Precalculus

Angles in standard position and coterminal angles

A radar sweep has turned through more than a full circle, or turned the other way. How do you say exactly where it points, and how far it has turned to get there?

You need

The coordinate planeGrades 5 to 6 Measuring anglesGrades 4 to 8

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Trig functions on the unit circlePrecalculus Plotting polar pointsPrecalculus

Where you are 4 in Angles and their measure

Trigonometry7

Angles and their measure4

Angles in standard position and coterminal angles

Radian measure Angular and linear speed Arc length and sector area in radians

Why this exists

In geometry an angle is a figure: two rays from one point, measured with a protractor somewhere between 0^\circ and 180^\circ (see Measuring angles). That is enough for a triangle. It is not enough for anything that turns.

A radar sweep starts pointing east and turns. After a while it has gone once around and a little more. Another sweep turns the other way. A protractor reading of the final position says where each one points, but not how it got there, and two sweeps pointing the same way may have turned through very different amounts.

The fix is to treat an angle as a rotation and to pin it to the coordinate plane. Every angle starts in the same place, the direction of the turn is recorded with a sign, and the amount can be any number of degrees at all. This is the picture the rest of trigonometry is built on.

Definition

An angle is in standard position when its vertex is at the origin and its initial side lies along the positive x-axis. The ray where the rotation ends is the terminal side.

Figure 1An angle in standard position. The vertex is at the origin and the turn starts on the positive x-axis.

The measure of the angle is the amount of rotation from the initial side to the terminal side, with a sign for the direction:

a positive angle turns counterclockwise;
a negative angle turns clockwise;
the angle 0^\circ does not turn at all, so its terminal side is the positive x-axis.

A full turn is 360^\circ, and nothing stops the rotation at one turn. An angle of 410^\circ goes all the way around and then 50^\circ more. Angles are often named with Greek letters such as \theta.

An angle in standard position lies in a quadrant when its terminal side lies in that quadrant. The quadrants are numbered I to IV counterclockwise, starting from the upper right.

An angle whose terminal side lies on one of the axes is a quadrantal angle. Its terminal side is on a boundary between quadrants, so a quadrantal angle lies in no quadrant. Between 0^\circ and 360^\circ the quadrantal angles are 0^\circ, 90^\circ, 180^\circ and 270^\circ.

Figure 2The quadrants, and the four quadrantal angles between 0^\circ and 360^\circ.

Two angles in standard position are coterminal when they have the same terminal side.

Intuition

Think of the terminal side as the sweep’s pointer and the angle as the sweep’s history. The pointer only knows where it is. The angle also knows how many times it went around and which way.

Adding 360^\circ to an angle sends the pointer once more around the circle, back to exactly where it was. Subtracting 360^\circ undoes one full turn. Either way the pointer ends in the same place, so the new angle is coterminal with the old one. That single observation is the whole theory of coterminal angles.

A negative angle is the same idea as a negative number: an amount measured in the opposite direction. Turning 90^\circ clockwise is the angle -90^\circ, and it ends on the negative y-axis, the same ray that 270^\circ reaches by turning the long way round.

Results

Coterminal angles. Two angles \alpha and \beta in standard position are coterminal exactly when they differ by a whole number of full turns:

\alpha - \beta = 360^\circ \cdot k \quad \text{for some integer } k

So the angles coterminal with \theta are

\theta + 360^\circ k, \qquad k = 0, \pm 1, \pm 2, \ldots

One representative. Every angle is coterminal with exactly one angle \theta_0 in the range 0^\circ \le \theta_0 < 360^\circ. To find it, add or subtract 360^\circ as many times as needed to land in that range.

Quadrants. An angle lies in the same quadrant as its representative \theta_0:

Representative \theta_0	Where the terminal side is
0^\circ < \theta_0 < 90^\circ	Quadrant I
90^\circ < \theta_0 < 180^\circ	Quadrant II
180^\circ < \theta_0 < 270^\circ	Quadrant III
270^\circ < \theta_0 < 360^\circ	Quadrant IV
0^\circ, 90^\circ, 180^\circ or 270^\circ	on an axis (quadrantal)

Quadrantal angles. An angle is quadrantal exactly when it is an integer multiple of 90^\circ.

Why it is true

Differing by full turns gives the same terminal side. Rotating by \beta and then by a further 360^\circ k adds k full turns (clockwise ones when k is negative). Each full turn returns a ray to its starting position, so the terminal side of \beta + 360^\circ k is the terminal side of \beta.

The same terminal side forces a difference of full turns. Suppose \alpha and \beta end on the same ray. Rotating that ray by \alpha - \beta carries the terminal side of \beta onto the terminal side of \alpha, which is the same ray. A rotation about the origin that leaves a ray where it was must be a whole number of full turns, so \alpha - \beta = 360^\circ k.

Exactly one representative. Write \theta = 360^\circ q + r with q an integer and 0 \le r < 360, which is division with remainder, and that pair (q, r) is unique. Then r is coterminal with \theta. If two angles in [0^\circ, 360^\circ) were coterminal, they would differ by a multiple of 360^\circ while being less than 360^\circ apart, so the multiple is 0 and they are equal.

Quadrantal means a multiple of 90^\circ. The representatives on the axes are 0^\circ, 90^\circ, 180^\circ, 270^\circ, which are 90^\circ times 0, 1, 2, 3. Adding 360^\circ k = 90^\circ \cdot 4k to any of them gives another multiple of 90^\circ, and every multiple of 90^\circ arises this way.

Worked examples

Example 1

More than a full turn

Draw 410^\circ in standard position and name the quadrant it lies in.

Solution

The angle is positive, so the turn is counterclockwise. It is more than 360^\circ, so take one full turn out of it:

410^\circ - 360^\circ = 50^\circ

The sweep goes once all the way around, back to the positive x-axis, and then turns 50^\circ more. Since 0^\circ < 50^\circ < 90^\circ, the terminal side is in Quadrant I, and 410^\circ is coterminal with 50^\circ.

Figure 3A turn of 410^\circ is one full turn and 50^\circ more.

The spiral in the figure is how a turn of more than 360^\circ is drawn: it shows the rotation, not just where it ends.

Example 2

A clockwise turn

Draw -235^\circ in standard position. Name its quadrant and the positive angle less than 360^\circ that is coterminal with it.

Solution

The sign is negative, so turn clockwise from the positive x-axis. A clockwise turn of 90^\circ reaches the negative y-axis and a clockwise turn of 180^\circ reaches the negative x-axis. That leaves 235^\circ - 180^\circ = 55^\circ more, still clockwise, which carries the terminal side up from the negative x-axis into Quadrant II.

To name a positive coterminal angle, add one full turn:

-235^\circ + 360^\circ = 125^\circ

Since 90^\circ < 125^\circ < 180^\circ, this confirms Quadrant II.

Figure 4Turning 125^\circ one way and 235^\circ the other way lands on the same terminal side.

The angles 125^\circ and -235^\circ end on the same ray but are different rotations. One is a short counterclockwise turn and the other a longer clockwise one, and they differ by exactly 360^\circ.

Example 3

Several turns at once

Find the angle \theta_0 with 0^\circ \le \theta_0 < 360^\circ that is coterminal with each angle, and name its quadrant.

(a) 865^\circ, (b) -1{,}310^\circ

Solution

Subtracting 360^\circ one turn at a time works, but dividing counts the turns at once.

(a) 865 \div 360 is about 2.4, so 865^\circ holds 2 full turns:

865^\circ - 2(360^\circ) = 865^\circ - 720^\circ = 145^\circ

That is between 90^\circ and 180^\circ, so 865^\circ lies in Quadrant II.

(b) -1310 \div 360 is about -3.6. Adding 3 full turns gives -1310 + 1080 = -230, still negative, so add 4:

-1{,}310^\circ + 4(360^\circ) = -1{,}310^\circ + 1{,}440^\circ = 130^\circ

So -1{,}310^\circ is coterminal with 130^\circ and lies in Quadrant II. For a negative angle the number of turns to add is the size of the quotient rounded up, here 4 rather than 3, because the result has to land at 0^\circ or above.

Example 4

Recognizing quadrantal angles

Decide whether each angle is quadrantal, and if it is, name the half-axis its terminal side lies on.

(a) 630^\circ, (b) -540^\circ, (c) 225^\circ

Solution

An angle is quadrantal exactly when it is an integer multiple of 90^\circ, so divide by 90 first, then find the representative to see which half-axis.

(a) 630 \div 90 = 7, an integer, so 630^\circ is quadrantal. Its representative is 630^\circ - 360^\circ = 270^\circ, so the terminal side is the negative y-axis.

(b) -540 \div 90 = -6, an integer, so -540^\circ is quadrantal. Adding two full turns, -540^\circ + 720^\circ = 180^\circ, so the terminal side is the negative x-axis. Read as a turn: one clockwise circle and then a clockwise half-turn.

(c) 225 \div 90 = 2.5, not an integer, so 225^\circ is not quadrantal. It lies strictly between 180^\circ and 270^\circ, in Quadrant III.

Example 5

The radar sweep

Chukwudi watches a radar sweep that starts pointing east, along the positive x-axis, and turns counterclockwise at 24^\circ per second.

(a) Through what angle has it turned after 40 seconds, and where does it point then?

(b) At what times in the first 30 seconds does it point due south, along the negative y-axis?

Solution

(a) The angle turned is rate times time:

24^\circ \times 40 = 960^\circ

Taking out two full turns, 960^\circ - 720^\circ = 240^\circ. The sweep has gone around twice and points into Quadrant III, 60^\circ past due west.

(b) Due south is the terminal side of 270^\circ, and of every angle coterminal with it: 270^\circ, 630^\circ, 990^\circ, and so on. The sweep reaches 270^\circ when 24t = 270, so t = 11.25 seconds, and reaches 630^\circ when 24t = 630, so t = 26.25 seconds. The next one, 990^\circ, would take 41.25 seconds, past the window. The answer is 11.25 and 26.25 seconds.

The angle in (a) is not just “240^\circ”. The sweep pointing at 240^\circ after one turn and after two turns are different moments, and only the full measure 960^\circ tells them apart.

Example 6

Testing two angles for coterminality

Ataahua has two angles, 1{,}195^\circ and -965^\circ. Are they coterminal? Is 1{,}195^\circ coterminal with -485^\circ?

Solution

Two angles are coterminal exactly when their difference is a multiple of 360^\circ.

1{,}195^\circ - (-965^\circ) = 2{,}160^\circ = 6 \cdot 360^\circ

So they are coterminal. As a check, both reduce to the same representative: 1{,}195^\circ - 1{,}080^\circ = 115^\circ and -965^\circ + 1{,}080^\circ = 115^\circ.

For the second pair,

1{,}195^\circ - (-485^\circ) = 1{,}680^\circ

and 1680 \div 360 = 4\tfrac{2}{3}, not an integer, so they are not coterminal. Indeed -485^\circ + 720^\circ = 235^\circ, which is in Quadrant III, not Quadrant II.

Example 7

An angle and five times itself

Felicjan asks for every angle \theta with 0^\circ < \theta < 360^\circ whose terminal side is the same as that of 5\theta. Find them all.

Solution

The condition is that \theta and 5\theta are coterminal, so their difference is a multiple of 360^\circ:

5\theta - \theta = 360^\circ k \qquad\Longrightarrow\qquad \theta = 90^\circ k

for some integer k. With 0^\circ < \theta < 360^\circ, the only choices are k = 1, 2, 3, so \theta is 90^\circ, 180^\circ or 270^\circ.

Check one: 5(90^\circ) = 450^\circ = 90^\circ + 360^\circ, the same terminal side as 90^\circ. The answers are exactly the quadrantal angles in the range, which is no accident: \theta = 90^\circ k is the test for a quadrantal angle.

Boundaries

Standard position needs coordinates. The words “initial side on the positive x-axis” only mean something once axes are fixed. An angle inside a triangle has no initial side and no sign.

Coterminal is not equal. 50^\circ and 410^\circ share a terminal side, but they are different rotations. Anything that depends only on the terminal side (where the sweep points, and later the sine and cosine) agrees on them; anything that depends on the turn itself (how far the tip of the sweep traveled, how long it took) does not.

The terminal side is a ray, not a line. 50^\circ and 230^\circ lie along the same line through the origin but point in opposite directions. They differ by 180^\circ, not by a multiple of 360^\circ, so they are not coterminal.

Quadrantal angles belong to no quadrant. 90^\circ is not in Quadrant I or Quadrant II; it is on the boundary between them. Any rule stated “for an angle in Quadrant II” says nothing about it.

Exactly one representative needs a half-open range. The range 0^\circ \le \theta_0 < 360^\circ includes 0^\circ and leaves out 360^\circ. Allow both ends and the positive x-axis would have two representatives.

Degrees for now. Every measure here is in degrees. The same ideas hold in radians, where a full turn is 2\pi instead of 360^\circ.

Where this goes next

Measuring an angle by how far its rotation travels around a circle leads to radian measure, where a full turn is 2\pi and coterminal angles differ by multiples of 2\pi. How fast a sweep turns is angular speed.

Put a circle of radius 1 at the origin and the terminal side of any angle meets it at exactly one point. The coordinates of that point are the cosine and sine of the angle, which is why coterminal angles share every trigonometric value and why the quadrant decides their signs.

Turning through an angle and then walking out a distance is also how a point is located in polar coordinates; see plotting polar points. A rotation of a whole figure, rather than a single ray, is a rotation in geometry, with the same counterclockwise-positive convention.

Exercises

Drawing angles and naming quadrants

Sketch each angle in standard position. Name the quadrant it lies in and the coterminal angle \theta_0 with 0^\circ \le \theta_0 < 360^\circ.

160^\circ
-70^\circ
500^\circ
-400^\circ
-1{,}000^\circ

Finding coterminal angles

Find one positive angle and one negative angle that are coterminal with 75^\circ.
Find the angle between 0^\circ and 360^\circ that is coterminal with 1{,}250^\circ.
Find the angle between 0^\circ and 360^\circ that is coterminal with -520^\circ.
Find the angle between 0^\circ and 360^\circ that is coterminal with -2{,}900^\circ.
Find the angle between -720^\circ and -360^\circ that is coterminal with 210^\circ.

Quadrantal angles and testing pairs

Which of these angles are quadrantal? For each one that is, name the half-axis its terminal side lies on: 450^\circ, -900^\circ, 990^\circ, -1{,}080^\circ, 135^\circ.
Describe every angle whose terminal side is the negative y-axis.
Are -100^\circ and 980^\circ coterminal?
Are 300^\circ and -300^\circ coterminal?

Turning and pointing

A radar sweep starts pointing east and turns clockwise at 15^\circ per second. What angle has it turned through after 50 seconds, and in which quadrant does it point?
Ataahua’s radar sweep starts pointing east and turns counterclockwise at 18^\circ per second. What are the first two times at which it points due north?
The terminal side of an angle in standard position passes through the point (-3, -3). Find the smallest positive such angle and the negative such angle closest to 0^\circ.
The terminal side of an angle passes through (0, -5). Is the angle quadrantal? Give one positive and one negative measure for it.

Reasoning

Explain why 50^\circ and 230^\circ are not coterminal, even though their terminal sides lie on the same line.
The angle \theta lies in Quadrant II. In which quadrant does \theta + 180^\circ lie? In which does -\theta lie? Explain.
Explain why every angle is coterminal with exactly one angle \theta_0 satisfying 0^\circ \le \theta_0 < 360^\circ.
Find every angle \theta with 0^\circ \le \theta < 360^\circ that is coterminal with 3\theta.
Show that if \theta is quadrantal, then 2\theta is coterminal with either 0^\circ or 180^\circ.
Felicjan says, “\theta is in Quadrant III, so \tfrac{\theta}{2} is in Quadrant II.” Explain why the quadrant of \tfrac{\theta}{2} is not decided by the quadrant of \theta, using \theta = 200^\circ and \theta = 560^\circ.

Answers

Quadrant II; \theta_0 = 160^\circ. It is already between 90^\circ and 180^\circ.
Quadrant IV; \theta_0 = 290^\circ. Turn 70^\circ clockwise, below the positive x-axis, and -70^\circ + 360^\circ = 290^\circ.
Quadrant II; \theta_0 = 140^\circ. One full turn and more: 500^\circ - 360^\circ = 140^\circ.
Quadrant IV; \theta_0 = 320^\circ. One clockwise turn and 40^\circ more: -400^\circ + 720^\circ = 320^\circ.
Quadrant I; \theta_0 = 80^\circ. Three full turns: -1{,}000^\circ + 1{,}080^\circ = 80^\circ.
For example 435^\circ and -285^\circ: 75^\circ + 360^\circ = 435^\circ and 75^\circ - 360^\circ = -285^\circ. Any 75^\circ + 360^\circ k works.
170^\circ. 1{,}250 \div 360 is about 3.5, so 1{,}250^\circ - 1{,}080^\circ = 170^\circ.
200^\circ. -520^\circ + 720^\circ = 200^\circ; one turn only reaches -160^\circ.
340^\circ. -2{,}900 \div 360 is about -8.06, so add 9 turns: -2{,}900^\circ + 3{,}240^\circ = 340^\circ.
-510^\circ. 210^\circ - 720^\circ = -510^\circ, which is between -720^\circ and -360^\circ.
450^\circ is quadrantal, on the positive y-axis (450^\circ - 360^\circ = 90^\circ). -900^\circ is quadrantal, on the negative x-axis (-900^\circ + 1{,}080^\circ = 180^\circ). 990^\circ is quadrantal, on the negative y-axis (990^\circ - 720^\circ = 270^\circ). -1{,}080^\circ is quadrantal, on the positive x-axis (exactly -3 full turns). 135^\circ is not quadrantal, since 135 \div 90 = 1.5; it lies in Quadrant II.
270^\circ + 360^\circ k for every integer k: these are the angles coterminal with 270^\circ, such as -90^\circ, 270^\circ and 630^\circ.
Yes. 980^\circ - (-100^\circ) = 1{,}080^\circ = 3 \cdot 360^\circ. Both reduce to 260^\circ.
No. 300^\circ - (-300^\circ) = 600^\circ, and 600 \div 360 is not an integer. Their representatives are 300^\circ and 60^\circ.
-750^\circ, pointing into Quadrant IV. Clockwise makes the angle negative: -15^\circ \times 50 = -750^\circ, and -750^\circ + 1{,}080^\circ = 330^\circ.
At 5 seconds and 25 seconds. Due north is 90^\circ or 450^\circ, and 90 \div 18 = 5, 450 \div 18 = 25.
225^\circ and -135^\circ. The point is in Quadrant III on the line y = x, halfway between the negative x-axis and the negative y-axis, so the angle is 180^\circ + 45^\circ; then 225^\circ - 360^\circ = -135^\circ.
Yes, the point is on the negative y-axis. For example 270^\circ and -90^\circ.
Their difference is 230^\circ - 50^\circ = 180^\circ, not a multiple of 360^\circ. The two terminal sides are opposite rays on one line: one points into Quadrant I, the other into Quadrant III. Coterminal angles must share the ray, not just the line.
\theta + 180^\circ is in Quadrant IV and -\theta is in Quadrant III. Take the representative, 90^\circ < \theta_0 < 180^\circ. Then 270^\circ < \theta_0 + 180^\circ < 360^\circ, Quadrant IV. And -180^\circ < -\theta_0 < -90^\circ; adding 360^\circ gives between 180^\circ and 270^\circ, Quadrant III. Adding a multiple of 360^\circ to \theta changes neither answer.
Divide \theta by 360 with remainder: \theta = 360^\circ q + r with q an integer and 0 \le r < 360. Then r^\circ differs from \theta by q full turns, so it is coterminal. Two angles in the range differ by less than 360^\circ, so if they are coterminal the multiple of 360^\circ between them is 0 and they are the same angle.
0^\circ and 180^\circ. Coterminal means 3\theta - \theta = 2\theta = 360^\circ k, so \theta = 180^\circ k, and in the range that gives k = 0 or k = 1.
Quadrantal means \theta = 90^\circ k for an integer k, so 2\theta = 180^\circ k. If k is even, k = 2m, then 2\theta = 360^\circ m, coterminal with 0^\circ. If k is odd, k = 2m + 1, then 2\theta = 180^\circ + 360^\circ m, coterminal with 180^\circ.
200^\circ and 560^\circ are coterminal, both in Quadrant III. But half of 200^\circ is 100^\circ, in Quadrant II, while half of 560^\circ is 280^\circ, in Quadrant IV. Halving \theta + 360^\circ adds 180^\circ to the result, sending it to the opposite ray, so the quadrant of \tfrac{\theta}{2} depends on which coterminal measure is halved, not only on where the terminal side of \theta is.