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Precalculus · F.TF.A.2

Trig functions on the unit circle

A right triangle gives an angle a sine and a cosine only when the angle is between 0^\circ and 90^\circ. What could \cos 140^\circ or \sin(-200^\circ) possibly mean?

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Why this exists

The sine, cosine and tangent of an angle were defined from a right triangle: opposite over hypotenuse, adjacent over hypotenuse, opposite over adjacent. A right triangle only has room for an acute angle, so those definitions say nothing about 140^\circ, about -200^\circ, or about the 410^\circ a turning arm can sweep through. Yet angles in standard position can be any size, and anything that turns (a wheel, a pendulum, a merry-go-round) needs a sine and a cosine for all of them.

The fix is to stop asking for a triangle and ask for a point instead. Put a circle of radius 1 at the origin. Every angle in standard position, whatever its size or sign, has a terminal side, and that ray meets the circle at exactly one point. The coordinates of that point become the cosine and the sine. For acute angles this gives back the triangle ratios exactly, so nothing already learned is lost; for every other angle it gives a value where there was none.

The lesson runs over two days. Part 1 (Examples 1 to 4) defines the terminal point and reads the three functions from it. Part 2 (Examples 5 to 7) shows why the new definition extends the old one, moves to circles of any radius, and reads the tangent as a slope.

Definition

The unit circle is the circle of radius 1 centered at the origin. By the equation of a circle its points are exactly those with

x^2 + y^2 = 1

Let \theta be an angle in standard position. The terminal point of \theta is the point where its terminal side meets the unit circle. If the terminal point is (x, y), then

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

In words: the cosine is the x-coordinate of the terminal point, the sine is its y-coordinate, and the tangent is the y-coordinate divided by the x-coordinate. When x = 0 the terminal side lies along the y-axis and \tan\theta is undefined.

x y
\theta
1
\cos\theta
\sin\theta
(\cos\theta, \sin\theta)
Figure 1The terminal point of \theta is where its terminal side meets the unit circle. Its coordinates are \cos\theta and \sin\theta.

The angle may be in degrees or in radians. In radians there is a second reading: on a circle of radius 1 an arc is as long as its angle, so \theta radians is the distance traveled around the circle from (1, 0) to the terminal point (counterclockwise for positive \theta). That reading makes sine and cosine functions of a real number, the subject of Sine, cosine and tangent as functions of a real number.

Intuition

Picture a merry-go-round seen from above, with its center at the origin and a horse fixed one meter out. The horse starts due east, at (1, 0). As the ride turns through \theta, the horse’s east position is \cos\theta meters and its north position is \sin\theta meters. A position west of the center is a negative east position, so a negative cosine is nothing strange: it says the horse is on the west side. A negative sine says it is on the south side.

As the ride turns from 0^\circ to 90^\circ, the horse moves from (1, 0) toward the top of the circle: the cosine shrinks from 1 toward 0 and the sine grows from 0 toward 1. Past 90^\circ the horse crosses to the west side and the cosine turns negative. Because the horse never leaves the circle, neither coordinate can ever be larger than 1 or smaller than -1.

The tangent is a different kind of number. It is \displaystyle \frac{y}{x}, rise over run from the origin to the terminal point, which is the Slope of the terminal side. A steep terminal side has a large tangent, a side close to the x-axis a small one, and a side that tips down to the right a negative one.

Results

The terminal point. For every angle \theta, the terminal point is

(\cos\theta, \sin\theta)

The squares add to one. Since the terminal point lies on the unit circle,

\cos^2\theta + \sin^2\theta = 1

for every angle \theta. This is the Pythagorean identity, studied on its own in The Pythagorean identities.

The range. For every angle, -1 \le \cos\theta \le 1 and -1 \le \sin\theta \le 1.

Tangent. \tan\theta = \dfrac{\sin\theta}{\cos\theta} whenever \cos\theta \ne 0, and this is the slope of the terminal side. The terminal side, extended, meets the vertical line x = 1 at the point (1, \tan\theta).

Any point on the terminal side. If (x, y) is any point other than the origin on the terminal side of \theta, and \displaystyle r = \sqrt{x^2 + y^2} is its distance from the origin, then

\begin{aligned} \cos\theta &= \frac{x}{r} \qquad \sin\theta = \frac{y}{r} \\ \tan\theta &= \frac{y}{x} \end{aligned}

Any circle. The point where the terminal side of \theta meets the circle of radius r centered at the origin is

(r\cos\theta,\ r\sin\theta)

Agreement. For an acute angle, the unit circle values equal the right triangle ratios.

Coterminal angles share a terminal point, so they share all three values.

Why it is true

The definition extends the triangle ratios. Let \theta be acute, with terminal point P = (x, y) in Quadrant I. Drop a perpendicular from P to the x-axis, meeting it at Q = (x, 0). Triangle OQP has a right angle at Q and the angle \theta at O. Its hypotenuse OP is a radius, so OP = 1; the side adjacent to \theta is OQ = x; the side opposite is QP = y. The triangle ratios give

\begin{aligned} \cos\theta &= \frac{OQ}{OP} = \frac{x}{1} = x \\ \sin\theta &= \frac{QP}{OP} = \frac{y}{1} = y \end{aligned}

so the old values and the new ones agree. Every other right triangle with the angle \theta is similar to OQP and gives the same ratios, which is why the triangle ratios belonged to the angle in the first place. For angles that are not acute there is no triangle to disagree with, and the coordinates are the only definition.

Any point on the terminal side. Let (x, y) be on the terminal side at distance r from the origin. Dividing both coordinates by the positive number r moves the point along the same ray (it is a positive multiple of the same direction) to the point \displaystyle \left(\frac{x}{r}, \frac{y}{r}\right), whose distance from the origin is \displaystyle \frac{1}{r}\sqrt{x^2 + y^2} = 1. A point on the terminal side at distance 1 is the terminal point, so \displaystyle \cos\theta = \frac{x}{r} and \displaystyle \sin\theta = \frac{y}{r}. Reading the same step backward, multiplying the terminal point by r gives the point on the circle of radius r, which is (r\cos\theta, r\sin\theta).

The squares add to one. The terminal point (x, y) satisfies x^2 + y^2 = 1, and x = \cos\theta, y = \sin\theta.

Tangent is a slope. The terminal side lies on the line through (0, 0) and (x, y), whose slope is \displaystyle \frac{y - 0}{x - 0} = \frac{y}{x} = \tan\theta. A line through the origin with slope m is y = mx, and at x = 1 its height is m; so the line meets x = 1 at (1, \tan\theta).

Worked examples

Example 1

Reading the three values off a point

The terminal point of an angle \theta is \displaystyle \left(-\frac{8}{17}, \frac{15}{17}\right). Check that this point is on the unit circle, then find \cos\theta, \sin\theta and \tan\theta and name the quadrant of \theta.

Solution

A point is on the unit circle exactly when the squares of its coordinates add to 1:

\begin{aligned} &\left(-\frac{8}{17}\right)^2 + \left(\frac{15}{17}\right)^2 \\ &\quad = \frac{64}{289} + \frac{225}{289} = 1 \end{aligned}

The cosine is the x-coordinate and the sine is the y-coordinate, so \displaystyle \cos\theta = -\frac{8}{17} and \displaystyle \sin\theta = \frac{15}{17}. The tangent is y over x:

\tan\theta = \frac{15/17}{-8/17} = -\frac{15}{8}

The point has x < 0 and y > 0, so it is up and to the left: \theta lies in Quadrant II.

x y
\theta
\left(-\tfrac{8}{17}, \tfrac{15}{17}\right)
Figure 2The terminal point in Example 1. Its x-coordinate is negative because the point is left of the y-axis.

This angle is more than 90^\circ, so no right triangle could hold it and the triangle definitions could not have produced a negative cosine. The coordinates do it without effort.

Example 2

A point that is not on the unit circle

The terminal side of \theta passes through (-24, -7). Find the terminal point of \theta and the three values.

Solution

The point (-24, -7) is on the terminal side but far outside the unit circle. Its distance from the origin is

r = \sqrt{(-24)^2 + (-7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25

Dividing both coordinates by 25 slides the point along the same ray to distance 1, so the terminal point is \displaystyle \left(-\frac{24}{25}, -\frac{7}{25}\right). Then

\begin{aligned} \cos\theta &= -\frac{24}{25} \qquad \sin\theta = -\frac{7}{25} \\ \tan\theta &= \frac{-7}{-24} = \frac{7}{24} \end{aligned}

The tangent needed no division by r at all: the slope from the origin to (-24, -7) is the same as the slope to the terminal point. It is positive because the ray, which points down and to the left into Quadrant III, lies on a line that rises to the right.

Example 3

One coordinate and the quadrant

An angle \theta lies in Quadrant IV and \displaystyle \cos\theta = \frac{39}{89}. Find \sin\theta and \tan\theta.

Solution

The terminal point is \displaystyle \left(\frac{39}{89}, y\right) for some y, and it is on the unit circle, so

\begin{aligned} \sin^2\theta &= 1 - \cos^2\theta = 1 - \frac{1521}{7921} = \frac{6400}{7921} \\ \sin\theta &= \pm\frac{80}{89} \end{aligned}

The circle equation allows two points, one above the x-axis and one below, since the vertical line \displaystyle x = \frac{39}{89} crosses the circle twice. A Quadrant IV point is below the x-axis, so \displaystyle \sin\theta = -\frac{80}{89}. Then

\tan\theta = \frac{-80/89}{39/89} = -\frac{80}{39}
Example 4

A terminal side on a given line

The terminal side of \theta lies along the line y = 3x, in Quadrant III. Find \cos\theta, \sin\theta and \tan\theta.

Solution

Choose any convenient point on the ray. In Quadrant III both coordinates are negative, and x = -1 gives y = 3(-1) = -3, so (-1, -3) is on the terminal side. Its distance from the origin is \displaystyle r = \sqrt{1 + 9} = \sqrt{10}. Dividing by r and clearing the radicals from the denominators,

\begin{aligned} \cos\theta &= \frac{-1}{\sqrt{10}} = -\frac{\sqrt{10}}{10} \\ \sin\theta &= \frac{-3}{\sqrt{10}} = -\frac{3\sqrt{10}}{10} \end{aligned}

and \displaystyle \tan\theta = \frac{-3}{-1} = 3, the slope of the line, as it must be.

The line y = 3x holds two rays from the origin. The other one, in Quadrant I, passes through (1, 3) and gives \displaystyle \cos = \frac{\sqrt{10}}{10} and \displaystyle \sin = \frac{3\sqrt{10}}{10}: both signs flip, but the tangent is 3 again. A line fixes the tangent; only the ray fixes the cosine and sine.

Example 5

Part 2: a horse three meters out

Saule rides a merry-go-round horse 3 meters from the center. Put the center at the origin with east along the positive x-axis. Her horse starts due east and the ride turns her 140^\circ counterclockwise. Where is she, to the nearest centimeter?

Solution

She ends on the terminal side of 140^\circ, 3 meters from the origin. The terminal point (\cos 140^\circ, \sin 140^\circ) is on that ray at distance 1, and multiplying it by 3 keeps the direction and makes the distance 3. So her position is (3\cos 140^\circ, 3\sin 140^\circ). With a calculator in degree mode,

\begin{aligned} x &= 3\cos 140^\circ \approx 3(-0.76604) \approx -2.30 \\ y &= 3\sin 140^\circ \approx 3(0.64279) \approx 1.93 \end{aligned}

so her position is about (-2.30, 1.93). She is about 2.30 meters west of the center and 1.93 meters north of it. As a check, 2.30^2 + 1.93^2 \approx 9.01, close to 3^2 = 9.

x y
140^\circ
P
Q
1
3
Figure 3The point Q on the circle of radius 3 is the terminal point P scaled by 3, so Q = (3\cos 140^\circ, 3\sin 140^\circ).
Example 6

Part 2: a distance around the ride

Zhanna’s horse is 2.5 meters from the center of the same ride, starting due east. The ride carries her 6 meters along her circle, counterclockwise. Find her position.

Solution

The distance is an arc, and radian measure is arc over radius:

\theta = \frac{s}{r} = \frac{6}{2.5} = 2.4 \text{ radians}

Since \displaystyle \frac{\pi}{2} \approx 1.571 < 2.4 < \pi \approx 3.142, she is in Quadrant II. With the calculator in radian mode,

\begin{aligned} x &= 2.5\cos 2.4 \approx 2.5(-0.73739) \approx -1.84 \\ y &= 2.5\sin 2.4 \approx 2.5(0.67546) \approx 1.69 \end{aligned}

Her position is about (-1.84, 1.69): she is about 1.84 meters west and 1.69 meters north of the center. Typing \cos 2.4 in degree mode would give \cos 2.4^\circ, a point barely past due east; the unit has to match the angle.

Example 7

Part 2: why it is called the tangent

The vertical line x = 1 touches the unit circle only at (1, 0): it is the tangent line to the circle there. The line containing the terminal side of \theta = 2 radians is extended until it meets x = 1. Find the meeting point, and explain why its height is \tan\theta.

Solution

The terminal side lies on the line through the origin with slope \tan\theta, which is y = (\tan\theta)\,x. At x = 1 its height is \tan\theta, so the meeting point is (1, \tan\theta). For \theta = 2,

(1, \tan 2) \approx (1, -2.19)

The angle 2 is between \displaystyle \frac{\pi}{2} and \pi, so the terminal side points up and to the left and never reaches x = 1 itself. It is the line through it, extended backward through the origin, that crosses x = 1, below the axis. That matches the sign: a ray rising to the left has a negative slope.

x y
\theta
(1, \tan\theta)
\tan\theta
(1, 0)
x = 1
Figure 4The terminal side meets the line x = 1 at height \tan\theta, since the slope of the terminal side is \tan\theta.

The name comes from this picture. For an acute angle the tangent is the length of a piece of the tangent line to the circle, the piece from (1, 0) up to where the terminal side reaches it.

Boundaries

Tangent needs x \ne 0. When the terminal side lies on the y-axis the terminal point is (0, 1) or (0, -1), the slope is undefined, and so is \tan\theta. The cosine and sine are defined for every angle. Secant, cosecant and cotangent, which divide by other coordinates, have their own gaps; see Secant, cosecant and cotangent on the unit circle.

The circle must be centered at the origin and the angle in standard position. The coordinates are the cosine and sine only for a circle of radius 1 centered at (0, 0), with the angle measured from the positive x-axis. On a circle of radius r they are r times the cosine and sine; on a circle centered elsewhere the center’s coordinates must be added on.

The point (x, y) on the terminal side cannot be the origin. The origin is on every terminal side, and r = 0 there, so it fixes no angle.

One coordinate does not fix the other’s sign. From \cos\theta alone, the circle equation gives \sin\theta only up to sign. The quadrant, or some other fact, has to choose; how the signs fall quadrant by quadrant is Signs of the trig functions by quadrant.

Degrees and radians must not be mixed. \cos 2 means 2 radians. A degree angle always carries its ^\circ.

Where this goes next

The coordinate definition opens most of the rest of trigonometry. Exact values on the axes and at 30^\circ, 45^\circ and 60^\circ are read off the circle in Exact values on the unit circle. Reflecting a terminal point across an axis is Unit circle symmetry and even-odd identities, and bringing any angle back to an acute one is reference angles. Because a full turn returns to the same point, every value repeats, which is periodicity.

Reading \theta as a distance around the circle turns sine and cosine into functions of a real number, and plotting the height \sin\theta against \theta gives the wave in Graphing sine and cosine. The point (r\cos\theta, r\sin\theta) is the bridge from Polar coordinates to x and y, worked out in Converting between polar and rectangular, and a point moving around a circle as time passes is the first example in parametric equations.

Exercises

Part 1: values from a point

For each exercise, find \cos\theta, \sin\theta and \tan\theta, and name the quadrant of \theta.

  1. The terminal point of \theta is \displaystyle \left(\frac{12}{13}, -\frac{5}{13}\right).
  2. The terminal point of \theta is (-0.28, -0.96).
  3. The terminal side of \theta passes through (-9, 40).
  4. The terminal side of \theta passes through (-2, 3). Give exact values with rational denominators.
  5. The terminal side of \theta passes through (3, -1). Give exact values with rational denominators.
  6. Is (0.5, 0.8) the terminal point of any angle? Explain.

Part 1: one value or a line given

  1. \theta is in Quadrant II and \displaystyle \sin\theta = \frac{13}{85}. Find \cos\theta and \tan\theta.
  2. \theta is in Quadrant III and \cos\theta = -0.6. Find \sin\theta and \tan\theta.
  3. \theta is in Quadrant III and \displaystyle \tan\theta = \frac{11}{60}. Find \cos\theta and \sin\theta.
  4. The terminal side of \theta lies along y = -2x in Quadrant IV. Find the three values exactly.

Part 2: circles of any radius

Put the center of the merry-go-round at the origin with east along the positive x-axis. Every rider starts due east. Give positions to the nearest centimeter.

  1. Timur rides 4 meters from the center and turns 75^\circ counterclockwise. Where is he?
  2. A rider 2 meters from the center turns 50^\circ clockwise. Where is she?
  3. A rider 3.5 meters from the center travels 9 meters counterclockwise along his circle. Find the angle in radians and his position.
  4. A horse sits at (-1.6, 6.3), in meters. How far is it from the center, and what are \cos\theta, \sin\theta and \tan\theta for the angle \theta of its position?

Part 2: tangent as a slope

  1. The line containing the terminal side of 160^\circ meets x = 1. Find the point, to three decimal places.
  2. Find \cos\theta and \sin\theta for every angle \theta with 0^\circ \le \theta < 360^\circ and \tan\theta = -0.75.
  3. Use a calculator to find the terminal point of -200^\circ to three decimal places, and name its quadrant.

Reasoning

  1. Explain why no angle has \sin\theta = 1.2.
  2. For an acute angle \theta, explain why \cos\theta from the unit circle equals adjacent over hypotenuse in every right triangle with angle \theta, not only in one with hypotenuse 1.
  3. Show that \displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta} wherever both sides are defined, and name the angles between 0^\circ and 360^\circ where neither side is.
  4. Timur says \sin 400^\circ = \sin 40^\circ. Explain why, using terminal points.
  5. Explain why (r\cos\theta, r\sin\theta) is on the terminal side of \theta and at distance r from the origin, for every r > 0.
  6. Zhanna says, “If \cos\theta = 0.6, then \sin\theta = 0.8.” What is missing? Give every possible value of \sin\theta.
  7. Explain why \theta and \theta + 180^\circ have the same tangent whenever it is defined, using the slope of the terminal side.
  8. Explain why no angle has \cos\theta = \sin\theta = 0.5.
Answers
  1. \displaystyle \cos\theta = \frac{12}{13}, \displaystyle \sin\theta = -\frac{5}{13}, \displaystyle \tan\theta = -\frac{5}{12}; Quadrant IV. The point checks: \displaystyle \frac{144}{169} + \frac{25}{169} = 1.
  2. \cos\theta = -0.28, \sin\theta = -0.96, \displaystyle \tan\theta = \frac{24}{7} \approx 3.43; Quadrant III. The point checks: 0.0784 + 0.9216 = 1.
  3. \displaystyle \cos\theta = -\frac{9}{41}, \displaystyle \sin\theta = \frac{40}{41}, \displaystyle \tan\theta = -\frac{40}{9}; Quadrant II. Here \displaystyle r = \sqrt{81 + 1600} = 41.
  4. \displaystyle \cos\theta = -\frac{2\sqrt{13}}{13}, \displaystyle \sin\theta = \frac{3\sqrt{13}}{13}, \displaystyle \tan\theta = -\frac{3}{2}; Quadrant II. Here \displaystyle r = \sqrt{13}.
  5. \displaystyle \cos\theta = \frac{3\sqrt{10}}{10}, \displaystyle \sin\theta = -\frac{\sqrt{10}}{10}, \displaystyle \tan\theta = -\frac{1}{3}; Quadrant IV. Here \displaystyle r = \sqrt{10}.
  6. No. 0.5^2 + 0.8^2 = 0.89 \ne 1, so the point is not on the unit circle. It is on the terminal side of some angle, but it is not that angle’s terminal point.
  7. \displaystyle \cos\theta = -\frac{84}{85}, \displaystyle \tan\theta = -\frac{13}{84}. \displaystyle \cos^2\theta = 1 - \frac{169}{7225} = \frac{7056}{7225}, and the cosine is negative left of the y-axis.
  8. \sin\theta = -0.8, \displaystyle \tan\theta = \frac{4}{3}. \sin^2\theta = 1 - 0.36 = 0.64, and a Quadrant III point is below the x-axis.
  9. \displaystyle \cos\theta = -\frac{60}{61}, \displaystyle \sin\theta = -\frac{11}{61}. The point (-60, -11) has slope \displaystyle \frac{11}{60} and lies in Quadrant III, and \displaystyle r = \sqrt{3600 + 121} = 61.
  10. \displaystyle \cos\theta = \frac{\sqrt{5}}{5}, \displaystyle \sin\theta = -\frac{2\sqrt{5}}{5}, \tan\theta = -2. The point (1, -2) is on the ray and \displaystyle r = \sqrt{5}.
  11. About (1.04, 3.86). 4\cos 75^\circ \approx 1.0353 and 4\sin 75^\circ \approx 3.8637.
  12. About (1.29, -1.53). Clockwise makes the angle -50^\circ: 2\cos(-50^\circ) \approx 1.2856 and 2\sin(-50^\circ) \approx -1.5321.
  13. \displaystyle \theta = \frac{9}{3.5} = \frac{18}{7} \approx 2.571 radians, and the position is about (-2.95, 1.89): \displaystyle 3.5\cos\frac{18}{7} \approx -2.9463 and \displaystyle 3.5\sin\frac{18}{7} \approx 1.8892.
  14. 6.5 meters; \displaystyle \cos\theta = -\frac{16}{65}, \displaystyle \sin\theta = \frac{63}{65}, \displaystyle \tan\theta = -\frac{63}{16}. \displaystyle r = \sqrt{2.56 + 39.69} = \sqrt{42.25} = 6.5, then divide each coordinate by 6.5.
  15. (1, -0.364). The height is \tan 160^\circ \approx -0.36397.
  16. Two angles: one in Quadrant II with \displaystyle \cos\theta = -\frac{4}{5}, \displaystyle \sin\theta = \frac{3}{5}, and one in Quadrant IV with \displaystyle \cos\theta = \frac{4}{5}, \displaystyle \sin\theta = -\frac{3}{5}. The terminal sides lie on y = -0.75x, through (-4, 3) and (4, -3), each at distance 5.
  17. About (-0.940, 0.342), in Quadrant II. \cos(-200^\circ) \approx -0.93969 and \sin(-200^\circ) \approx 0.34202; -200^\circ is coterminal with 160^\circ.
  18. \sin\theta is the y-coordinate of a point on the unit circle, and x^2 + y^2 = 1 forces y^2 \le 1, so -1 \le y \le 1. The value 1.2 is out of reach.
  19. Any right triangle with the acute angle \theta is similar to the triangle OQP formed by the terminal point, so its sides are those of OQP multiplied by one factor k. Adjacent over hypotenuse is then \displaystyle \frac{kx}{k \cdot 1} = x = \cos\theta.
  20. With terminal point (x, y), \displaystyle \frac{\sin\theta}{\cos\theta} = \frac{y}{x} = \tan\theta whenever x \ne 0. Both sides are undefined exactly when x = 0, at 90^\circ and 270^\circ.
  21. 400^\circ - 360^\circ = 40^\circ, so the two angles are coterminal. They have the same terminal side, hence the same terminal point, hence the same y-coordinate.
  22. It is the terminal point (\cos\theta, \sin\theta) multiplied by the positive number r, which keeps the direction from the origin, so it is on the same ray. Its distance is \displaystyle \sqrt{r^2\cos^2\theta + r^2\sin^2\theta} = r\sqrt{1} = r.
  23. The sign is missing. \sin^2\theta = 1 - 0.36 = 0.64, so \sin\theta = 0.8 or \sin\theta = -0.8. The line x = 0.6 meets the unit circle twice; without the quadrant both are possible.
  24. Adding 180^\circ turns the terminal side to the opposite ray on the same line through the origin. One line has one slope, so the tangent is the same. Algebraically the terminal point (x, y) becomes (-x, -y), and \displaystyle \frac{-y}{-x} = \frac{y}{x}.
  25. The terminal point would be (0.5, 0.5), but 0.5^2 + 0.5^2 = 0.5 \ne 1, so that point is not on the unit circle.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0