In short
A function f is periodic with period T > 0 if f(x + T) = f(x) for every x in its domain. The smallest such T is the fundamental period. Trigonometric functions are the most familiar examples: \sin x and \cos x have fundamental period 2\pi, while \tan x has fundamental period \pi. The period changes predictably under horizontal scaling, but finding the period of a sum or product of periodic functions requires care — it is not always the LCM of the individual periods.
A train on the Delhi Metro runs every 5 minutes during peak hours. If you arrive at the platform at 8:00 AM and see a train, you will see another train at 8:05, another at 8:10, and so on. The pattern of arrivals repeats with a fixed interval of 5 minutes. If you model the train's position as a function of time, that function has the same shape every 5 minutes — it is periodic with a period of 5 minutes.
The same kind of repetition appears everywhere: the phases of the moon repeat roughly every 29.5 days, the vibration of a guitar string repeats hundreds of times per second, and the voltage in your home's AC power supply completes one full cycle 50 times per second (50 Hz). Whenever a process repeats, the function describing it is periodic.
In mathematics, periodicity is not just a property — it is a tool. If you know one full cycle of a periodic function, you know the entire function. Every value, every slope, every area — all information about the function can be extracted from a single period.
The formal definition
Periodic function
A function f is periodic if there exists a positive number T such that
Any such T is called a period of f.
The condition says: shifting the input by T does not change the output. The graph of f looks exactly the same if you slide it T units to the left.
If T is a period, then 2T, 3T, 4T, and in general nT (for any positive integer n) are also periods. Proof: f(x + 2T) = f((x + T) + T) = f(x + T) = f(x). The same argument extends by induction. So a periodic function has infinitely many periods.
Fundamental period
The fundamental period (or simply the period) of a periodic function is the smallest positive value of T for which f(x + T) = f(x) for all x.
Not every periodic function has a fundamental period. The constant function f(x) = 5 satisfies f(x + T) = 5 = f(x) for every positive T, so every positive number is a period. There is no smallest positive period. But among non-constant periodic functions, the fundamental period almost always exists.
Periods of the standard functions
The trigonometric functions are the most important periodic functions. Here are their fundamental periods:
| Function | Fundamental period |
|---|---|
| \sin x | 2\pi |
| \cos x | 2\pi |
| \tan x | \pi |
| \cot x | \pi |
| \sec x | 2\pi |
| \csc x | 2\pi |
| \sin^2 x | \pi |
| $ | \sin x |
The last two deserve explanation. For \sin^2 x: use the identity \sin^2 x = \frac{1 - \cos 2x}{2}. Since \cos 2x has period \frac{2\pi}{2} = \pi, so does \sin^2 x. For |\sin x|: the absolute value folds the negative half-waves upward, so the pattern that was a full sine wave over [0, 2\pi] now repeats over [0, \pi].
Period after horizontal scaling
If f(x) has fundamental period T, what is the period of f(ax) where a > 0?
Replace x with x + \frac{T}{a} in the argument:
So \frac{T}{a} is a period of f(ax). If T was the fundamental period of f, then \frac{T}{a} is the fundamental period of f(ax).
The rule: f(ax) has period \frac{T}{a}.
This is why \sin 2x has period \frac{2\pi}{2} = \pi, and \cos \frac{x}{3} has period \frac{2\pi}{1/3} = 6\pi.
Vertical scaling and vertical shifting do not change the period. The function 5\sin x + 3 has the same period 2\pi as \sin x — you are stretching and shifting the graph vertically, which does not affect the horizontal repetition interval.
Period of a sum of periodic functions
If f has period T_1 and g has period T_2, when is f + g periodic, and what is its period?
The natural guess is: the period of f + g is \operatorname{lcm}(T_1, T_2) — the least common multiple. This is true when T_1 and T_2 are both rational multiples of some common unit, because then there exists a T that is a multiple of both T_1 and T_2, and we have:
Proof that \operatorname{lcm}(T_1, T_2) is a period. Let T = \operatorname{lcm}(T_1, T_2). Then T = m \cdot T_1 for some positive integer m, and T = n \cdot T_2 for some positive integer n. So:
Adding: (f + g)(x + T) = f(x) + g(x) = (f + g)(x).
So \operatorname{lcm}(T_1, T_2) is a period. But is it the fundamental period? Not necessarily. The fundamental period of f + g could be smaller if there is cancellation.
Example of cancellation. Take f(x) = \sin x + \cos x. Here \sin x has period 2\pi and \cos x has period 2\pi. The LCM is 2\pi. And indeed, \sin x + \cos x has fundamental period 2\pi.
But now take f(x) = |\sin x| + |\cos x|. Both |\sin x| and |\cos x| have period \pi. The LCM is \pi. But |\sin x| + |\cos x| actually has fundamental period \frac{\pi}{2}, because the sum has extra symmetry: shifting by \frac{\pi}{2} swaps the roles of sine and cosine, leaving the sum unchanged.
When the ratio T_1/T_2 is irrational, the function f + g is generally not periodic. The function \sin x + \sin(\sqrt{2}\, x) is not periodic, because there is no common multiple of 2\pi and \frac{2\pi}{\sqrt{2}} = \pi\sqrt{2}. The periods 2\pi and \pi\sqrt{2} have an irrational ratio, so no finite T can be a multiple of both.
Period of a product
The product f \cdot g of two periodic functions does not follow a simple universal rule. But for specific cases, you can work it out from identities.
Example. \sin x \cdot \cos x = \frac{1}{2}\sin 2x, which has period \frac{2\pi}{2} = \pi. The original factors both have period 2\pi, but their product has period \pi — the product's period is smaller than either factor's.
Example. \sin^2 x = \frac{1 - \cos 2x}{2}, which has period \frac{2\pi}{2} = \pi.
The key technique: convert the product into a sum using trigonometric identities, then determine the period of the sum.
Two worked examples
Example 1: Find the fundamental period of $f(x) = \sin 3x + \cos 5x$
Step 1. Find the period of each term.
\sin 3x has period \frac{2\pi}{3}. \cos 5x has period \frac{2\pi}{5}.
Why: the rule f(ax) has period \frac{T}{a} gives \frac{2\pi}{3} for \sin 3x and \frac{2\pi}{5} for \cos 5x.
Step 2. Compute \operatorname{lcm}\!\left(\frac{2\pi}{3}, \frac{2\pi}{5}\right).
For fractions \frac{a}{b} and \frac{c}{d}, \operatorname{lcm}\!\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\operatorname{lcm}(a, c)}{\gcd(b, d)}.
Here: \operatorname{lcm}(2\pi, 2\pi) = 2\pi and \gcd(3, 5) = 1. So \operatorname{lcm}\!\left(\frac{2\pi}{3}, \frac{2\pi}{5}\right) = \frac{2\pi}{1} = 2\pi.
Why: we need the smallest T that is simultaneously a multiple of \frac{2\pi}{3} and a multiple of \frac{2\pi}{5}. That means T = k \cdot \frac{2\pi}{3} = m \cdot \frac{2\pi}{5} for positive integers k, m. The smallest solution is k = 3, m = 5, giving T = 2\pi.
Step 3. Verify. Check that no smaller period works. At T = 2\pi: \sin(3(x + 2\pi)) = \sin(3x + 6\pi) = \sin 3x and \cos(5(x + 2\pi)) = \cos(5x + 10\pi) = \cos 5x. Both terms repeat, so f(x + 2\pi) = f(x).
Could T = \pi work? \sin(3(x + \pi)) = \sin(3x + 3\pi) = -\sin 3x \neq \sin 3x in general. So \pi is not a period.
Why: a candidate period must work for both terms simultaneously. Failing for one term is enough to reject it.
Step 4. The fundamental period is 2\pi.
Result: The fundamental period of \sin 3x + \cos 5x is 2\pi.
The visual check confirms: the wave pattern to the right of x = 2\pi is an exact copy of the pattern on [0, 2\pi).
Example 2: Find the fundamental period of $f(x) = \tan\!\left(\frac{\pi x}{3}\right) + \sin\!\left(\frac{\pi x}{4}\right)$
Step 1. Find the period of each term.
\tan\!\left(\frac{\pi x}{3}\right): tangent has base period \pi. With argument \frac{\pi x}{3}, the coefficient of x is \frac{\pi}{3}, so the period is \frac{\pi}{\pi/3} = 3.
\sin\!\left(\frac{\pi x}{4}\right): sine has base period 2\pi. With argument \frac{\pi x}{4}, the coefficient of x is \frac{\pi}{4}, so the period is \frac{2\pi}{\pi/4} = 8.
Why: for \tan(ax), the period is \frac{\pi}{a}. For \sin(ax), the period is \frac{2\pi}{a}.
Step 2. Compute \operatorname{lcm}(3, 8).
Since \gcd(3, 8) = 1, we get \operatorname{lcm}(3, 8) = 3 \times 8 = 24.
Why: the periods are both integers with no common factors, so the LCM is simply their product.
Step 3. Verify. At T = 24: \tan\!\left(\frac{\pi(x+24)}{3}\right) = \tan\!\left(\frac{\pi x}{3} + 8\pi\right) = \tan\!\left(\frac{\pi x}{3}\right) (since \tan has period \pi and 8\pi is 8 full periods). And \sin\!\left(\frac{\pi(x+24)}{4}\right) = \sin\!\left(\frac{\pi x}{4} + 6\pi\right) = \sin\!\left(\frac{\pi x}{4}\right) (since 6\pi is 3 full periods of 2\pi). Both terms repeat.
Could T = 12 work? For \tan: \frac{\pi \cdot 12}{3} = 4\pi, which is a multiple of \pi. For \sin: \frac{\pi \cdot 12}{4} = 3\pi, which is an odd multiple of \pi, giving \sin\!\left(\frac{\pi x}{4} + 3\pi\right) = -\sin\!\left(\frac{\pi x}{4}\right) \neq \sin\!\left(\frac{\pi x}{4}\right) in general. So T = 12 fails.
Why: the tangent piece repeats every 3 units, so at T = 12 it has completed 4 full periods. But the sine piece needs 8 units for one period, so at T = 12 it has completed 1.5 periods — exactly out of phase.
Step 4. The fundamental period is 24.
Result: The fundamental period of \tan\!\left(\frac{\pi x}{3}\right) + \sin\!\left(\frac{\pi x}{4}\right) is 24.
The diagram shows why 24 is the answer: it is the first point where both the tangent cycles (groups of 3) and the sine cycles (groups of 8) complete a whole number of repetitions. At x = 12, the tangent has completed 4 full cycles, but the sine has only done 1.5 — so the sum at x = 12 is not the same as at x = 0.
Common confusions
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"The period of f + g is always \operatorname{lcm}(T_1, T_2)." The LCM is always a period, but it might not be the fundamental period. The fundamental period can be smaller if cancellation occurs. And if T_1/T_2 is irrational, f + g may not be periodic at all.
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"A constant function has period 0." A period must be a positive number. The constant function f(x) = c has every positive number as a period, but no fundamental period. We say f is periodic but does not have a fundamental period.
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"f(x) = x - [x] has period 1." This is the fractional part function \{x\}, and yes, it has period 1. But students sometimes write f(x) = x - [x] = \{x\} and then forget that this equals the fractional part. Recognise the connection: \{x\} is periodic precisely because subtracting the floor function removes the integer part, and the leftover repeats every unit.
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"Periodic functions must be bounded." \tan x is periodic with period \pi, but it is unbounded — it shoots off to \pm\infty at each asymptote. Periodicity says the shape repeats; it does not say the shape stays within finite bounds.
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"If f has period T, then f^2 has period T." T is certainly a period of f^2, but the fundamental period of f^2 might be smaller. For example, \sin x has period 2\pi, but \sin^2 x has period \pi. Squaring can halve the period because it removes the sign changes.
Going deeper
If you can find the fundamental period of any trigonometric expression, determine whether a sum of periodic functions is periodic, and apply the horizontal scaling rule, you have the tools for the next chapter. The material below explores subtler points.
A proof that the LCM works only for rational ratios
Let f have period T_1 and g have period T_2. If f + g is periodic with period T, then T must satisfy T = m T_1 = n T_2 for positive integers m, n. This gives \frac{T_1}{T_2} = \frac{n}{m}, which is rational.
So if T_1/T_2 is irrational, no such T exists, and f + g is not periodic (assuming the individual functions are "genuinely" periodic in the sense that no simpler relation links them).
This is why \sin x + \sin(\sqrt{2}\,x) is not periodic. The ratio of the periods is \frac{2\pi}{2\pi/\sqrt{2}} = \sqrt{2}, which is irrational. You can prove rigorously that no T > 0 satisfies (f+g)(x+T) = (f+g)(x) for all x.
An interactive period explorer
Drag the point to change the frequency multiplier n in \sin(nx). The graph updates in real time, and the period \frac{2\pi}{n} shrinks as n increases.
Period and composition
If f is periodic with period T and g is any function, then f \circ g is periodic if g(x + S) = g(x) + T for some S. In particular, f(g(x)) where g(x) = ax + b gives period T/a (the horizontal scaling rule is a special case of this).
But if g is not linear, f \circ g may not be periodic at all. For instance, \sin(x^2) is not periodic, even though \sin is — the squaring function does not translate periods into periods.
Where this leads next
Periodicity is the mathematical framework behind waves, oscillations, and cycles. The articles below use periodic functions in more specific settings.
- Trigonometric Functions — the canonical periodic functions, with their exact values, identities, and graphs.
- Even and Odd Functions — the other kind of function symmetry. Many trigonometric functions are both periodic and even (like \cos) or both periodic and odd (like \sin).
- Functions — Definition and Notation — the foundational definitions that all of this builds on.
- Graphs of Basic Functions — the visual library of function shapes, including the periodic ones.
- Composition of Functions — how composing with a periodic function can create or destroy periodicity.