Periodic functions: the values repeat every lap

Explain it like I am five

Picture the hour hand of a clock: every 12 hours it is back in the same spot. A periodic function works the same way. Slide the input forward by a fixed step k and the output repeats: f(t + k) = f(t), read 'f of t plus k equals f of t'. The smallest such step is the period.

On the unit circle one full lap is 360°, or 2π (two pi). A full lap returns the point to the same spot, so all six values repeat. Angles that end in the same place, such as 60° and 420°, are coterminal; they share every value, so you can trade a big angle for a small one you know.

Example: cos 1140°. Count laps: 1140 ÷ 360 ≈ 3.17 (≈ means about), so remove 3 × 360° = 1080°, leaving 60°. So cos 1140° = cos 60° = $1 over 2$ .

Tangent and cotangent repeat sooner, every half lap (180°, or π). A half turn sends the point (x, y) to (−x, −y), and $−y over −x$ = $\frac{y}{x}$ because the two minus signs cancel.

In plain words

Think of a calendar. Seven days after a Monday, the weekday is Monday again. A periodic function repeats its outputs after a fixed positive change in its input. On the unit circle, turning one full lap brings the rider to the same position. The coordinates, and every function built from them, repeat. One lap is 2π radians or 360°. Sine, cosine, secant and cosecant need that full lap for their shortest repeat, called the period. Tangent and cotangent repeat after half a lap, π radians or 180°, because both coordinates reverse sign together and their ratio stays the same.

30° and 390° end on the same side, so every trig function has the same value at both; 750° adds one more lap.

Reminder

Common denominators. For sixths of π, 2π = $12 π over 6$ , so $25 π over 6$ − 2π = $13 π over 6$ .
Signs in a quotient. $−3 over −2$ = $3 over 2$ because two negative signs cancel.
Integer multiples. Three periods of tangent in radians are 3π; three full laps are 6π.

30° and 390° end on the same side, so every trig function has the same value at both; 750° adds one more lap.

Coordinate functions and their reciprocals repeat after a full lap; ratios repeat after a half lap.

Read the period in the column under the function.

Why it works. The circle's circumference, its distance around the rim, is 2π × 1 = 2π, so a full lap restores both coordinates and every allowed division built from them. A half lap sends (x, y) to (−x, −y). Thus = and = , proving the shorter tangent and cotangent repeats. The other four change sign after half a lap, so that repeat fails. To prove these are the shortest periods, compare repeated maximum values or repeated zero values, as the individual parts below show.

RuleA positive repeat k preserves the domain in both directions and satisfies f(t + k) = f(t) wherever defined. A function with such a repeat is periodic. If there is a smallest positive repeat, it is the period. For any integer n: sin(t ± 2nπ) = sin t, and the same for cos, sec and csc; tan(t ± nπ) = tan t, and the same for cot. Degree repeats are 360° and 180°, respectively. In degrees, sin t = sin(t ± n·360°), with integer n; likewise cos, sec and csc.

The same idea, five ways

Say it

A periodic function repeats after a fixed positive change in its input.

Write it

Adding or subtracting whole periods preserves the value and keeps permitted inputs permitted.

In math

f(t + k) = f(t) for a positive repeat k.
sin(t ± 2nπ) = sin t, with integer n; likewise cos, sec and csc.
tan(t ± nπ) = tan t, with integer n; likewise cot.
Coordinate and reciprocal periods: 2π = 360°.
Ratio periods: π = 180°.
On a graph, the same shape returns after a horizontal shift of one period.

A weekday repeats after seven days even while the date number changes.

See it

The sine output pattern returns after a horizontal change of 2π.

The same idea, other ways

As a calendar

Adding 7 days preserves the weekday. Adding 360° preserves a point on the wheel. The extra travel changes the input while the output repeats.

An extra full lap preserves the terminal point.

As coordinates

A half lap changes (x, y) into (−x, −y). Sine and cosine change sign, but tangent and cotangent divide two negatives and keep their values.

The point opposite 60° has both coordinates negated.

With numbers

Sine at 30°, 390° and 750° is $1 over 2$ each time. Tangent at 60° and 240° is $square root of 3$ ; the half-lap ratio does not change.

The same output repeats after each full lap.

Why shortest matters

An output matching at one pair of inputs does not establish a period. The change must work at every permitted input. A constant function, such as f(t) = 3, repeats after every positive change, so it is periodic without a smallest positive period.

A constant output repeats after every positive input change, so no smallest repeat exists.

Function	sin t	cos t	tan t	cot t	sec t	csc t
Period	2π (360°)	2π (360°)	π (180°)	π (180°)	2π (360°)	2π (360°)

.1Sine repeats height

A full lap returns the rider to the same height. Half a lap usually puts the rider at the opposite height.

sin(t + 2π) = sin t.
Period: 2π, or 360°.
Any sine period k must give sin k = sin 0 = 0, so k is a multiple of π. The candidate π fails at t = $π over 2$ , so 2π is the smallest positive repeat.

Removing full laps preserves this function's output.

Worked exampleA repeat of sin

The question asks for the unchanged output after removing periods. Find sin 1110°.

Removing full laps preserves this function's output.

1110° − 3 × 360° = 30°.Remove 3 whole periods of 360°; the output is preserved.
sin 1110° = sin 30° = $1 over 2$ .sin 30° = $1 over 2$

Answer

sin 1110° = .

Check The angle difference is 1080°, exactly 3 periods of 360°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A matching output at one pair of inputs proves the period.

A period must preserve every allowed input's output and domain.

✓ Instead: Use the full coordinate or ratio argument and the least-period test.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.2Cosine repeats horizontal position

A full lap returns the rider to the same horizontal position. A half lap reverses that position.

cos(t + 2π) = cos t.
Period: 2π, or 360°.
At t = 0, cosine is 1. The next angle with cosine 1 is 2π, so no smaller positive shift can repeat every cosine value.

Removing full laps preserves this function's output.

Worked exampleA repeat of cos

The question asks for the unchanged output after removing periods. Find cos 1500°.

Removing full laps preserves this function's output.

1500° − 4 × 360° = 60°.Remove 4 whole periods of 360°; the output is preserved.
cos 1500° = cos 60° = $1 over 2$ .cos 60° = $1 over 2$

Answer

cos 1500° = .

Check The angle difference is 1440°, exactly 4 periods of 360°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A matching output at one pair of inputs proves the period.

A period must preserve every allowed input's output and domain.

✓ Instead: Use the full coordinate or ratio argument and the least-period test.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.3Tangent repeats after a half lap

A half lap reverses both coordinates. Height divided by horizontal position keeps its value because both signs reverse.

tan(t + π) = tan t wherever defined.
Period: π, or 180°.
At t = 0, tangent is 0. Its next zero is π, so a smaller positive shift cannot repeat all tangent values.

The ratio pattern repeats after each horizontal change of π.

Worked exampleA repeat of tan

The question asks for the unchanged output after removing periods. Find tan 1140°.

The ratio pattern repeats after each horizontal change of π.

1140° − 6 × 180° = 60°.Remove 6 whole periods of 180°; the output is preserved.
tan 1140° = tan 60° = $square root of 3$ .tan 60° = ( $square root of 3 over 2$ ) ÷ $1 over 2$ = $square root of 3$

Answer

tan 1140° = .

Check The angle difference is 1080°, exactly 6 periods of 180°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A matching output at one pair of inputs proves the period.

A period must preserve every allowed input's output and domain.

✓ Instead: Use the full coordinate or ratio argument and the least-period test.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.4Cotangent repeats after a half lap

Horizontal position divided by height also keeps its value after both coordinates reverse sign.

cot(t + π) = cot t wherever defined.
Period: π, or 180°.
At t = $π over 2$ , cotangent is 0. Its next zero is $3 π over 2$ , a distance π later, so no smaller positive shift repeats all its values.

The ratio pattern repeats after each horizontal change of π.

Worked exampleA repeat of cot

The question asks for the unchanged output after removing periods. Find cot 1305°.

The ratio pattern repeats after each horizontal change of π.

1305° − 7 × 180° = 45°.Remove 7 whole periods of 180°; the output is preserved.
cot 1305° = cot 45° = 1.cot 45° = ( $square root of 2 over 2$ ) ÷ $square root of 2 over 2$ = 1

Answer

cot 1305° = 1.

Check The angle difference is 1260°, exactly 7 periods of 180°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A matching output at one pair of inputs proves the period.

A period must preserve every allowed input's output and domain.

✓ Instead: Use the full coordinate or ratio argument and the least-period test.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.5Secant repeats its horizontal reciprocal

A full lap restores cosine, so it restores secant. A half lap changes cosine's sign and therefore secant's sign.

sec(t + 2π) = sec t wherever defined.
Period: 2π, or 360°.
At t = 0, secant is 1. That requires cosine 1, which next occurs at 2π. A smaller positive shift cannot repeat all secant values.

Removing full laps preserves this function's output.

Worked exampleA repeat of sec

The question asks for the unchanged output after removing periods. Find sec 1500°.

Removing full laps preserves this function's output.

1500° − 4 × 360° = 60°.Remove 4 whole periods of 360°; the output is preserved.
sec 1500° = sec 60° = 2.sec 60° = 1 ÷ $1 over 2$ = 2

Answer

sec 1500° = 2.

Check The angle difference is 1440°, exactly 4 periods of 360°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A matching output at one pair of inputs proves the period.

A period must preserve every allowed input's output and domain.

✓ Instead: Use the full coordinate or ratio argument and the least-period test.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.6Cosecant repeats its height reciprocal

A full lap restores sine, so it restores cosecant. A half lap reverses the sign of both sine and cosecant.

csc(t + 2π) = csc t wherever defined.
Period: 2π, or 360°.
At t = $π over 2$ , cosecant is 1. Sine next returns to 1 at $5 π over 2$ , a distance 2π later, so no smaller positive shift repeats all its values.

Removing full laps preserves this function's output.

Worked exampleA repeat of csc

The question asks for the unchanged output after removing periods. Find csc 1110°.

Removing full laps preserves this function's output.

1110° − 3 × 360° = 30°.Remove 3 whole periods of 360°; the output is preserved.
csc 1110° = csc 30° = 2.csc 30° = 1 ÷ $1 over 2$ = 2

Answer

csc 1110° = 2.

Check The angle difference is 1080°, exactly 3 periods of 360°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A matching output at one pair of inputs proves the period.

A period must preserve every allowed input's output and domain.

✓ Instead: Use the full coordinate or ratio argument and the least-period test.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

Strategy: step by step

1. Identify the requested function and its period: 2π (360°) for sin, cos, sec and csc; π (180°) for tan and cot.
2. Add whole periods to a negative angle, or subtract them from an angle at least one period, until 0 ≤ reduced angle < period.
3. Use a known special or axis value if the reduced angle is one you know. Other angles may permit a sign decision without a familiar exact value.
4. The original expression has exactly the same output, sign included. If the reduced expression is undefined, the original one is also undefined.
Read a reference-table entry in the column under its input or category in the matching picture.

Strategy

Reducing an angle by a function's period

Is the function tangent or cotangent?

YesUse π radians or 180°.

NoUse 2π radians or 360°.

↓

Is the input negative?

YesAdd periods until 0 ≤ input.

NoKeep the input and consider whether it is too large.

↓

Is the input at least one period?

YesSubtract periods until the input is below one period.

NoUse the resulting input as it stands.

1. Identify the function and choose its period in the input's unit.
2. For a negative angle, add whole periods until the result is at least 0.
3. For an angle at least one period, subtract whole periods until the result is below one period.
4. Evaluate the reduced angle if its value is known. If it is not a known exact angle, the reduction still helps locate its sign.

Worked exampleReducing a negative radian angle and a large degree angle by whole periods

Find the exact value of each expression without a calculator. (a) sin(− $17 π over 6$ ) (b) tan(1110°)

The angles 30° and 1110° share a terminal side, because 1110° = 30° + 3·360°. So tan(1110°) = tan 30° = /3.

(a) The function is sine, so the period is 2π = $12 π over 6$ .sin(t ± 2nπ) = sin t for every integer n, and 2π is the smallest positive repeat of sine.
(a) The angle is negative, so add whole periods. Add 2π once: − $17 π over 6$ + $12 π over 6$ = − $5 π over 6$ . This is still negative, so add 2π again: − $5 π over 6$ + $12 π over 6$ = $7 π over 6$ . Now 0 ≤ $7 π over 6$ < 2π.Adding two whole periods (n = 2) moves the angle into the interval from 0 to one period without changing the sine value.
(a) sin( $7 π over 6$ ) = − $1 over 2$ . $7 π over 6$ is the special angle $π over 6$ past π, in Quadrant III. There sine is negative and the reference value is sin( $π over 6$ ) = $1 over 2$ .
(a) So sin(− $17 π over 6$ ) = sin( $7 π over 6$ ) = − $1 over 2$ .The original expression has exactly the same output as the reduced one, sign included.
(b) The function is tangent, so the period is 180°.tan(t ± n·180°) = tan t for every integer n. Tangent repeats every half lap, not every full lap.
(b) The angle is at least one period, so subtract whole periods. 1110° ÷ 180° is 6 with remainder 30°, so subtract 6 periods: 1110° − 6·180° = 1110° − 1080° = 30°. Now 0° ≤ 30° < 180°.Subtracting six whole periods (n = 6) gives the reduced angle in the interval from 0° up to one period.
(b) tan(30°) = $1 over square root of 3$ = $square root of 3 over 3$ , so tan(1110°) = $square root of 3 over 3$ .30° is a special angle with tan 30° = $sin 30° over cos 30°$ = $1/2 over square root of 3 /2$ . The original expression has the same value and the same sign.

Answer

(a) sin(−) = − (b) tan(1110°) =

Check (a) Convert to degrees: − = −510°. Adding 2·360° gives −510° + 720° = 210°, which is in Quadrant III, and sin 210° = −. This matches. (b) Using full laps instead: 1110° − 3·360° = 30°, so 1110° is coterminal with 30° and tan(1110°) = tan 30° = . This agrees with the half-lap reduction.

Work to write

sine has period 2π; tangent has period 180°
− $17 π over 6$ + 2·2π = $7 π over 6$
sin(− $17 π over 6$ ) = sin( $7 π over 6$ ) = − $1 over 2$
1110° − 6·180° = 30°
tan(1110°) = tan 30° = $square root of 3 over 3$

(a) sin(− $17 π over 6$ ) = − $1 over 2$ (b) tan(1110°) = $square root of 3 over 3$

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: remove one full lap

The question asks for sine after removing one full lap. Find sin 390°.

One extra lap leaves sine unchanged.

390° − 360° = 30°.Sine repeats every 360°.
sin 390° = sin 30° = $1 over 2$ .The special-angle sine is $1 over 2$ .

Answer

sin 390° = .

Check The terminal point is exactly the same as at 30°.

Rung 2Rung 2: use the shorter period

The question asks for tangent after removing half-lap periods. Find tan 1140°.

After reduction, the ratio is sine 60° divided by cosine 60°.

1140° − 6 × 180° = 1140° − 1080° = 60°.Tangent repeats every 180°.
tan 1140° = tan 60° = $square root of 3$ .The ratio at 60° is ( $square root of 3 over 2$ ) ÷ $1 over 2$ = $square root of 3$ .

Answer

tan 1140° = .

Check Removing three full laps, 1080°, also reaches 60°, so the coordinate-ratio result agrees.

Rung 3Rung 3: reduce a fractional radian angle

The question asks for sine after removing whole laps in radians. Find sin $37 π over 6$ .

1110° is , three full turns beyond 30°.

Three full laps are 6π = $36 π over 6$ .Sine's period is 2π, so three periods equal 6π.
$37 π over 6$ − $36 π over 6$ = $π over 6$ .The denominator 6 means the pieces match.
sin $37 π over 6$ = sin $π over 6$ = $1 over 2$ . $π over 6$ is the 30° special angle.

Answer

sin = .

Check The difference 6π is three full turns, so the terminal point agrees.

Rung 4Rung 4: move a negative angle forward

The question asks for sine after adding full laps to a negative input. Find sin(− $47 π over 6$ ).

−1410° is −, four full turns before 30°.

Four sine periods are 4 × 2π = 8π = $48 π over 6$ .Sine's period is 2π, and sixths give a common denominator.
− $47 π over 6$ + $48 π over 6$ = $π over 6$ .Adding four full laps preserves the stopping point and its sine output.
sin(− $47 π over 6$ ) = sin $π over 6$ = $1 over 2$ .The reduced angle is the 30° special angle, whose height is $1 over 2$ .

Answer

sin(−) = .

Check The two angles differ by 8π, exactly four full laps. Both finish at the unit-circle point (, ).

Rung 5Rung 5: same-angle reciprocal cancellation

The question asks for a product whose two angles must first be matched. Find 7 cos $π over 3$ sec $37 π over 3$ .

cos =

sec = sec = 2

7 × × 2 = 7

Reduce the secant angle before pairing its reciprocal with cosine.

$37 π over 3$ − 6(2π) = $37 π over 3$ − $36 π over 3$ = $π over 3$ .Secant's period is 2π, and six periods remove 12π.
7 cos $π over 3$ sec $π over 3$ = 7( $1 over 2$ )(2) = 7.Cosine and secant are reciprocal at this defined angle.
The denominator cos $π over 3$ = $1 over 2$ is nonzero.This confirms that reciprocal multiplication is allowed.

Answer

Check Direct values give 7 × × 2 = 7.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Sine has period π because sin 0 = sin π = 0.

A period must preserve every output. At , adding π changes sine from 1 to −1.

✓ Instead: Sine's period is 2π. One matching pair of values is insufficient.

✗ Not this: 60° and 240° are coterminal because their tangent values are equal.

The angles differ by half a turn and end on opposite sides. Equal function outputs do not require equal terminal points.

✓ Instead: They share a tangent value because tangent has period 180°. Coterminal angles differ by whole turns.

Tips and tricks

Know the pattern: ratios repeat after half a lap; coordinates and their reciprocals repeat after a full lap.
When reducing, remove a whole-number multiple of the requested function's period. Keep radians and degrees separate.

Trap. Removing a single π from a sine, cosine, secant or cosecant angle. Half a lap flips their sign: sin(t + π) = −sin t, so sin 210° is −, not sin 30° = . Only tangent and cotangent may drop π at a time; the others must drop 2π (360°).

Keep in mind

Drop only whole periods: removing 180° from cosine flips its sign, so cos 240° = −cos 60° = − $1 over 2$ , not $1 over 2$ .
To count laps, divide by 360 and keep only the whole-number part, never rounding up: 1070 ÷ 360 ≈ 2.97, so remove 2 laps, 720°, leaving 350°.
Periodic means the whole pattern repeats, not one lucky match: sin 30° and sin 150° are both $1 over 2$ , yet 120° is not a period of sine, since sin 90° = 1 but sin 210° = − $1 over 2$ .

Memory hookTangent is in a hurry: tan and cot repeat every half lap (180°, π); sin, cos, sec and csc need the full lap (360°, 2π).

Flash cards: say the answer out loud, then flip

What is a period?

The smallest positive step after which every output repeats.

What are coterminal angles?

Angles whose turns stop in the same place, on the same terminal side. They differ by whole laps (360° or 2π) and share all six values.

Period of sec t? Period of cot t?

sec t: 2π (360°).
cot t: π (180°).

sin 1470° = ?

: 1470° − 4 × 360° = 30°, and sin 30° = .

tan 585° = ?

1: 585° − 3 × 180° = 45°, and tan 45° = 1.

Can you write cos 200° = cos 20° by removing 180°?

No. 180° is half of cosine's period: cos 200° = −cos 20°.