Precalculus · F.TF.A.2

Sine, cosine and tangent as functions of a real number

A calculator gives \sin 10 \approx -0.544 even though nobody said degrees, or said angle at all. What is the 10 the input of?

You need

Trig functions on the unit circlePrecalculus

Where you are 8 in The unit circle

Trigonometry7

The unit circle8

▴ 3 above Secant, cosecant and cotangent on the unit circle Signs of the trig functions by quadrant

Sine, cosine and tangent as functions of a real number

Unit circle symmetry and even-odd identities Exact values on the unit circle

Why this exists

In Trig functions on the unit circle, the cosine and sine belonged to an angle: turn the terminal side, find where it meets the unit circle, read the coordinates. Many of the places sine and cosine appear have no angle in sight. A runner’s position depends on the distance run, a tide’s height on the hour, a sound’s pressure on the time. To graph y = \sin t or to write h(t) = 3 + 2\cos t in Function notation, the input has to be a plain real number, like the x in f(x) = x^2.

Radian measure is what makes the step possible. On a circle of radius 1, an arc is exactly as long as the angle it spans, measured in radians. So a real number t can be read as a distance traveled around the circle, with no angle mentioned, and the cosine and sine become functions whose input is any real number and whose output is a real number.

Definition

Picture the number line standing upright, touching the unit circle at (1, 0), with its 0 at that point and its positive numbers going up. Now wrap the line around the circle: the positive half winds counterclockwise and the negative half winds clockwise, with no stretching. Each real number t lands on one point of the circle, called P(t).

Figure 1Wrapping the number line around the unit circle. Each number t lands at one point P(t); the numbers 1, 2 and 3 go counterclockwise, the numbers -1 and -2 clockwise.

Stated without the picture: for a real number t, the point P(t) is reached by starting at (1, 0) and traveling a distance |t| along the unit circle, counterclockwise if t > 0 and clockwise if t < 0; P(0) = (1, 0). If P(t) = (x, y), then

\begin{aligned} \cos t &= x \\ \sin t &= y \\ \tan t &= \frac{y}{x} \quad (x \ne 0) \end{aligned}

These are the cosine function, the sine function and the tangent function. Their input t is a real number; \sin t is shorthand for \sin(t), just as f(t) names the output of a function f.

The domain of a function is the set of inputs it accepts, and its range is the set of outputs it actually produces.

Intuition

Take a circular running track, seen from above, with its center at the origin and a radius of 1 unit; on the real track one unit is 50 meters. Every runner starts at the east point (1, 0). A runner who has gone a distance t counterclockwise is at P(t): \cos t is how far east of the center she is and \sin t how far north, both in units. Running clockwise is a negative distance.

One lap is the circumference, 2\pi \approx 6.283 units. That fixes four landmarks every runner passes, and they sort any t into a quadrant.

Distance run	Fraction of a lap	Point
\displaystyle \frac{\pi}{2} \approx 1.571	one quarter	(0, 1), the north point
\pi \approx 3.142	one half	(-1, 0), the west point
\displaystyle \frac{3\pi}{2} \approx 4.712	three quarters	(0, -1), the south point
2\pi \approx 6.283	one full lap	(1, 0), back at the start

Nothing stops a runner from going on: t = 10 or t = 100 is simply a longer run, and a negative t is a run the other way. That is why every real number is an input. Yet however far the runner goes she stays on the track, so her east and north positions never leave the interval from -1 to 1. The tangent is different: it is the slope from the center to the runner, and near the north and south points that slope is steep without limit.

Results

Same values as the angle. P(t) is the terminal point of the angle of t radians in standard position, so \cos t, \sin t and \tan t agree with the values of that angle.

Sine and cosine. The domain of each is all real numbers. The range of each is

-1 \le \sin t \le 1 \qquad -1 \le \cos t \le 1

and every number in [-1, 1] is reached.

Tangent. \tan t = \dfrac{\sin t}{\cos t}. Its domain is every real number except

t = \frac{\pi}{2} + k\pi, \quad k \text{ an integer}

and its range is all real numbers.

Laps. P(t + 2\pi) = P(t) for every t, so the values repeat after 2\pi; this is periodicity.

Why it is true

The wrapped point is the terminal point. An arc of length s on a circle of radius r spans \displaystyle \frac{s}{r} radians (Arc length and sector area in radians). With r = 1, traveling a distance t counterclockwise sweeps an angle of exactly t radians, so the trip ends at the terminal point of that angle. Clockwise travel sweeps a negative angle in the same way.

Every real number is an input. Any distance can be traveled, in either direction, so P(t) exists for every real t, and so do \cos t and \sin t.

The range of sine and cosine is [-1, 1]. P(t) lies on x^2 + y^2 = 1, so y^2 \le 1 and x^2 \le 1, which keeps both coordinates between -1 and 1. Every such value is reached: as t runs from 0 to 2\pi, the point travels once around the whole circle and passes every point on it. For any y in [-1, 1], the point \displaystyle \left(\sqrt{1 - y^2}, y\right) is on the circle, so it is P(t) for some t, and that t has \sin t = y. The same argument with \displaystyle \left(x, \sqrt{1 - x^2}\right) works for cosine.

Where tangent fails. \tan t is undefined exactly when \cos t = 0, that is, when P(t) is (0, 1) or (0, -1). The north point is reached at \displaystyle \frac{\pi}{2} plus any integer number of laps, and the south point half a lap later, so between them they are reached once every half lap, at \displaystyle \frac{\pi}{2} + k\pi for every integer k.

The range of tangent is every real number. Given any number m, the point

\left(\frac{1}{\sqrt{1 + m^2}}, \frac{m}{\sqrt{1 + m^2}}\right)

has squares adding to \displaystyle \frac{1 + m^2}{1 + m^2} = 1, so it is on the circle, and its y over x is m. It is P(t) for some t, and \tan t = m.

Worked examples

Example 1

The point of t = 2.5

Find the quadrant of P(2.5), then \cos 2.5, \sin 2.5 and \tan 2.5 to five decimal places.

Solution

Compare 2.5 with the landmarks: \displaystyle \frac{\pi}{2} \approx 1.571 < 2.5 < \pi \approx 3.142. The trip passes the north point but stops before the west point, so P(2.5) is in Quadrant II. It has gone 2.5 - 1.571 \approx 0.929 past the north point and stops 3.142 - 2.5 \approx 0.642 short of the west point, so it sits closer to (-1, 0): expect a cosine larger in size than the sine.

With a calculator in radian mode,

\cos 2.5 \approx -0.80114 \qquad \sin 2.5 \approx 0.59847

The signs match Quadrant II, and 0.80114 is the larger, as predicted. Then

\tan 2.5 = \frac{\sin 2.5}{\cos 2.5} \approx -0.74702

As a check, the squares add to one:

\begin{aligned} &(-0.80114)^2 + 0.59847^2 \\ &\quad \approx 0.64183 + 0.35817 = 1.00000 \end{aligned}

Figure 2Example 1. The arc from (1, 0) has length 2.5 and ends in Quadrant II.

Example 2

Aigerim runs 500 meters

Aigerim starts at the east point of the track and runs 500 meters counterclockwise. Where is she, in meters from the center, to the nearest centimeter?

Solution

One unit is 50 meters, so she runs \displaystyle t = \frac{500}{50} = 10 units. A lap is 2\pi \approx 6.283 units, and 10 is more than one lap but less than two. After one full lap she is back at (1, 0) with

10 - 2\pi \approx 3.717

units still to run. Since \displaystyle \pi \approx 3.142 < 3.717 < \frac{3\pi}{2} \approx 4.712, she ends past the west point and before the south point, in Quadrant III. In radian mode,

\cos 10 \approx -0.83907 \qquad \sin 10 \approx -0.54402

Both negative, as Quadrant III requires. These are her coordinates in units; multiplying by 50 gives meters:

\begin{aligned} x &= 50\cos 10 \approx -41.95 \\ y &= 50\sin 10 \approx -27.20 \end{aligned}

She is about 41.95 meters west of the center and 27.20 meters south of it.

Figure 3Example 2. After one full lap of length 2\pi, the remaining 3.72 of the trip ends in Quadrant III.

Example 3

Yerlan runs the other way

Yerlan starts at the east point and runs 200 meters clockwise. Find his quadrant and his position in meters.

Solution

Clockwise is the negative direction, so \displaystyle t = -\frac{200}{50} = -4. Going clockwise the landmarks come in the order south (\displaystyle -\frac{\pi}{2} \approx -1.571), west (-\pi \approx -3.142), north (\displaystyle -\frac{3\pi}{2} \approx -4.712). Since -4.712 < -4 < -3.142, he has passed the west point and is heading up toward the north point: Quadrant II.

\cos(-4) \approx -0.65364 \qquad \sin(-4) \approx 0.75680

so, multiplying by 50, his position is about (-32.68, 37.84) meters.

A second way to see it: 200 meters clockwise leaves him at the same spot as 2\pi - 4 \approx 2.283 units counterclockwise, since the two trips together make one lap, and 2.283 lies between 1.571 and 3.142: Quadrant II again.

Example 4

Where the tangent function has no value

Find every t in the interval [0, 10] where \tan t is undefined, and state the domain of the tangent function.

Solution

\displaystyle \tan t = \frac{\sin t}{\cos t} fails exactly when \cos t = 0, which happens when P(t) is the north or the south point. Those are reached at \displaystyle t = \frac{\pi}{2} + k\pi for integers k. Listing from k = 0:

\frac{\pi}{2} \approx 1.571, \quad \frac{3\pi}{2} \approx 4.712, \quad \frac{5\pi}{2} \approx 7.854

The next one, \displaystyle \frac{7\pi}{2} \approx 10.996, is past 10, and k = -1 gives a negative number. So there are three values in [0, 10]. The domain of tangent is every real number except \displaystyle \frac{\pi}{2} + k\pi, k an integer.

Example 5

The range of a new function

Find the domain and range of g(t) = 3 - 2\cos t.

Solution

Cosine accepts every real number, so g does too: the domain is all real numbers. For the range, start from what cosine can produce and follow each operation:

\begin{aligned} -1 &\le \cos t \le 1 \\ 2 &\ge -2\cos t \ge -2 \\ 5 &\ge 3 - 2\cos t \ge 1 \end{aligned}

Multiplying by -2 reverses both inequalities, and adding 3 keeps them. So 1 \le g(t) \le 5. Every value in between is produced, because \cos t takes every value from -1 to 1. The ends are reached too: g(0) = 3 - 2(1) = 1 and g(\pi) = 3 - 2(-1) = 5. The range is [1, 5].

Example 6

A tangent of 50

Show that some t has \tan t = 50, find P(t) for one such t in Quadrant I, and estimate t.

Solution

Following the argument for the range of tangent, take

P = \left(\frac{1}{\sqrt{2501}}, \frac{50}{\sqrt{2501}}\right) \approx (0.02000, 0.99980)

since 1 + 50^2 = 2501. Its coordinates square to \displaystyle \frac{1}{2501} + \frac{2500}{2501} = 1, so it is on the unit circle, and y over x is 50. The point is a hair to the right of the north point, so t is just under \displaystyle \frac{\pi}{2} \approx 1.57080. This t is an acute angle, so the inverse tangent key of Inverse trigonometric ratios finds it; in radian mode it gives t \approx 1.55080, and indeed \tan 1.55080 \approx 50.0. The key returns only this one t; the others with \tan t = 50 come from the circle, by adding half laps.

The trip from 1.55080 to \displaystyle \frac{\pi}{2} is only 0.02 units, yet over it the tangent climbs from 50 with no upper limit, because x shrinks to 0 while y stays near 1. That is what the graph in Graphing tangent and the reciprocal functions shows as a vertical asymptote.

Example 7

Sibusiso, position as a function of time

Sibusiso runs counterclockwise from the east point at a steady 5 meters per second. Write his north position, in meters, as a function of the time u in seconds, and find where he is after 75 seconds.

Solution

After u seconds he has run 5u meters, which is \displaystyle \frac{5u}{50} = 0.1u units, so he is at P(0.1u). Scaling the unit circle up to meters,

N(u) = 50\sin(0.1u)

and his east position is E(u) = 50\cos(0.1u). Here sine takes in a number of units run, which came from a number of seconds; no angle was ever named.

At u = 75 the input is t = 7.5. One lap is 6.283, and 7.5 - 2\pi \approx 1.217 is less than 1.571, so he is in Quadrant I of his second lap:

\begin{aligned} E(75) &= 50\cos 7.5 \approx 17.33 \\ N(75) &= 50\sin 7.5 \approx 46.90 \end{aligned}

He is about 17.33 meters east and 46.90 meters north of the center.

Boundaries

The input is a radian count. \sin 10 means the sine of 10 units of distance on the unit circle, which is 10 radians. The degree function is a different function: \sin 10^\circ \approx 0.17365, while \sin 10 \approx -0.54402. A calculator must be in radian mode for \sin t.

The circle has radius 1. On a track of radius 50 meters, a distance d meters becomes \displaystyle t = \frac{d}{50}, and the cosine and sine give positions in units, to be multiplied by 50.

Tangent has gaps. \tan t does not exist at \displaystyle t = \frac{\pi}{2} + k\pi. The full ranges stated above hold on the full domains; on a smaller interval of inputs the outputs may cover less, for example \sin t for 0 \le t \le 1 only reaches 0 to \sin 1 \approx 0.841.

A point does not fix t. Each point of the circle is P(t) for infinitely many t, one for every whole number of laps added or removed, so knowing \cos t and \sin t never pins down t alone.

Where this goes next

Plotting \sin t against t turns the runner’s north position into the wave of Graphing sine and cosine, and plotting \tan t gives the branches and vertical lines of Graphing tangent and the reciprocal functions. The repeat after each lap is periodicity; reflecting P(t) to P(-t) is Unit circle symmetry and even-odd identities; and cutting a long trip down to a short one is reference angles. Exact values at the landmarks are in Exact values on the unit circle.

Exercises

Where t lands

For each value, name the quadrant of P(t) and give \cos t and \sin t to three decimal places.

t = 1.2
t = 5
t = 3
t = -1.5
t = 20
t = -8
t = 100

Tangent and domain

Name the quadrant of P(5.5) and find \tan 5.5 to three decimal places.
Which of \displaystyle t = \frac{3\pi}{2}, t = 2\pi, \displaystyle t = -\frac{\pi}{2} and t = 4 are outside the domain of the tangent function?
Find every t in [-5, 5] where \tan t is undefined.
Find every t in [0, 12] where \sin t = 0.
State the domain and range of h(t) = 4\sin t.

Ranges

Find the range of g(t) = 5 + 3\cos t.
Find the range of k(t) = 1 - \sin t.
Find the range of q(t) = \sin^2 t.
Which of these equations have a solution: \sin t = -0.9, \cos t = 1.01, \tan t = -300?
State the domain and range of f(t) = 2\tan t + 1.

On the track

The track has radius 50 meters, every runner starts at the east point, and positions are in meters from the center, east and north positive. Round positions to the nearest centimeter.

Sibusiso runs 300 meters counterclockwise. Find his quadrant and position.
Aigerim runs 900 meters clockwise. Find her quadrant and position.
Yerlan stops at (-30, 40). What is the shortest counterclockwise distance he could have run, to the nearest centimeter, and what are all the counterclockwise distances that end there?
Sibusiso runs counterclockwise at 4 meters per second. Find his quadrant and position after 2 minutes.

Reasoning

Explain why sine and cosine accept every real number as input but tangent does not.
Explain why \sin t is never more than 1 while \tan t = 1000 has a solution.
Aigerim says \sin 10 and \sin 10^\circ are the same number. Explain what each one means and why they differ.
Explain why, as t runs over [0, \pi], \cos t takes each value in [-1, 1] exactly once.

Answers

Quadrant I; \cos 1.2 \approx 0.362, \sin 1.2 \approx 0.932. 0 < 1.2 < 1.571.
Quadrant IV; \cos 5 \approx 0.284, \sin 5 \approx -0.959. 4.712 < 5 < 6.283.
Quadrant II; \cos 3 \approx -0.990, \sin 3 \approx 0.141. 1.571 < 3 < 3.142, just short of the west point.
Quadrant IV; \cos(-1.5) \approx 0.071, \sin(-1.5) \approx -0.997. Clockwise, 1.5 is short of the south point at -1.571.
Quadrant I; \cos 20 \approx 0.408, \sin 20 \approx 0.913. Three laps are 6\pi \approx 18.850, and 20 - 6\pi \approx 1.150.
Quadrant III; \cos(-8) \approx -0.146, \sin(-8) \approx -0.989. Adding two laps, -8 + 4\pi \approx 4.566, between 3.142 and 4.712.
Quadrant IV; \cos 100 \approx 0.862, \sin 100 \approx -0.506. Fifteen laps are 30\pi \approx 94.248, and 100 - 30\pi \approx 5.752.
Quadrant IV, since 4.712 < 5.5 < 6.283; \tan 5.5 \approx -0.996, from \displaystyle \frac{-0.70554}{0.70867}.
\displaystyle \frac{3\pi}{2} and \displaystyle -\frac{\pi}{2}. Both are \displaystyle \frac{\pi}{2} + k\pi (with k = 1 and k = -1), where P(t) is (0, -1). At 2\pi the point is (1, 0), and \cos 4 \approx -0.654 \ne 0.
\displaystyle t = -\frac{3\pi}{2}, \displaystyle -\frac{\pi}{2}, \displaystyle \frac{\pi}{2}, \displaystyle \frac{3\pi}{2}, about \pm 1.571 and \pm 4.712. The next ones, \displaystyle \pm\frac{5\pi}{2} \approx \pm 7.854, are outside.
t = 0, \pi, 2\pi, 3\pi. \sin t = 0 at the east and west points, every half lap; 4\pi \approx 12.566 is past 12.
Domain all real numbers; range [-4, 4]. Multiplying -1 \le \sin t \le 1 by 4.
[2, 8]. -3 \le 3\cos t \le 3, then add 5.
[0, 2]. -1 \le -\sin t \le 1, then add 1; k = 0 at \displaystyle t = \frac{\pi}{2}, k = 2 at \displaystyle t = \frac{3\pi}{2}.
[0, 1]. \sin t covers [-1, 1], and squaring sends that onto [0, 1].
\sin t = -0.9 and \tan t = -300 have solutions; \cos t = 1.01 does not, since 1.01 > 1. Tangent’s range is all real numbers.
Domain: every real number except \displaystyle \frac{\pi}{2} + k\pi; range: all real numbers. 2\tan t takes every real value, and so does adding 1.
t = 6, Quadrant IV; about (48.01, -13.97). 50\cos 6 \approx 48.009, 50\sin 6 \approx -13.971.
t = -18, Quadrant I, since -18 + 6\pi \approx 0.850; about (33.02, 37.55). 50\cos(-18) \approx 33.016, 50\sin(-18) \approx 37.549.
About 110.71 meters; then 110.71 + 314.16k meters for k = 0, 1, 2, \ldots The point is P(t) with (\cos t, \sin t) = (-0.6, 0.8), in Quadrant II. The acute angle between the radius and the negative x-axis has tangent \displaystyle \frac{0.8}{0.6}, and the inverse tangent key (Inverse trigonometric ratios) in radian mode gives 0.92730, so t \approx \pi - 0.92730 \approx 2.21430 and 50t \approx 110.71. Each lap adds 100\pi \approx 314.16 meters.
\displaystyle t = \frac{4 \cdot 120}{50} = 9.6, Quadrant III, since 9.6 - 2\pi \approx 3.317; about (-49.23, -8.72). 50\cos 9.6 \approx -49.234, 50\sin 9.6 \approx -8.716.
Any distance, either way round, can be traveled, so P(t) and its coordinates exist for every t. Tangent divides by the x-coordinate, which is 0 at the north and south points, reached at \displaystyle t = \frac{\pi}{2} + k\pi.
\sin t is a coordinate of a point on x^2 + y^2 = 1, so it is at most 1. Tangent is y over x, and dividing (1, 1000) by its distance \displaystyle \sqrt{1000001} from the origin gives a point on the circle with that same ratio 1000.
\sin 10 is the y-coordinate after traveling 10 units around the unit circle (10 radians, more than a lap), about -0.544. \sin 10^\circ is for a small turn of 10 degrees, about 0.174. Different inputs, different points.
As t goes from 0 to \pi, P(t) moves along the upper half of the circle from (1, 0) to (-1, 0) without turning back, so its x-coordinate falls steadily from 1 to -1. Each x in [-1, 1] has exactly one point \displaystyle \left(x, \sqrt{1 - x^2}\right) on that upper half.