Precalculus · F.TF.A.2

Reference angles

A skater spins through 1{,}020^\circ, or turns \displaystyle \frac{17\pi}{6} the other way. Is there a short road from an angle that size to its sine and cosine, without drawing every turn?

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Trig functions on the unit circlePrecalculus

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Exact values on the unit circlePrecalculus The ambiguous casePrecalculus Solving trigonometric equationsGrade 11

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)

When you reach for it

On the unit circle, \cos\theta and \sin\theta are the coordinates of the terminal point of \theta (Trig functions on the unit circle). That definition works for every angle, but it does not say how to find the coordinates when the angle is 1{,}020^\circ, \displaystyle -\frac{17\pi}{6} or 5 radians. The exact values you know by heart belong to acute angles: 30^\circ, 45^\circ and 60^\circ from the special right triangles.

A reference angle is the bridge. Any angle whose terminal side is off the axes makes an acute angle with the x-axis, and the terminal point sits at the same distances from the two axes as the terminal point of that acute angle. So the values of the big angle are the values of the acute one, up to sign. You reach for it whenever a trig value is wanted at an angle that is negative, larger than a quarter turn, or more than a full turn, and whenever you need every angle that shares a value.

The procedure

The reference angle. Let \theta be an angle in standard position whose terminal side is not on an axis. Its reference angle \theta' is the acute angle between the terminal side and the x-axis. It is always measured to the x-axis, the positive half or the negative half, whichever is nearer, and never to the y-axis.

Figure 1The reference angle \theta' is the acute angle between the terminal side and the x-axis. Here \displaystyle \theta = \frac{7\pi}{6} and \displaystyle \theta' = \frac{\pi}{6}; the legs of the triangle are \cos\theta' and \sin\theta'.

Bring the angle inside one turn. Adding or subtracting a full turn does not move the terminal side, so \theta and \theta \pm 360^\circ (or \theta \pm 2\pi) are coterminal and have the same terminal point, as Angles in standard position and coterminal angles showed. Add or subtract whole turns until the angle lands in 0^\circ \le \theta_0 < 360^\circ (or 0 \le \theta_0 < 2\pi). Nothing is lost: \theta_0 and \theta share every trig value. For a negative angle, add turns; for a large one, subtract them.

Measure to the x-axis. Inside one turn the quadrant of \theta_0 says which half of the x-axis is nearer, and the gap to it is the reference angle:

Quadrant of \theta_0	Degrees	Radians
I	\theta' = \theta_0	\theta' = \theta_0
II	\theta' = 180^\circ - \theta_0	\theta' = \pi - \theta_0
III	\theta' = \theta_0 - 180^\circ	\theta' = \theta_0 - \pi
IV	\theta' = 360^\circ - \theta_0	\theta' = 2\pi - \theta_0

In Quadrant II the terminal side has passed the positive x-axis by \theta_0 and is short of the negative x-axis (at 180^\circ) by the rest. In Quadrant III it has passed the negative x-axis by \theta_0 - 180^\circ. In Quadrant IV it is short of a full turn, which ends on the positive x-axis, by 360^\circ - \theta_0.

Why the values match up to sign. Let the terminal point of \theta be (x, y), and drop a perpendicular from it to the x-axis. The triangle formed with the origin has a right angle at the foot, hypotenuse 1 (a radius of the unit circle), and the angle \theta' at the origin. Its legs are the distances of the point from the axes: |x| along the x-axis and |y| up or down. In that right triangle, adjacent over hypotenuse and opposite over hypotenuse give

\begin{aligned} |x| &= \cos\theta' \\ |y| &= \sin\theta' \end{aligned}

and dividing, \displaystyle \left|\frac{y}{x}\right| = \tan\theta'. Since x = \cos\theta and y = \sin\theta,

\begin{aligned} |\cos\theta| &= \cos\theta' \\ |\sin\theta| &= \sin\theta' \\ |\tan\theta| &= \tan\theta' \end{aligned}

Take the sign from the quadrant. The reference angle gives the size; where the terminal point sits gives the sign. Cosine is the x-coordinate, negative left of the y-axis (Quadrants II and III). Sine is the y-coordinate, negative below the x-axis (Quadrants III and IV). Tangent is y over x, positive when the two agree in sign (Quadrants I and III). These sign patterns, and reading a quadrant back from two signs, are the subject of Signs of the trig functions by quadrant.

For the special acute angles, the values the reference angle supplies are

\theta'	\cos\theta'	\sin\theta'	\tan\theta'
\displaystyle 30^\circ = \frac{\pi}{6}	\displaystyle \frac{\sqrt{3}}{2}	\displaystyle \frac{1}{2}	\displaystyle \frac{\sqrt{3}}{3}
\displaystyle 45^\circ = \frac{\pi}{4}	\displaystyle \frac{\sqrt{2}}{2}	\displaystyle \frac{\sqrt{2}}{2}	1
\displaystyle 60^\circ = \frac{\pi}{3}	\displaystyle \frac{1}{2}	\displaystyle \frac{\sqrt{3}}{2}	\displaystyle \sqrt{3}

The full circle of exact values, including the angles on the axes, is laid out in Exact values on the unit circle.

Worked examples

Example 1

One angle in each quadrant

Find the reference angle of 145^\circ, of 220^\circ and of 310^\circ.

Solution

All three are already inside one turn, so only the quadrant and the nearer half of the x-axis matter.

145^\circ is between 90^\circ and 180^\circ, in Quadrant II. The terminal side is short of the negative x-axis by \theta' = 180^\circ - 145^\circ = 35^\circ.

220^\circ is between 180^\circ and 270^\circ, in Quadrant III. It has passed the negative x-axis by \theta' = 220^\circ - 180^\circ = 40^\circ.

310^\circ is between 270^\circ and 360^\circ, in Quadrant IV. It is short of the positive x-axis by \theta' = 360^\circ - 310^\circ = 50^\circ.

Figure 2The reference angle is always measured to the nearer half of the x-axis, never to the y-axis.

For 310^\circ the y-axis (at 270^\circ) is 40^\circ away, closer than the x-axis. That gap is not the reference angle: the reference angle is defined to the x-axis, because that is the leg whose length is the cosine.

Example 2

Khanyisile’s spin of 1,020 degrees

A figure skater, Khanyisile, starts a spin with her right hand stretched out due east, 0.7 meters from her spin axis. She turns 1{,}020^\circ counterclockwise. Put the axis at the origin with east along the positive x-axis. Find \cos 1{,}020^\circ and \sin 1{,}020^\circ exactly, and the position of her hand.

Solution

Two full turns are 720^\circ, and three would be 1{,}080^\circ, too many. So

\theta_0 = 1{,}020^\circ - 720^\circ = 300^\circ

and 1{,}020^\circ ends where 300^\circ does. That is between 270^\circ and 360^\circ, Quadrant IV, so the reference angle is \theta' = 360^\circ - 300^\circ = 60^\circ.

The sizes come from 60^\circ: \displaystyle \cos 60^\circ = \frac{1}{2} and \displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}. The signs come from Quadrant IV, right of the y-axis and below the x-axis: cosine positive, sine negative. So

\begin{aligned} \cos 1{,}020^\circ &= \frac{1}{2} \\ \sin 1{,}020^\circ &= -\frac{\sqrt{3}}{2} \end{aligned}

Her hand is on the circle of radius 0.7, so its position is 0.7 times the terminal point:

\begin{aligned} &\left(0.7 \cdot \tfrac{1}{2},\ -0.7 \cdot \tfrac{\sqrt{3}}{2}\right) \\ &\quad = \left(0.35,\ -0.35\sqrt{3}\right) \end{aligned}

which is about (0.35, -0.61): 0.35 meters east and 0.61 meters south of the axis.

Example 3

Lwazi turns the other way

Lwazi spins \displaystyle \frac{17\pi}{6} radians clockwise. Find the cosine, sine and tangent of \displaystyle -\frac{17\pi}{6}.

Solution

Clockwise turns are negative, so the angle is \displaystyle -\frac{17\pi}{6}. A full turn is \displaystyle 2\pi = \frac{12\pi}{6}. One turn added gives \displaystyle -\frac{5\pi}{6}, still negative, so add a second:

\begin{aligned} \theta_0 &= -\frac{17\pi}{6} + 4\pi \\ &= -\frac{17\pi}{6} + \frac{24\pi}{6} = \frac{7\pi}{6} \end{aligned}

Since \displaystyle \pi = \frac{6\pi}{6} < \frac{7\pi}{6} < \frac{9\pi}{6} = \frac{3\pi}{2}, the angle is in Quadrant III. It has passed the negative x-axis by

\theta' = \frac{7\pi}{6} - \pi = \frac{\pi}{6}

the angle pictured in the first figure. From \displaystyle \frac{\pi}{6} the sizes are \displaystyle \frac{\sqrt{3}}{2} for cosine and \displaystyle \frac{1}{2} for sine. In Quadrant III both coordinates are negative, so

\begin{aligned} \cos\left(-\tfrac{17\pi}{6}\right) &= -\frac{\sqrt{3}}{2} \\ \sin\left(-\tfrac{17\pi}{6}\right) &= -\frac{1}{2} \end{aligned}

and the tangent, a negative over a negative, is positive:

\begin{aligned} \tan\left(-\tfrac{17\pi}{6}\right) &= \frac{-1/2}{-\sqrt{3}/2} \\ &= \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{aligned}

Example 4

An angle that is not special

Write \cos(-1{,}250^\circ) and \sin(-1{,}250^\circ) as a trig value of an acute angle, with a sign.

Solution

The angle is negative, so add turns. Three turns give -170^\circ, still negative; four turns, 1{,}440^\circ, give

\theta_0 = -1{,}250^\circ + 1{,}440^\circ = 190^\circ

in Quadrant III, with \theta' = 190^\circ - 180^\circ = 10^\circ. There is no exact value for 10^\circ to read off, and none is needed: the reference angle still fixes the size. Both coordinates are negative in Quadrant III, so

\begin{aligned} \cos(-1{,}250^\circ) &= -\cos 10^\circ \\ \sin(-1{,}250^\circ) &= -\sin 10^\circ \end{aligned}

A calculator in degree mode agrees: \cos(-1{,}250^\circ) \approx -0.9848 and \cos 10^\circ \approx 0.9848; \sin(-1{,}250^\circ) \approx -0.1736 and \sin 10^\circ \approx 0.1736.

Example 5

A reference angle in radians

Find the reference angle of 5 radians, and write \sin 5 and \cos 5 in terms of it.

Solution

No \pi appears, so compare 5 with decimal values of the quarter turns: \displaystyle \frac{3\pi}{2} \approx 4.712 and 2\pi \approx 6.283. Since 4.712 < 5 < 6.283, the angle is in Quadrant IV, already inside one turn. It is short of the full turn by

\theta' = 2\pi - 5 \approx 1.283

which is acute, as it must be: \displaystyle 1.283 < \frac{\pi}{2} \approx 1.571. In Quadrant IV the cosine is positive and the sine negative, so

\begin{aligned} \sin 5 &= -\sin(2\pi - 5) \\ \cos 5 &= \cos(2\pi - 5) \end{aligned}

In radian mode, \sin 5 \approx -0.9589 and \sin 1.283 \approx 0.9589, while \cos 5 \approx 0.2837. The reference angle of a radian angle is in radians; subtracting 5 from 180 would mix the units.

Example 6

Every angle with a given sine

Find every \theta with 0 \le \theta < 2\pi and \displaystyle \sin\theta = -\frac{\sqrt{3}}{2}.

Solution

The size of the sine is \displaystyle \frac{\sqrt{3}}{2}, and the acute angle with that sine is \displaystyle \frac{\pi}{3}. So every solution has reference angle \displaystyle \theta' = \frac{\pi}{3}. The sine is negative, so the terminal point is below the x-axis: Quadrant III or Quadrant IV. Reading the table backward, an angle in Quadrant III with reference angle \theta' is \pi + \theta', and one in Quadrant IV is 2\pi - \theta':

\begin{aligned} \theta &= \pi + \frac{\pi}{3} = \frac{4\pi}{3} \\ \theta &= 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \end{aligned}

Inside one turn there are exactly four angles with reference angle \displaystyle \frac{\pi}{3}, one per quadrant, and the sign keeps two of them. Solving equations this way over every turn is Solving trigonometric equations.

Example 7

Working backward to the size of a spin

Nandi spins counterclockwise more than two full turns but fewer than three. Her outstretched hand stops in Quadrant II, at a reference angle of 15^\circ. Through how many degrees did she spin?

Solution

Inside one turn, a Quadrant II angle sits \theta' short of the negative x-axis, so

\theta_0 = 180^\circ - 15^\circ = 165^\circ

Her spin is coterminal with 165^\circ, so it is 165^\circ plus a whole number of turns. Two turns (720^\circ) give 885^\circ, and three would give 1{,}245^\circ, more than three turns. Only one choice lies between 720^\circ and 1{,}080^\circ:

165^\circ + 720^\circ = 885^\circ

Check: 885^\circ - 720^\circ = 165^\circ, in Quadrant II, and 180^\circ - 165^\circ = 15^\circ.

Boundaries

Angles on an axis have no reference angle. When the terminal side lies on an axis, its angle with the x-axis is 0 or a right angle, not acute. Those angles need no bridge: the terminal point is (\pm 1, 0) or (0, \pm 1), read directly in Exact values on the unit circle.

The reference angle gives the size only. \theta and \theta' have equal cosines, sines and tangents only in Quadrant I. Elsewhere at least one value changes sign, and only the quadrant says which.

Always to the x-axis. The angle to the y-axis is the complement of \theta', and using it swaps the roles of sine and cosine.

Units stay matched. A radian angle has a radian reference angle, found with \pi and 2\pi; a degree angle uses 180^\circ and 360^\circ.

Where this goes next

Reference angles turn the four quadrants into one: every value on the unit circle is a first-quadrant value with a sign. The reflections behind that, across the axes and through the origin, are written as identities in Unit circle symmetry and even-odd identities, and the fact that adding a turn changes nothing is periodicity. Running the move backward, as in the last two examples, is how Solving trigonometric equations find every angle with a given value, and why an equation such as \sin\theta = 0.6 has two answers between 0^\circ and 180^\circ, the source of the two possible triangles in the ambiguous case.

Exercises

Reducing and measuring

For each angle, find the coterminal angle \theta_0 inside one turn, its quadrant, and its reference angle \theta'.

155^\circ
245^\circ
820^\circ
-130^\circ
-1{,}100^\circ
\displaystyle \frac{11\pi}{4}
\displaystyle -\frac{5\pi}{3}
\displaystyle \frac{23\pi}{6}
-3 radians (give \theta' exactly and to three decimal places)

Exact values

Find each value exactly.

\sin 1{,}110^\circ
\cos(-585^\circ)
\tan 945^\circ
\displaystyle \cos\frac{29\pi}{6}
\displaystyle \sin\left(-\frac{11\pi}{4}\right)
\displaystyle \tan\left(-\frac{13\pi}{3}\right)

Values of angles that are not special

Write each as a value of an acute angle with a sign, then check with a calculator.

\sin 2{,}000^\circ
\cos(-700^\circ)
\cos 4 (radians)

Spins and solutions

Lwazi spins 1{,}305^\circ clockwise, starting with his hand due east, 0.6 meters from his spin axis. Find the exact position of his hand, and round it to the nearest centimeter.
Find every \theta with 0^\circ \le \theta < 360^\circ and \tan\theta = -1.
Khanyisile spins counterclockwise more than three full turns but fewer than four, and stops in Quadrant IV at a reference angle of 70^\circ. Through how many degrees did she spin?

Reasoning

Explain why a reference angle is always strictly between 0^\circ and 90^\circ, and why an angle whose terminal side is on an axis has none.
Explain why \theta and -\theta always have the same reference angle.
Nandi says \cos 840^\circ = \cos 60^\circ, because 840^\circ has reference angle 60^\circ. What is right and what is wrong in her reasoning? Give the correct value.
Explain why a reference angle and a quadrant together fix exactly one angle \theta_0 with 0^\circ \le \theta_0 < 360^\circ.

Answers

\theta_0 = 155^\circ, Quadrant II, \theta' = 25^\circ. It is 180^\circ - 155^\circ short of the negative x-axis.
\theta_0 = 245^\circ, Quadrant III, \theta' = 65^\circ, from 245^\circ - 180^\circ.
\theta_0 = 100^\circ, Quadrant II, \theta' = 80^\circ. 820^\circ - 720^\circ = 100^\circ and 180^\circ - 100^\circ = 80^\circ.
\theta_0 = 230^\circ, Quadrant III, \theta' = 50^\circ. -130^\circ + 360^\circ = 230^\circ.
\theta_0 = 340^\circ, Quadrant IV, \theta' = 20^\circ. -1{,}100^\circ + 1{,}440^\circ = 340^\circ and 360^\circ - 340^\circ = 20^\circ.
\displaystyle \theta_0 = \frac{3\pi}{4}, Quadrant II, \displaystyle \theta' = \frac{\pi}{4}. \displaystyle \frac{11\pi}{4} - \frac{8\pi}{4} = \frac{3\pi}{4} and \displaystyle \pi - \frac{3\pi}{4} = \frac{\pi}{4}.
\displaystyle \theta_0 = \frac{\pi}{3}, Quadrant I, \displaystyle \theta' = \frac{\pi}{3}. \displaystyle -\frac{5\pi}{3} + \frac{6\pi}{3} = \frac{\pi}{3}.
\displaystyle \theta_0 = \frac{11\pi}{6}, Quadrant IV, \displaystyle \theta' = \frac{\pi}{6}. \displaystyle \frac{23\pi}{6} - \frac{12\pi}{6} = \frac{11\pi}{6} and \displaystyle 2\pi - \frac{11\pi}{6} = \frac{\pi}{6}.
\theta_0 = 2\pi - 3 \approx 3.283, Quadrant III, \theta' = \pi - 3 \approx 0.142. Since \displaystyle \pi < 3.283 < \frac{3\pi}{2}, subtract \pi: (2\pi - 3) - \pi = \pi - 3.
\displaystyle \frac{1}{2}. 1{,}110^\circ - 1{,}080^\circ = 30^\circ, in Quadrant I.
\displaystyle -\frac{\sqrt{2}}{2}. -585^\circ + 720^\circ = 135^\circ, Quadrant II, \theta' = 45^\circ; cosine is negative left of the y-axis.
1. 945^\circ - 720^\circ = 225^\circ, Quadrant III, \theta' = 45^\circ; both coordinates are negative, so the tangent is positive.
\displaystyle -\frac{\sqrt{3}}{2}. \displaystyle \frac{29\pi}{6} - 4\pi = \frac{5\pi}{6}, Quadrant II, \displaystyle \theta' = \frac{\pi}{6}.
\displaystyle -\frac{\sqrt{2}}{2}. \displaystyle -\frac{11\pi}{4} + 4\pi = \frac{5\pi}{4}, Quadrant III, \displaystyle \theta' = \frac{\pi}{4}; sine is negative below the x-axis.
\displaystyle -\sqrt{3}. \displaystyle -\frac{13\pi}{3} + 6\pi = \frac{5\pi}{3}, Quadrant IV, \displaystyle \theta' = \frac{\pi}{3}; x and y have opposite signs, so the tangent is negative.
\sin 2{,}000^\circ = -\sin 20^\circ \approx -0.342. 2{,}000^\circ - 1{,}800^\circ = 200^\circ, Quadrant III, \theta' = 20^\circ.
\cos(-700^\circ) = \cos 20^\circ \approx 0.940. -700^\circ + 720^\circ = 20^\circ, Quadrant I, so no sign changes.
\cos 4 = -\cos(4 - \pi) \approx -0.654. \pi \approx 3.142 < 4 < 4.712, so Quadrant III and \theta' = 4 - \pi \approx 0.858.
\displaystyle \left(-0.3\sqrt{2},\ 0.3\sqrt{2}\right) \approx (-0.42, 0.42). The angle is -1{,}305^\circ, and -1{,}305^\circ + 1{,}440^\circ = 135^\circ, Quadrant II with \theta' = 45^\circ; the terminal point is \displaystyle \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right), scaled by 0.6.
135^\circ and 315^\circ. |\tan\theta| = 1 gives \theta' = 45^\circ; tangent is negative where x and y have opposite signs, Quadrants II and IV: 180^\circ - 45^\circ and 360^\circ - 45^\circ.
1{,}370^\circ. In Quadrant IV, \theta_0 = 360^\circ - 70^\circ = 290^\circ, and three turns add 1{,}080^\circ: 290^\circ + 1{,}080^\circ = 1{,}370^\circ, which is between 1{,}080^\circ and 1{,}440^\circ.
A terminal side off the axes lies inside one quadrant, a wedge of 90^\circ bounded by a half of the x-axis and a half of the y-axis. Its angle to that half of the x-axis is more than 0^\circ (it is not on the x-axis) and less than 90^\circ (it is not on the y-axis). On an axis the angle to the x-axis is 0^\circ or 90^\circ, neither acute, so there is no reference angle.
-\theta is the same turn made clockwise, so its terminal side is the reflection of the terminal side of \theta across the x-axis: the terminal point (x, y) becomes (x, -y). Reflection across the x-axis keeps the angle with the x-axis, so the reference angle is unchanged.
The reference angle is right: 840^\circ - 720^\circ = 120^\circ and 180^\circ - 120^\circ = 60^\circ. The sign is wrong: 120^\circ is in Quadrant II, left of the y-axis, where cosine is negative. So \displaystyle \cos 840^\circ = -\frac{1}{2}, not \displaystyle \frac{1}{2}.
The rays at angle \theta' to the x-axis are four, one in each quadrant, at \theta', 180^\circ - \theta', 180^\circ + \theta' and 360^\circ - \theta'. These are different angles, since \theta' is strictly between 0^\circ and 90^\circ, so naming the quadrant picks exactly one.