Precalculus · F.TF.A.4 · F.TF.A.3

Periodicity and the periodicity identities

A paddle wheel can turn through \displaystyle \frac{29\pi}{3} radians, many times around. Does finding the sine of an angle that large need anything new, or has the circle already answered it?

You need

Trig functions on the unit circlePrecalculus Angles and their measureGrades 9 to 12

Unlocks

Periodic functionsPrecalculus

Where you are 8 in The unit circle

Trigonometry7

The unit circle8

Trig functions on the unit circle

Periodicity and the periodicity identities

Reference angles Secant, cosecant and cotangent on the unit circle Signs of the trig functions by quadrant ▾ Show all 8 (3 more)

Why this exists

A paddle wheel on a riverboat turns all day. If a paint mark on one paddle starts level with the axle and the wheel has turned through \displaystyle \frac{29\pi}{3} radians, how high is the mark? The angle is more than four full turns, and none of the special angles in a table is that large.

The circle has already answered the question. After every full turn the mark is back where it started, so the wheel’s position after \displaystyle \frac{29\pi}{3} radians is its position after whatever is left once the full turns are taken away. Since sine and cosine are the coordinates of the terminal point, they come back too. This lesson states that fact as six identities, one for each trigonometric function, and finds one surprise: tangent and cotangent repeat after only half a turn.

Definition

Two angles in standard position are coterminal when they differ by an integer number of full turns, as in Angles in standard position and coterminal angles. In radians a full turn is 2\pi, so t and t + 2\pi n are coterminal for every integer n (positive, negative or zero; Angles and their measure has radians and degrees side by side).

A function f repeats after a shift p > 0 when

f(t + p) = f(t)

for every t in its domain, with t + p in the domain exactly when t is. The periodicity identities say which shifts the six trigonometric functions repeat after:

\begin{aligned} \sin(t + 2\pi n) &= \sin t \\ \cos(t + 2\pi n) &= \cos t \\ \csc(t + 2\pi n) &= \csc t \\ \sec(t + 2\pi n) &= \sec t \\ \tan(t + \pi n) &= \tan t \\ \cot(t + \pi n) &= \cot t \end{aligned}

for every integer n and every t where the right side is defined. In degrees, 2\pi becomes 360^\circ and \pi becomes 180^\circ.

Intuition

Watch one point on the paddle wheel. Turning it through t and then one more full turn puts the point exactly where turning through t alone did. Nothing about the point remembers how many times it went around, so no value read from it can tell t from t + 2\pi.

Figure 1Turning through t + 2\pi is turning through t plus one full turn. The terminal side, and so the terminal point, is the same.

A half turn is different. It carries the point straight across the center, from (x, y) to (-x, -y). The sine and the cosine both change sign, so neither repeats after \pi. But the tangent is the ratio \displaystyle \frac{y}{x}, the slope of the line through the origin and the point, and the new point is on that same line. Two sign changes cancel in a ratio, so the tangent does repeat. The cotangent \displaystyle \frac{x}{y} is a ratio of the same two numbers and repeats with it.

Figure 2A half turn sends (x, y) to (-x, -y), on the same line through the origin. Both coordinates change sign, so their ratio does not.

Results

Full turns. Sine, cosine, cosecant and secant are unchanged by adding or subtracting any whole number of full turns, 2\pi n or 360^\circ n.

Half turns. Tangent and cotangent are unchanged by adding or subtracting any whole number of half turns, \pi n or 180^\circ n.

Half turns for the others. For every t,

\begin{aligned} \sin(t + \pi) &= -\sin t \\ \cos(t + \pi) &= -\cos t \end{aligned}

and so \csc(t + \pi) = -\csc t and \sec(t + \pi) = -\sec t wherever they are defined.

No smaller shift works. 2\pi is the smallest positive shift after which sine, cosine, cosecant and secant repeat, and \pi is the smallest for tangent and cotangent. The smallest such shift is the period, named and used for graphs in Periodic functions.

Reduction. Every real t can be written as t = t_0 + 2\pi n with 0 \le t_0 < 2\pi and n an integer, so the value of any of the six functions at t is its value at t_0.

Why it is true

Full turns. The angles t and t + 2\pi n are coterminal, so they have the same terminal side and the same terminal point (x, y). Sine is y, cosine is x, and the other four are built from x and y alone: \displaystyle \tan t = \frac{y}{x}, \displaystyle \cot t = \frac{x}{y}, \displaystyle \sec t = \frac{1}{x}, \displaystyle \csc t = \frac{1}{y}. Equal inputs give equal outputs, and a value undefined at one angle (because x = 0 or y = 0) is undefined at the other.

Half turns. Adding \pi turns the terminal side to the opposite ray. The opposite ray is the first ray reflected through the origin, so the terminal point (x, y) becomes (-x, -y). Then

\tan(t + \pi) = \frac{-y}{-x} = \frac{y}{x} = \tan t

and in the same way \displaystyle \cot(t + \pi) = \frac{-x}{-y} = \cot t. Applying this n times (or undoing it, for negative n) gives \tan(t + \pi n) = \tan t. The same point (-x, -y) gives \sin(t + \pi) = -y = -\sin t and \cos(t + \pi) = -x = -\cos t.

No smaller shift for sine. Suppose \sin(t + p) = \sin t for every t, with p > 0. At \displaystyle t = \frac{\pi}{2} this says \displaystyle \sin\left(\frac{\pi}{2} + p\right) = 1. The only point of the unit circle with y = 1 is (0, 1), the terminal point of \displaystyle \frac{\pi}{2}, so \displaystyle \frac{\pi}{2} + p is coterminal with \displaystyle \frac{\pi}{2} and p is a whole number of full turns. The smallest positive one is 2\pi. Cosine, secant and cosecant go the same way (Exercise 23 does cosine).

No smaller shift for tangent. If \tan(t + p) = \tan t for every t, then at t = 0, \tan p = 0. That needs y = 0, so the terminal point of p is (1, 0) or (-1, 0) and p is a multiple of \pi. The smallest positive one is \pi. Cotangent has the same argument at \displaystyle t = \frac{\pi}{2}.

Reduction. Let n be the largest integer with 2\pi n \le t. Then t_0 = t - 2\pi n satisfies 0 \le t_0, and t_0 < 2\pi because otherwise n + 1 would also qualify.

Worked examples

Example 1

Four turns and some

Find \displaystyle \sin\frac{29\pi}{3}, \displaystyle \cos\frac{29\pi}{3} and \displaystyle \tan\frac{29\pi}{3} exactly.

Solution

Count the full turns. One turn is \displaystyle 2\pi = \frac{6\pi}{3}, and \displaystyle \frac{29}{6} = 4\frac{5}{6}, so the angle is four full turns and five sixths of another. Taking away the four turns,

\frac{29\pi}{3} - 8\pi = \frac{29\pi}{3} - \frac{24\pi}{3} = \frac{5\pi}{3}

By the full-turn identities the values at \displaystyle \frac{29\pi}{3} are those at \displaystyle \frac{5\pi}{3}, whose terminal point is \displaystyle \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) in Quadrant IV. So

\begin{aligned} \sin\frac{29\pi}{3} &= -\frac{\sqrt{3}}{2} \\ \cos\frac{29\pi}{3} &= \frac{1}{2} \\ \tan\frac{29\pi}{3} &= -\sqrt{3} \end{aligned}

For the tangent alone, half turns are enough: \displaystyle \frac{29\pi}{3} - 9\pi = \frac{2\pi}{3}, and \displaystyle \tan\frac{2\pi}{3} = -\sqrt{3} as well.

Example 2

A negative angle and all six values

Find all six trigonometric values of \displaystyle -\frac{41\pi}{4}.

Solution

The angle turns clockwise \displaystyle \frac{41}{8} = 5\frac{1}{8} turns. To land between 0 and 2\pi, add full turns until the result is no longer negative. Five turns (10\pi) give \displaystyle -\frac{\pi}{4}, still negative, so add six:

-\frac{41\pi}{4} + 12\pi = -\frac{41\pi}{4} + \frac{48\pi}{4} = \frac{7\pi}{4}

Here n = -6 in t = t_0 + 2\pi n, which is why the identities are stated for every integer. The terminal point of \displaystyle \frac{7\pi}{4} is \displaystyle \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right), so

\begin{aligned} \sin\left(-\frac{41\pi}{4}\right) &= -\frac{\sqrt{2}}{2} \\ \cos\left(-\frac{41\pi}{4}\right) &= \frac{\sqrt{2}}{2} \\ \tan\left(-\frac{41\pi}{4}\right) &= -1 \\ \cot\left(-\frac{41\pi}{4}\right) &= -1 \\ \sec\left(-\frac{41\pi}{4}\right) &= \sqrt{2} \\ \csc\left(-\frac{41\pi}{4}\right) &= -\sqrt{2} \end{aligned}

The secant is \displaystyle \frac{1}{\sqrt{2}/2} = \frac{2}{\sqrt{2}} = \sqrt{2}, and the cosecant is its negative.

Example 3

Half turns in degrees

Find \tan 2{,}010^\circ and \cot 2{,}010^\circ exactly, first with half turns and then with full turns.

Solution

Half turns are 180^\circ, and 2{,}010 = 11 \cdot 180 + 30, so taking away eleven half turns leaves

2{,}010^\circ - 1{,}980^\circ = 30^\circ

Then \displaystyle \tan 2{,}010^\circ = \tan 30^\circ = \frac{\sqrt{3}}{3} and \displaystyle \cot 2{,}010^\circ = \cot 30^\circ = \sqrt{3}.

With full turns, 2{,}010 = 5 \cdot 360 + 210, so the angle is coterminal with 210^\circ, in Quadrant III, with terminal point \displaystyle \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right). Its tangent is

\frac{-1/2}{-\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

the same answer. The two routes end at 30^\circ and 210^\circ, opposite points on one line through the origin, which is exactly the picture of the half-turn identity. Sine would not allow the shortcut: \displaystyle \sin 210^\circ = -\frac{1}{2}, not \sin 30^\circ.

Example 4

Bojana’s paddle wheel

Bojana watches a riverboat’s paddle wheel of radius 3 meters from the side. Put the axle at the origin, with x to the right and y up. A paint mark on a paddle tip starts at (3, 0) and the wheel turns counterclockwise, as she sees it, through 40 radians. How many full turns has the mark made, in which quadrant does it stop, and where is it, to the nearest centimeter?

Solution

Full turns are 2\pi \approx 6.2832 radians, and \displaystyle \frac{40}{2\pi} \approx 6.37, so the mark has made 6 full turns. What is left is

40 - 12\pi \approx 2.3009 \text{ radians}

Since \displaystyle \frac{\pi}{2} \approx 1.5708 < 2.3009 < \pi \approx 3.1416, the mark stops in Quadrant II, up and to the left of the axle.

As in Trig functions on the unit circle, a point at angle t on a circle of radius 3 is (3\cos t, 3\sin t). With a calculator in radian mode,

\begin{aligned} x &= 3\cos 40 \approx 3(-0.66694) \approx -2.00 \\ y &= 3\sin 40 \approx 3(0.74511) \approx 2.24 \end{aligned}

so the mark is at about (-2.00, 2.24), about 2.24 meters above the axle. The calculator never counts turns, and it does not need to: by the periodicity identities, \cos 40 = \cos(40 - 12\pi) and \sin 40 = \sin(40 - 12\pi), so the six turns change nothing.

Example 5

Simplifying with shifts

Simplify the expression below, and say for which t the result holds.

\begin{aligned} &\sin(t + 8\pi) - \sin(t - 6\pi) \\ &\quad + \tan(t + 5\pi)\cot(t - 3\pi) \end{aligned}

Solution

8\pi and -6\pi are integer multiples of 2\pi (n = 4 and n = -3), so both sines equal \sin t and cancel. 5\pi and -3\pi are integer multiples of \pi, so \tan(t + 5\pi) = \tan t and \cot(t - 3\pi) = \cot t. The expression is

\begin{aligned} &\sin t - \sin t + \tan t \cot t \\ &\quad = \tan t \cdot \frac{1}{\tan t} = 1 \end{aligned}

This holds where every piece is defined. \tan t needs \cos t \ne 0 and \cot t needs \sin t \ne 0, so t must not be a multiple of \displaystyle \frac{\pi}{2}. The shifted functions have the same gaps, since shifting by a multiple of \pi moves a gap onto a gap. So the expression equals 1 for every t except \displaystyle t = \frac{\pi}{2}k with k an integer, where it has no value.

Example 6

Every angle with a given tangent

Find every t with 0 \le t < 6\pi and \displaystyle \tan t = \sqrt{3}.

Solution

Start with one half turn, 0 \le t < \pi. The tangent is the slope of the terminal side, and a slope of \displaystyle \sqrt{3} rising into Quadrant I belongs to \displaystyle t = \frac{\pi}{3}, terminal point \displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right). In Quadrant II the terminal side falls to the right and its slope is negative, so \displaystyle \frac{\pi}{3} is the only answer in [0, \pi).

Every other answer differs from \displaystyle \frac{\pi}{3} by a whole number of half turns, because the half-turn identity says \tan repeats after \pi and the interval [0, \pi) already holds one copy of every slope. So the answers are \displaystyle t = \frac{\pi}{3} + \pi n:

\begin{aligned} &\frac{\pi}{3},\ \frac{4\pi}{3},\ \frac{7\pi}{3}, \\ &\frac{10\pi}{3},\ \frac{13\pi}{3},\ \frac{16\pi}{3} \end{aligned}

The next one, \displaystyle \frac{19\pi}{3}, is past \displaystyle 6\pi = \frac{18\pi}{3}. There are six, two for each full turn, one in Quadrant I and one in Quadrant III each time.

Example 7

A product that repeats sooner

Let f(t) = \sin t \cos t. Show that f(t + \pi) = f(t) for every t, and that no positive shift smaller than \pi works.

Solution

By the half-turn results, \sin(t + \pi) = -\sin t and \cos(t + \pi) = -\cos t, so

f(t + \pi) = (-\sin t)(-\cos t) = \sin t \cos t = f(t)

Neither factor repeats after \pi, but their product does, because the two sign changes cancel.

Now suppose 0 < p < \pi and f(t + p) = f(t) for every t. At t = 0, f(p) = f(0) = 0, so \sin p = 0 or \cos p = 0. Between 0 and \pi only \displaystyle p = \frac{\pi}{2} does that. Test it at \displaystyle t = \frac{\pi}{4}:

\begin{aligned} f\left(\frac{\pi}{4}\right) &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} \\ f\left(\frac{3\pi}{4}\right) &= \frac{\sqrt{2}}{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{1}{2} \end{aligned}

These differ, so \displaystyle \frac{\pi}{2} fails and \pi is the smallest positive shift that works.

Boundaries

n must be an integer. \displaystyle \sin\left(t + \frac{\pi}{2}\right) is \cos t, not \sin t; only whole turns (or whole half turns, for tangent and cotangent) leave the values alone.

Half turns work only for tangent and cotangent. For sine, cosine, secant and cosecant a half turn flips the sign, so \pi is never a repeat shift for them.

Undefined stays undefined. The identities hold where the functions are defined. A shift by a multiple of \pi carries an angle where tangent is undefined, such as \displaystyle \frac{\pi}{2}, to another such angle, so the gaps repeat as well; Secant, cosecant and cotangent on the unit circle lists them.

Degrees and radians must match. Remove 360^\circ from an angle in degrees and 2\pi from an angle in radians. Subtracting 2\pi from 40^\circ is meaningless.

Where this goes next

Reducing an angle to one turn is the first move in finding any value exactly. The second is the reference angle, and the sign changes under reflection are Unit circle symmetry and even-odd identities. With t read as a distance along the circle, as in Sine, cosine and tangent as functions of a real number, the periodicity identities say sine and cosine are functions whose outputs repeat forever.

Plotted against t, that repetition is the wave of the sine and cosine graphs, and the smallest repeat shift becomes the period in Periodic functions. Every model of something that comes back around (tides, sound, a turning wheel) starts from these identities.

Exercises

Removing whole turns

Find each value exactly.

\displaystyle \cos\frac{49\pi}{6}
\displaystyle \sin\left(-\frac{19\pi}{3}\right)
\displaystyle \tan\frac{37\pi}{4}
\displaystyle \sec\frac{23\pi}{3}
\displaystyle \csc\left(-\frac{35\pi}{6}\right)
\displaystyle \cot\frac{43\pi}{6}
\sin 1{,}830^\circ
\cos(-1{,}560^\circ)
\tan 2{,}295^\circ
\cos 15\pi and \tan 15\pi

All six, and the paddle wheel

Find all six trigonometric values of \displaystyle \frac{53\pi}{6}.
Which of the six functions are undefined at \displaystyle \frac{31\pi}{2}? Give the values of the others.
Milica’s paddle wheel has radius 2.5 meters, axle at the origin, y up. A mark starts at (2.5, 0) and the wheel turns 25 radians counterclockwise. How many full turns does the mark make, in which quadrant does it stop, and where is it, to the nearest centimeter?
Ananya’s paddle wheel has radius 3 meters and turns counterclockwise at 1.2 radians per second. A mark starts at (3, 0). After 60 seconds, how many full turns has it made, and how high above or below the axle is it, to the nearest centimeter?
On a wheel of radius 3 meters, a paddle tip is at (1.8, -2.4). Where is the tip directly opposite it? Find the cosine, sine and tangent of the angle of each tip.

Simplifying and solving

Simplify \cos(\theta - 6\pi) + \cos(\theta + 4\pi).
Simplify \tan(x + 7\pi) - \tan(x - 2\pi), and say where it holds.
Simplify \sin(t + 3\pi).
Simplify \sec(\theta + 12\pi)\cos(\theta - 2\pi), and say where it holds.
Find every t with 0 \le t < 4\pi and \cot t = -1.

Reasoning

Explain why \sin(t + 2\pi n) = \sin t holds for negative integers n as well as positive ones.
Show that \sec(t + \pi) = -\sec t wherever \sec t is defined, and give one t showing that \pi is not a repeat shift for secant.
Show that no positive shift smaller than 2\pi leaves \cos t unchanged for every t.
Bojana says tangent repeats every 2\pi. Milica says it repeats every \pi. Who is right?
Does g(t) = \sin t + \tan t repeat after \pi? After 2\pi? Explain.

Answers

\displaystyle \frac{\sqrt{3}}{2}. \displaystyle \frac{49\pi}{6} - 8\pi = \frac{\pi}{6}, and \displaystyle \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}.
\displaystyle -\frac{\sqrt{3}}{2}. \displaystyle -\frac{19\pi}{3} + 8\pi = \frac{5\pi}{3}, in Quadrant IV.
1. \displaystyle \frac{37\pi}{4} - 9\pi = \frac{\pi}{4}, nine half turns.
2. \displaystyle \frac{23\pi}{3} - 6\pi = \frac{5\pi}{3}, where \displaystyle \cos = \frac{1}{2}.
2. \displaystyle -\frac{35\pi}{6} + 6\pi = \frac{\pi}{6}, where \displaystyle \sin = \frac{1}{2}.
\displaystyle \sqrt{3}. \displaystyle \frac{43\pi}{6} - 7\pi = \frac{\pi}{6}, seven half turns, and \displaystyle \cot\frac{\pi}{6} = \sqrt{3}.
\displaystyle \frac{1}{2}. 1{,}830^\circ - 1{,}800^\circ = 30^\circ.
\displaystyle -\frac{1}{2}. -1{,}560^\circ + 1{,}800^\circ = 240^\circ, in Quadrant III.
-1. 2{,}295^\circ - 2{,}160^\circ = 135^\circ, twelve half turns.
\cos 15\pi = -1 and \tan 15\pi = 0. 15\pi - 14\pi = \pi, terminal point (-1, 0).
\displaystyle \sin = \frac{1}{2}, \displaystyle \cos = -\frac{\sqrt{3}}{2}, \displaystyle \tan = -\frac{\sqrt{3}}{3}, \displaystyle \cot = -\sqrt{3}, \displaystyle \sec = -\frac{2\sqrt{3}}{3}, \csc = 2. \displaystyle \frac{53\pi}{6} - 8\pi = \frac{5\pi}{6}, terminal point \displaystyle \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).
Tangent and secant are undefined. \displaystyle \frac{31\pi}{2} - 14\pi = \frac{3\pi}{2}, terminal point (0, -1): \sin = -1, \cos = 0, \cot = 0, \csc = -1.
3 full turns; Quadrant IV; about (2.48, -0.33). \displaystyle \frac{25}{2\pi} \approx 3.98, and 25 - 6\pi \approx 6.1504 lies between \displaystyle \frac{3\pi}{2} \approx 4.7124 and 2\pi \approx 6.2832. 2.5\cos 25 \approx 2.4780 and 2.5\sin 25 \approx -0.3309.
11 full turns; about 0.76 meters above the axle. The angle is 1.2 \cdot 60 = 72 radians, \displaystyle \frac{72}{2\pi} \approx 11.46, and 72 - 22\pi \approx 2.8850 is in Quadrant II. 3\sin 72 \approx 0.7615.
(-1.8, 2.4). First tip: \cos = 0.6, \sin = -0.8, \displaystyle \tan = -\frac{4}{3}. Opposite tip: \cos = -0.6, \sin = 0.8, \displaystyle \tan = -\frac{4}{3}. Divide by the radius 3; the opposite tip is a half turn away, so cosine and sine flip sign and tangent does not.
2\cos\theta. Both shifts are whole turns.
0, wherever \tan x is defined: \displaystyle x \ne \frac{\pi}{2} + \pi k for integers k. Both shifts are whole half turns.
-\sin t. 3\pi = 2\pi + \pi: the full turn changes nothing and the half turn flips the sign.
1, wherever \cos\theta \ne 0, that is \displaystyle \theta \ne \frac{\pi}{2} + \pi k. The product is \sec\theta\cos\theta = 1.
\displaystyle \frac{3\pi}{4}, \displaystyle \frac{7\pi}{4}, \displaystyle \frac{11\pi}{4}, \displaystyle \frac{15\pi}{4}. In [0, \pi) only \displaystyle \frac{3\pi}{4} has \cot = -1; then add half turns.
For negative n, t + 2\pi n is t turned |n| full turns clockwise. A clockwise full turn also returns to the same terminal side, so the angles are coterminal and share a terminal point.
t + \pi has terminal point (-x, -y), so where x \ne 0, \displaystyle \sec(t + \pi) = \frac{1}{-x} = -\frac{1}{x} = -\sec t; where x = 0 both sides are undefined. At t = 0, \sec 0 = 1 but \sec\pi = -1.
If \cos(t + p) = \cos t for all t, then at t = 0, \cos p = 1. The only unit-circle point with x = 1 is (1, 0), so p is coterminal with 0, a multiple of 2\pi; the smallest positive one is 2\pi.
Both statements are true: 2\pi = 2 \cdot \pi is a whole number of half turns. Milica’s says more, since \pi is the smallest positive shift that works.
Not after \pi: g(t + \pi) = -\sin t + \tan t, which differs from g(t) whenever \sin t \ne 0; for example \displaystyle g\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + 1 but \displaystyle g\left(\frac{5\pi}{4}\right) = 1 - \frac{\sqrt{2}}{2}. After 2\pi, yes: both terms repeat after a full turn.