factorise the denominator first.    3 marks

Stuck on 4.12? Denominator = $\sin x(\sin x - 1)$; need $\sin x \neq 0$ and $\sin x \neq 1$.

5. Self-check the easy 3

Tick the first three once you've verified your method works.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Domains and ranges

$\sin x$: domain all real $x$, range $[-1, 1]$.   $\tan x$: domain all real $x$ except $x = \frac{\pi}{2} + n\pi$, range all real $y$.   $\sec x$: domain all real $x$ except $x = \frac{\pi}{2} + n\pi$, range $(-\infty, -1] \cup [1, \infty)$.

Q1.2 — Why $\tan x$ is undefined at zeros of cos

$\tan x = \frac{\sin x}{\cos x}$; when $\cos x = 0$ the denominator is zero, so the ratio is undefined (division by zero).

Q1.3 — Bounded/unbounded

A function is bounded if its outputs stay between two fixed values. $\sin x$ is bounded ($[-1, 1]$). $\tan x$ is unbounded (extends to $\pm \infty$).

Q3 — Faded example: range of $3 \sin x + 1$

Step 1: $-1 \leq \sin x \leq 1$.
Step 2: $-3 \leq 3 \sin x \leq 3$.
Step 3: $-2 \leq 3 \sin x + 1 \leq 4$.
Step 4: Range = $[-2, 4]$.
Conclusion: Range = $\mathbf{[-2, 4]}$.

Q4.1 — $\cos x$

Domain: all real $x$.   Range: $[-1, 1]$.

Q4.2 — $\sec x$

Domain: all real $x$ except $x = \frac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$.   Range: $(-\infty, -1] \cup [1, \infty)$.

Q4.3 — $\cot x$

Domain: all real $x$ except $x = n\pi$, $n \in \mathbb{Z}$.   Range: all real $y$.

Q4.4 — $\csc x$

Domain: all real $x$ except $x = n\pi$, $n \in \mathbb{Z}$.   Range: $(-\infty, -1] \cup [1, \infty)$.

Q4.5 — Range of $2 \sin x - 3$

$-1 \leq \sin x \leq 1 \Rightarrow -2 \leq 2\sin x \leq 2 \Rightarrow -5 \leq 2 \sin x - 3 \leq -1$. Range = $\mathbf{[-5, -1]}$.

Q4.6 — Range of $4 - 2\cos x$

$-1 \leq \cos x \leq 1 \Rightarrow -2 \leq -2\cos x \leq 2$ (multiplying by $-2$ reverses the inequalities, then re-ordering: $-2 \leq -2\cos x \leq 2$). Add 4: $2 \leq 4 - 2\cos x \leq 6$. Range = $\mathbf{[2, 6]}$.

Q4.7 — Where $\csc x$ is undefined on $[0, 2\pi]$

$\csc x = \frac{1}{\sin x}$, undefined when $\sin x = 0$. On $[0, 2\pi]$: $x = \mathbf{0, \pi, 2\pi}$.

Q4.8 — Domain of $\tan 2x$

$\tan 2x$ undefined when $2x = \frac{\pi}{2} + n\pi$, i.e. $x = \frac{\pi}{4} + \frac{n\pi}{2}$, $n \in \mathbb{Z}$. On $[0, \pi]$: $x = \mathbf{\frac{\pi}{4}, \frac{3\pi}{4}}$.

Q4.9 — Domain of $\frac{1}{\sin x - 1}$

Need $\sin x - 1 \neq 0$, i.e. $\sin x \neq 1$. Solutions of $\sin x = 1$: $x = \frac{\pi}{2} + 2n\pi$. Domain: all real $x$ except $x = \mathbf{\frac{\pi}{2} + 2n\pi}$, $n \in \mathbb{Z}$.

Q4.10 — Critique of "$\csc x \in [-1, 1]$"

This commits Trap 2: cosecant is the reciprocal of sine, so it can never take values in $(-1, 1)$. Correct range: $\mathbf{(-\infty, -1] \cup [1, \infty)}$.

Q4.11 — Domain and range of $\sec\left(x - \frac{\pi}{4}\right) + 2$

$\sec$ undefined when its argument equals $\frac{\pi}{2} + n\pi$: $x - \frac{\pi}{4} = \frac{\pi}{2} + n\pi \Rightarrow x = \frac{3\pi}{4} + n\pi$. Domain: all real $x$ except $\mathbf{x = \frac{3\pi}{4} + n\pi}$, $n \in \mathbb{Z}$.
Range of inner $\sec(\cdot)$: $(-\infty, -1] \cup [1, \infty)$. Add 2: $(-\infty, 1] \cup [3, \infty)$. Range = $\mathbf{(-\infty, 1] \cup [3, \infty)}$.

Q4.12 — Domain of $\frac{\cos x}{\sin^2 x - \sin x}$

Denominator: $\sin^2 x - \sin x = \sin x(\sin x - 1)$. Need both $\sin x \neq 0$ and $\sin x \neq 1$. $\sin x = 0$: $x = n\pi$. $\sin x = 1$: $x = \frac{\pi}{2} + 2n\pi$. Domain: all real $x$ except $x = n\pi$ and $x = \frac{\pi}{2} + 2n\pi$, $n \in \mathbb{Z}$.