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.
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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.
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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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