Target Audience: Pre-calculus / Calculus I students
Prerequisites: Basic understanding of functions, domain/range, and trigonometry fundamentals
A function \(f\) is called periodic if there exists a positive real number \(T\) such that
The smallest such \(T \gt 0\) is called the fundamental period of \(f\).
The sine function \(f(\theta) = \sin\theta\) has fundamental period \(T = 2\pi\):
This is not a curiosity — it is the defining structural fact about sine. The function traces the same output values over and over as \(\theta\) increases without bound. Every output value in \([-1, 1]\) is achieved infinitely many times.
Intuition: Think of wrapping the real number line around a unit circle of circumference \(2\pi\). Every time you complete one full revolution, you are back at the same point — and \(\sin\theta\) measures the same \(y\)-coordinate again.
| Property | Value |
|---|---|
| Domain | \(\mathbb{R}\) — all real numbers |
| Codomain / Range | \([-1,\, 1]\) |
| Period | \(2\pi\) |
| Zeros | \(\theta = n\pi,\quad n \in \mathbb{Z}\) |
| Maximum value 1 | Achieved at \(\theta = \dfrac{\pi}{2} + 2n\pi\) |
| Minimum value \(-1\) | Achieved at \(\theta = -\dfrac{\pi}{2} + 2n\pi\) |
These two tests answer entirely different questions about a graph.
Question it answers: Is this graph the graph of a function?
A graph represents a function if and only if every vertical line \(x = c\) intersects the graph at most once. If any vertical line hits the graph twice, the relation assigns two outputs to a single input — violating the definition of a function.
Question it answers: Is this function one-to-one (injective)?
A function \(f\) is one-to-one if and only if every horizontal line \(y = k\) intersects the graph at most once. Failing this test means at least two distinct inputs produce the same output.
This is one of the most important conceptual separations in introductory analysis.
| Property | Test | Formal Statement | Failure Means |
|---|---|---|---|
| Is a function | Vertical Line Test | \(\forall x,\; f(x)\) is unique | One input → multiple outputs |
| Is one-to-one | Horizontal Line Test | \(f(a) = f(b) \Rightarrow a = b\) | Two inputs → same output |
These are independent properties. A relation can:
Important: "One-to-one" does not mean "each output comes from exactly one input" in some vague sense — it is the precise logical statement
or equivalently, \(a \neq b \Rightarrow f(a) \neq f(b)\).
Here is the subtlety that trips up many students:
When \(\sin\theta\) is evaluated at \(\theta = \frac{\pi}{6}\) and at \(\theta = \frac{5\pi}{6}\), these are two genuinely distinct real numbers — there is no ambiguity, no repetition, no confusion in the inputs themselves. Yet:
The inputs \(\frac{\pi}{6} \approx 0.524\) and \(\frac{5\pi}{6} \approx 2.618\) are distinct elements of \(\mathbb{R}\). The sine function maps both of them to the same real number \(\frac{1}{2}\). This is not a failure of sine to be a function — it is a failure of sine to be injective.
For any value \(y_0 \in (-1, 1)\), the equation \(\sin\theta = y_0\) has infinitely many solutions:
Every horizontal line \(y = y_0\) (for \(y_0 \in (-1,1)\)) intersects the sine curve infinitely many times — a spectacular failure of the HLT.
The horizontal line \(y = \frac{1}{2}\) cuts through \(\sin\theta\) at:
Every one of those \(\theta\) values is a distinct real number. Sine is a perfectly well-defined function — it just isn't injective.
The parabola \(f(x) = x^2\) provides an excellent finite analogy.
For any \(k \gt 0\), the horizontal line \(y = k\) intersects \(y = x^2\) at exactly two points:
For example, \(f(3) = 9 = f(-3)\). The inputs \(3\) and \(-3\) are distinct, but they share the output \(9\).
| Feature | \(\sin\theta\) | \(f(x) = x^2\) |
|---|---|---|
| Type of failure | Periodic (infinitely many-to-one) | Symmetric (two-to-one for \(k \gt 0\)) |
| Number of inputs per output | Countably infinite | 2 (for \(k \gt 0\)), 1 (for \(k = 0\)) |
| Cause | Periodicity: \(2\pi\) symmetry | Even function: \(f(-x) = f(x)\) |
| Domain restriction fix | Restrict to \([-\pi/2,\, \pi/2]\) | Restrict to \([0, \infty)\) |
| Resulting inverse | \(\arcsin x\) | \(\sqrt{x}\) |
The following well-known functions all pass the VLT (they are genuine functions) but fail the HLT (they are not injective on their natural domains):
| Function | Natural Domain | Why It Fails HLT | Standard Restriction for Inverse |
|---|---|---|---|
| \(\sin\theta\) | \(\mathbb{R}\) | Periodic, period \(2\pi\) | \([-\pi/2,\, \pi/2]\) |
| \(\cos\theta\) | \(\mathbb{R}\) | Periodic, period \(2\pi\) | \([0,\, \pi]\) |
| \(\tan\theta\) | \(\mathbb{R} \setminus \{(2n+1)\pi/2\}\) | Periodic, period \(\pi\) | \((-\pi/2,\, \pi/2)\) |
| \(x^2\) | \(\mathbb{R}\) | Even: \(f(-x) = f(x)\) | \([0,\,\infty)\) |
| \(|x|\) | \(\mathbb{R}\) | Symmetric about \(y\)-axis | \([0,\,\infty)\) |
| \(x^4\) | \(\mathbb{R}\) | Even function | \([0,\,\infty)\) |
| \(\sin^2\theta\) | \(\mathbb{R}\) | Even and periodic | \([0,\, \pi/2]\) |
A non-injective function is not broken — it is a perfectly valid function for most purposes: computing values, graphing, integrating, differentiating. The HLT becomes critical only when you ask:
Can we define an inverse?
The inverse function \(f^{-1}\) is defined by:
For this to define a function (not merely a relation), the right-hand side must determine \(x\) uniquely given \(y\). If \(f\) fails the HLT, then for some \(y\) there are multiple candidates for \(x\), and \(f^{-1}(y)\) would be multivalued — not a function.
Theorem.
A function \(f : A \to B\) has an inverse function \(f^{-1} : B \to A\) if and only if \(f\) is bijective — both injective (one-to-one) and surjective (onto).
For our purposes (defining inverses on restricted domains), surjectivity is handled by adjusting the codomain to equal the range. The hard constraint is injectivity, which is exactly what the HLT tests.
Rather than abandoning the inverse, we restrict the domain of the original function to a region where it is injective. On that restricted domain, the HLT is satisfied, and the inverse function is well-defined.
The restriction must be chosen carefully:
Problem: \(\sin\theta\) is not injective on \(\mathbb{R}\).
Fix: Restrict the domain of sine to \(\left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right]\).
On this interval:
Definition of \(\arcsin\):
Question: Why is \(\arcsin\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}\) and not \(\dfrac{5\pi}{6}\)?
Answer: Both \(\dfrac{\pi}{6}\) and \(\dfrac{5\pi}{6}\) satisfy \(\sin\theta = \dfrac{1}{2}\). However:
By the definition of \(\arcsin\), we select only the value in the principal range, so \(\arcsin\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}\).
The same logic applies to the parabola.
Problem: \(f(x) = x^2\) is not injective on \(\mathbb{R}\) because \(f(-3) = f(3) = 9\).
Fix: Restrict the domain to \([0, \infty)\), where \(f(x) = x^2\) is strictly increasing.
Definition of \(\sqrt{x}\):
This is why \(\sqrt{9} = 3\), not \(\pm 3\). The positive square root is a convention enforced by domain restriction, not an arbitrary choice — it is the canonical inverse of \(x^2\) on \([0,\infty)\).
This tutorial is part of a series on real analysis and function theory. Next: continuity, limits, and the formal ε-δ definition.