Domains and Ranges of Trigonometric Functions
Not every angle can be plugged into every trig function. Tangent blows up at $90^\circ$, cosecant is undefined at $0^\circ$, and cosine never exceeds 1. In this lesson you will systematically determine the domain and range of all six trigonometric functions — an essential foundation for graphing, solving equations, and working with inverse trig functions.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
The function $y = \sin x$ has a maximum value of 1 and a minimum value of $-1$. But $y = \tan x$ has no maximum or minimum — it extends to $+\infty$ and $-\infty$. Why do you think $\sin x$ is bounded while $\tan x$ is not? And why is $\tan x$ undefined at $90^\circ$?
The range restrictions for sine and cosine come from the unit circle ($x^2 + y^2 = 1$), where $x$ and $y$ cannot exceed 1 in magnitude. The domain restriction for tangent comes from division by zero when $\cos x = 0$.
Sine and cosine are defined for all real inputs but can only output values between $-1$ and $1$. Tangent can output any real number but is undefined where cosine is zero. Reciprocal functions swap these features: cosecant and secant have restricted domains and restricted ranges.
sin x, cos x: Domain = all real $x$; Range = $[-1, 1]$; tan x: Domain = all real $x$, $x \neq \frac{\pi}{2} + n\pi$; Range = all real $y$
Pause — copy the domain/range table: $\sin x$ and $\cos x$ have range $[-1,1]$ (all real domain); $\tan x$ has all-real range but domain excludes $x = \frac{\pi}{2} + n\pi$ into your book.
True or false: The function $y = \sin x$ is defined for all real values of $x$.
Key facts
- The domain and range of all six trig functions
- Why division by zero creates domain restrictions
- How the unit circle determines range
Concepts
- How the unit circle determines the range of sine and cosine
- Why tangent is unbounded while sine and cosine are bounded
- How reciprocal functions swap domain and range features
Skills
- State the domain and range of all six trig functions
- Find values of $x$ where a trig function is undefined
- Determine the domain of combinations of trig functions
We just saw that $\sin x$ and $\cos x$ are bounded between $-1$ and $1$, while $\tan x$ has vertical asymptotes where $\cos x = 0$. That raises a question: what do "bounded" and "vertical asymptote" actually mean precisely? This card answers it → pinning down the vocabulary you need to discuss domain and range rigorously.
Bounded: outputs confined between two fixed values — sine and cosine are bounded by $\pm 1$; Vertical asymptote: where the denominator of a fraction equals zero (e.g. $\tan x$ at $x = \frac{\pi}{2} + n\pi$)
Pause — copy the two key terms: "bounded" (output confined to $[-1,1]$ for sine/cosine) and "vertical asymptote" (where denominator = 0, e.g. $\tan x$ at $x = \frac{\pi}{2} + n\pi$) into your book.
Quick check: Which trig function has the range $|y| \geq 1$?
We just saw that bounded means outputs stay in $[-1,1]$ and asymptotes appear where a trig ratio's denominator is zero. That raises a question: how do these properties play out across all six trig functions, not just $\sin$, $\cos$, $\tan$? This card answers it → using the unit circle to derive domain and range for all six, including the reciprocals.
On the unit circle, $\sin \theta = y$ and $\cos \theta = x$. Since every point on the unit circle has $x^2 + y^2 = 1$, both $x$ and $y$ must lie between $-1$ and $1$ inclusive. The angle $\theta$ can be any real number — we can rotate infinitely many times around the circle.
$\tan x = \frac{\sin x}{\cos x}$ is undefined whenever $\cos x = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. As $x$ approaches these values, $\tan x$ approaches $\pm\infty$.
The reciprocal functions inherit restrictions from their denominators:
- $\csc x = \frac{1}{\sin x}$: undefined at $x = n\pi$; range $|y| \geq 1$
- $\sec x = \frac{1}{\cos x}$: undefined at $x = \frac{\pi}{2} + n\pi$; range $|y| \geq 1$
- $\cot x = \frac{\cos x}{\sin x}$: undefined at $x = n\pi$; range all real $y$
Unit circle: $\sin\theta = y$-coord, $\cos\theta = x$-coord, both between $-1$ and $1$; $\tan x = \frac{\sin x}{\cos x}$ — undefined when $\cos x = 0$, i.e. $x = \frac{\pi}{2} + n\pi$
Pause — copy the unit circle link ($\sin\theta = y$-coord, $\cos\theta = x$-coord, both in $[-1,1]$) and the $\tan x = \frac{\sin x}{\cos x}$ undefined-when-$\cos x = 0$ rule into your book.
Fill the blanks: drag each token to the matching gap.
The range of $\sin x$ is ___. Tangent is undefined when ___. As $x$ approaches a vertical asymptote, $\tan x$ approaches ___. Cosecant is undefined when ___.
We just saw that $\tan x = \frac{\sin x}{\cos x}$ is undefined whenever $\cos x = 0$, which creates vertical asymptotes. That raises a question: how do we find all such values systematically within a given interval? This card answers it → solve $\cos x = 0$ on $[0, 2\pi]$: solutions are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.
Find all values of $x$ in $[0, 2\pi]$ where $\tan x$ is undefined.
To find where $\tan x$ is undefined: solve $\cos x = 0$; In $[0, 2\pi]$: $\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$
Pause — copy the procedure (find where $\tan x$ is undefined: solve $\cos x = 0$) and the two solutions on $[0, 2\pi]$: $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ into your book.
Check: At which value is $\tan x$ undefined in $[0, 2\pi]$?
We just saw that $\tan x$ is undefined at exactly the values where $\cos x = 0$. That raises a question: what if we have a more complex fraction like $\frac{1}{1 + \cos x}$, where the denominator is not just $\cos x$ alone? This card answers it → set the whole denominator to zero: $1 + \cos x = 0 \Rightarrow \cos x = -1$, giving $x = \pi + 2n\pi$.
State the domain of $f(x) = \frac{\sin x}{1 + \cos x}$.
For combined trig fractions: set the denominator equal to zero and solve; $1 + \cos x = 0 \Rightarrow \cos x = -1 \Rightarrow x = \pi + 2n\pi$
Pause — copy the combined-denominator strategy (set entire denominator = 0, solve for $x$) and the worked result $1 + \cos x = 0 \Rightarrow x = \pi + 2n\pi$ into your book.
In your own words: Explain the first step you take to find the domain of a fraction involving trig functions.
We just saw that domain is found by excluding values that make the denominator zero. That raises a question: if domain tells us what goes in, how do we find the range — what outputs are actually achievable for a transformed trig function? This card answers it → start from $-1 \leq \sin x \leq 1$ and apply transformations step by step across the entire inequality.
Find the range of $y = 3\sin x + 1$.
Start with $-1 \leq \sin x \leq 1$, then apply transformations step by step; Multiply all parts of the inequality by the amplitude (keep direction unless multiplying by negative)
Pause — copy the inequality-chain strategy ($-1 \leq \sin x \leq 1$, then multiply all parts by amplitude) and the caution (flip inequality direction when multiplying by a negative) into your book.
Apply it: What is the range of $y = 2\cos x - 1$?
We just saw the inequality-chain method for range and the domain exclusion method for domain. That raises a question: where do students most often go wrong applying these procedures under exam conditions? This card answers it → Trap 1: forgetting $\tan x$ has domain restrictions; Trap 2: assuming reciprocal functions have range $[-1,1]$ when actually $|y| \geq 1$.
Students sometimes say tangent is defined for all real numbers. It is not — it has vertical asymptotes where cosine is zero. Always remember $\tan x = \frac{\sin x}{\cos x}$.
Cosecant is the reciprocal of sine, so it can never take values between $-1$ and $1$. Its range is $y \leq -1$ or $y \geq 1$. For reciprocal trig functions, the range excludes the interval $(-1, 1)$.
When stating the domain, there are infinitely many excluded values. You must use $+ n\pi$ or $+ 2n\pi$ notation. Writing $x \neq \frac{\pi}{2}, \frac{3\pi}{2}$ is incomplete — use $x \neq \frac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$.
Trap 1 fix: $\tan x$ has restricted domain — always check if denominator can be zero; Trap 2 fix: reciprocal of small number is large — $\csc x, \sec x$ have range $|y| \geq 1$
Pause — copy both traps: Trap 1 ($\tan x$ has restricted domain — always check denominator) and Trap 2 (reciprocals $\csc x, \sec x$ have range $|y| \geq 1$, not $[-1,1]$) into your book.
Odd one out: Which statement is INCORRECT?
We just saw the two traps — forgetting $\tan$ domain restrictions and confusing reciprocal ranges. That raises a question: can you identify domain and range quickly under time pressure? This card answers it → rapid-fire practice across six functions to build the pattern recognition you need for exam conditions.
State the domain and range of each function.
$y = \cos x$
$y = \sec x$
$y = \cot x$
$y = 2\sin x - 3$
Find all $x \in [0, 2\pi]$ where $\csc x$ is undefined.
$\csc x = \frac{1}{\sin x}$ is undefined when $\sin x = 0$.
For any transformed sine/cosine: Domain = all real $x$; find Range using inequality chain; For $y = 2\sin x - 3$: $-2 \leq 2\sin x \leq 2$, then $-5 \leq 2\sin x - 3 \leq -1$
Pause — copy the shortcut for transformed sine/cosine range: apply the inequality chain to $-1 \leq \sin x \leq 1$, e.g. $y = 2\sin x - 3$ gives range $[-5, -1]$ into your book.
Match up: connect each function to its range.
Return to your original answer from Section 01. $\sin x$ represents the $y$-coordinate on the unit circle, which can never be larger than 1 or smaller than $-1$. So sine is bounded. But $\tan x = \frac{\sin x}{\cos x}$ is a ratio. As $x$ approaches $90^\circ$, $\cos x$ approaches 0, making the ratio blow up to $\pm\infty$. At exactly $90^\circ$, $\cos x = 0$, so division by zero makes $\tan x$ undefined.
Did your initial thinking distinguish between boundedness (a property of the output) and undefined points (a property of the input)?
Domain of a transformed tangent
(a) State the domain of $y = \tan 2x$. (b) Find all values of $x$ in $[0, \pi]$ where $\tan 2x$ is undefined.
View comprehensive answer
(a) $2x \neq \frac{\pi}{2} + n\pi \Rightarrow x \neq \frac{\pi}{4} + \frac{n\pi}{2}$.
(b) In $[0, \pi]$: $x = \frac{\pi}{4}, \frac{3\pi}{4}$.
Range of a transformed cosine
Find the range of $y = 4 - 2\cos x$. Show your reasoning.
View comprehensive answer
Working:
$-1 \leq \cos x \leq 1$
Multiply by $-2$ (flip inequalities): $-2 \leq -2\cos x \leq 2$
Add 4: $2 \leq 4 - 2\cos x \leq 6$
Answer: Range is $\mathbf{[2, 6]}$
Careful: multiplying by a negative number reverses the inequality signs.
Explain the range of secant
Explain why the range of $y = \sec x$ is $y \leq -1$ or $y \geq 1$. Your explanation should reference the range of $\cos x$.
View comprehensive answer
$\sec x = \frac{1}{\cos x}$.
Since $-1 \leq \cos x \leq 1$ and $\cos x \neq 0$, the reciprocal satisfies $|\sec x| \geq 1$.
When $\cos x$ is positive, $\sec x \geq 1$. When $\cos x$ is negative, $\sec x \leq -1$.
Therefore the range is $y \leq -1$ or $y \geq 1$.
The key insight: taking the reciprocal of a number in $(-1, 1)$ (excluding 0) produces a number outside $(-1, 1)$.
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