Precalculus · F.TF.A.3 · F.TF.A.4

Unit circle symmetry and even-odd identities

Once you know where one angle lands on the unit circle, what do you already know about the angles that land on its mirror images?

You need

Trig functions on the unit circlePrecalculus Even and odd functionsAlgebra 2 Line symmetryGrade 4

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, every angle t in standard position got a terminal point (\cos t, \sin t) on the unit circle x^2 + y^2 = 1. That circle is an unusually symmetric figure. As in Even and odd functions, replacing x by -x, y by -y, or both, leaves its equation unchanged, so the circle is its own mirror image across the x-axis, across the y-axis, and through the origin (Line symmetry).

So each mirror image of a terminal point is again a point of the circle, and therefore the terminal point of some other angle. Knowing one point gives three more for free. Reading their coordinates turns the cosine or sine of a negative angle, or of an angle on the left half of the circle, into the cosine or sine of t itself, with at most a change of sign.

Definition

Let t be an angle in standard position with terminal point P = (x, y) = (\cos t, \sin t). The three mirror images of P are:

Move	Image of (x, y)	Terminal point of	In degrees
reflect across the x-axis	(x, -y)	-t	-\theta
reflect across the y-axis	(-x, y)	\pi - t	180^\circ - \theta
half turn about the origin	(-x, -y)	\pi + t	180^\circ + \theta

A function f is even when f(-t) = f(t) for every t in its domain and odd when f(-t) = -f(t), exactly as in Even and odd functions. The even and odd identities below say which of the six trig functions are which.

Intuition

Figure 1The three mirror images of the terminal point of t. Each one makes the same angle t with the x-axis, so its coordinates differ from (\cos t, \sin t) only in sign.

Picture a weathervane seen from above, with the pivot at the origin, east along the positive x-axis and north along the positive y-axis. Say the tip of its arrow is one foot from the pivot, so when the arrow makes the angle t counterclockwise from east, the tip is at (\cos t, \sin t).

If the wind swings the arrow to -t, it has turned the same amount clockwise instead of counterclockwise. The tip is as far east as before, and as far south as it used to be north. If the arrow points to \pi - t, it makes the angle t with west instead of east: the tip is as far north as before, now on the west side. If it points to \pi + t, it points exactly the opposite way, and both readings flip.

In every case the image sits at the same distance from each axis as P; only the side changes. So the four angles t, -t, \pi - t, \pi + t have cosines of one size and sines of one size, and which sign each gets is read off the side of the axis its point sits on.

Results

Negative angles (reflection across the x-axis).

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

Even and odd identities. Cosine and secant are even; sine, tangent, cosecant and cotangent are odd:

\begin{aligned} \sec(-t) &= \sec t \\ \csc(-t) &= -\csc t \\ \cot(-t) &= -\cot t \end{aligned}

Reflection across the y-axis.

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

Half turn about the origin.

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

In degrees, write 180^\circ for \pi. The rules for sine and cosine hold for every real t; the others hold wherever both sides are defined.

Why it is true

Across the x-axis. The map (x, y) \to (x, -y) keeps every point on the unit circle, since x^2 + (-y)^2 = x^2 + y^2 = 1. It leaves the positive x-axis, where every angle starts, in place, and it reverses the direction of turning. So the ray at angle t counterclockwise from the positive x-axis goes to the ray at angle t clockwise from it, which is the angle -t. In radians the same thing reads as arc length: the arc from (1, 0) to P has length t, and its mirror image is an arc of the same length run the other way. Hence the terminal point of -t is (\cos t, -\sin t), and reading its coordinates gives \cos(-t) = \cos t and \sin(-t) = -\sin t.

Across the y-axis. The map (x, y) \to (-x, y) also keeps the circle. It sends the positive x-axis to the negative x-axis, the ray at angle \pi, and reverses turning. The ray t counterclockwise from the positive x-axis goes to the ray t clockwise from the angle \pi, which is the angle \pi - t. Its terminal point is (-\cos t, \sin t).

The half turn. The map (x, y) \to (-x, -y) is one reflection followed by the other, and it is a rotation by \pi about the origin. A rotation by \pi adds \pi to every angle, so the terminal point of \pi + t is (-\cos t, -\sin t).

The other four functions. Each is built from sine and cosine, so its rule follows by substitution:

\begin{aligned} \tan(-t) &= \frac{\sin(-t)}{\cos(-t)} = \frac{-\sin t}{\cos t} = -\tan t \\ \sec(-t) &= \frac{1}{\cos(-t)} = \frac{1}{\cos t} = \sec t \end{aligned}

and in the same way \displaystyle \csc(-t) = \frac{1}{-\sin t} = -\csc t and \displaystyle \cot(-t) = \frac{\cos t}{-\sin t} = -\cot t. For the tangent rules with \pi \pm t: \displaystyle \tan(\pi - t) = \frac{\sin t}{-\cos t} = -\tan t and \displaystyle \tan(\pi + t) = \frac{-\sin t}{-\cos t} = \tan t. Since P and its images have their zero coordinates in the same places, each side is undefined exactly when the other is.

Worked examples

Example 1

Three mirror images of one point

The terminal point of an angle t is \displaystyle \left(\frac{20}{101}, \frac{99}{101}\right). Find the terminal points of -t, \pi - t and \pi + t, and the cosine, sine and tangent of all four angles.

Solution

First, the point is on the unit circle: 20^2 + 99^2 = 400 + 9{,}801 = 10{,}201 = 101^2, so the squares of its coordinates add to 1. Each image comes from a change of sign, so the terminal points of -t, \pi - t and \pi + t are

\begin{aligned} &\left(\frac{20}{101}, -\frac{99}{101}\right),\ \left(-\frac{20}{101}, \frac{99}{101}\right), \\ &\left(-\frac{20}{101}, -\frac{99}{101}\right) \end{aligned}

Figure 2Example 1. The terminal points of t (upper right), \pi - t (upper left), \pi + t (lower left) and -t (lower right).

The cosine is the x-coordinate, the sine the y-coordinate, and the tangent is y over x, so \displaystyle \tan t = \frac{99/101}{20/101} = \frac{99}{20}.

Angle	\cos	\sin	\tan
t	\displaystyle \frac{20}{101}	\displaystyle \frac{99}{101}	\displaystyle \frac{99}{20}
-t	\displaystyle \frac{20}{101}	\displaystyle -\frac{99}{101}	\displaystyle -\frac{99}{20}
\pi - t	\displaystyle -\frac{20}{101}	\displaystyle \frac{99}{101}	\displaystyle -\frac{99}{20}
\pi + t	\displaystyle -\frac{20}{101}	\displaystyle -\frac{99}{101}	\displaystyle \frac{99}{20}

One sign change flips the tangent; two cancel, so \pi + t has the same tangent as t. That is the half turn keeping the terminal side on the same line through the origin, and a line has one slope.

Example 2

All six functions of a negative angle

An angle t has \displaystyle \cos t = -\frac{36}{85} and \displaystyle \sin t = -\frac{77}{85}. Find all six trig functions of -t.

Solution

Cosine is even and sine is odd, so \displaystyle \cos(-t) = \cos t = -\frac{36}{85} and \displaystyle \sin(-t) = -\sin t = \frac{77}{85}. The other four are built from these two, as in secant, cosecant and cotangent on the unit circle:

\begin{aligned} \tan(-t) &= \frac{77/85}{-36/85} = -\frac{77}{36} \\ \sec(-t) &= \frac{1}{\cos(-t)} = -\frac{85}{36} \\ \csc(-t) &= \frac{1}{\sin(-t)} = \frac{85}{77} \\ \cot(-t) &= \frac{\cos(-t)}{\sin(-t)} = -\frac{36}{77} \end{aligned}

Check against the identities: \displaystyle \tan t = \frac{-77/85}{-36/85} = \frac{77}{36}, and \tan(-t) is its opposite, as an odd function requires; \displaystyle \sec t = -\frac{85}{36} equals \sec(-t), as an even one requires. The point (\cos t, \sin t) is in Quadrant III, and its mirror image across the x-axis is in Quadrant II, where cosine is negative and sine positive, matching the signs found.

Example 3

Simplifying with the symmetry rules

Simplify (a) \sin(-t)\cos(-t) + \sin t\cos t; (b) \cos(\pi + t) - \cos(-t); (c) \sin(t - \pi).

Solution

(a) Sine is odd and cosine is even, so the first product is (-\sin t)(\cos t), and

-\sin t\cos t + \sin t\cos t = 0

(b) The half turn gives \cos(\pi + t) = -\cos t, and \cos(-t) = \cos t, so the expression is -\cos t - \cos t = -2\cos t.

\sin(t - \pi) = -\sin(\pi - t) = -\sin t

Check with \displaystyle t = \frac{\pi}{2}: \displaystyle \sin\left(-\frac{\pi}{2}\right) = -1 = -\sin\frac{\pi}{2}. The angle t - \pi is also coterminal with t + \pi, and the half-turn rule gives the same answer.

Example 4

Other angles with the same sine, cosine or tangent

Let t = 0.9. Between 0 and 2\pi, find the other angle with the same sine as t, the other with the same cosine, and the other with the same tangent, to three decimal places.

Solution

Since \displaystyle 0 < 0.9 < \frac{\pi}{2} \approx 1.571, the terminal point P is in Quadrant I. Points of the circle with the same sine have the same y-coordinate, so they lie on the horizontal line through P. That line meets the circle at P and at its mirror image across the y-axis, the terminal point of

\pi - 0.9 \approx 2.242

Points with the same cosine lie on the vertical line through P, which meets the circle at P and at its mirror image across the x-axis, the terminal point of -0.9. That angle is negative; adding a full turn gives a coterminal angle in range, 2\pi - 0.9 \approx 5.383.

Points with the same tangent lie on the line through the origin and P, which meets the circle at P and its half-turn image, the terminal point of \pi + 0.9 \approx 4.042. Each line meets the circle in exactly two points here, so each answer is the only other one. Check: \sin 2.242 \approx 0.783 \approx \sin 0.9.

Example 5

Svetozar’s weathervane

Svetozar’s weathervane arrow is 1.3 feet from pivot to tip. Seen from above, with the pivot at the origin, east along the positive x-axis and north along the positive y-axis, the tip is at (0.5, 1.2) at noon, when the arrow makes the angle t. Where is the tip when the arrow points at -t, at \pi - t, and at \pi + t? Which of these points the arrow exactly opposite to its noon direction?

Solution

On a circle of radius r centered at the origin, the point at angle t is (r\cos t, r\sin t). Here r = 1.3, \displaystyle \cos t = \frac{0.5}{1.3} = \frac{5}{13} and \displaystyle \sin t = \frac{1.2}{1.3} = \frac{12}{13}. At -t the tip is

\begin{aligned} &\big(1.3\cos(-t),\ 1.3\sin(-t)\big) \\ &\quad = (1.3\cos t,\ -1.3\sin t) = (0.5, -1.2) \end{aligned}

In the same way, at \pi - t it is (-0.5, 1.2) and at \pi + t it is (-0.5, -1.2). Every coordinate is 1.3 times a unit circle coordinate, so the sign rules carry over unchanged to a circle of any radius centered at the origin. The angle \pi + t is a half turn from t, so that arrow points exactly opposite: at noon the tip is east and north of the pivot, and then it is west and south.

Example 6

Even, odd or neither

Decide whether each function is even, odd or neither: f(t) = t^2 + \cos t, g(t) = \sin t + \tan t, h(t) = \cos t + \sin t.

Solution

f is defined for every t, and f(-t) = (-t)^2 + \cos(-t) = t^2 + \cos t = f(t), so f is even.

g is defined where \cos t \ne 0; that domain is symmetric, because \cos(-t) = \cos t. Then g(-t) = -\sin t - \tan t = -g(t), so g is odd.

h(-t) = \cos t - \sin t, which matches neither h(t) nor -h(t). Single inputs prove it. \displaystyle h\left(\frac{\pi}{2}\right) = 0 + 1 = 1 but \displaystyle h\left(-\frac{\pi}{2}\right) = 0 - 1 = -1, so h is not even. An odd function defined at 0 has h(0) = -h(0), so h(0) = 0; here h(0) = 1, so h is not odd. It is neither: an even function plus an odd one, as with x^2 + x.

Boundaries

Only these three mirrors give pure sign changes. The unit circle is symmetric about every line through the origin, but reflecting across y = x swaps the coordinates, (x, y) \to (y, x), and lands on the terminal point of \displaystyle \frac{\pi}{2} - t. That leads to the Cofunction identities, not to a sign rule. Other lines mix the coordinates.
The circle must be centered at the origin. A circle centered elsewhere is not carried onto itself by reflection across the axes.
The tangent, secant, cosecant and cotangent rules hold only where defined. At \displaystyle t = \frac{\pi}{2}, \tan(-t) = -\tan t says nothing: both sides are undefined.
The units must match: \pi - t for t in radians, 180^\circ - \theta for \theta in degrees. \cos(180 - 0.5) is not -\cos 0.5.

Where this goes next

The mirror images say that every angle shares the size of its cosine and sine with an angle whose terminal point is in Quadrant I. Reference angles turns that into a way to evaluate any angle, and Signs of the trig functions by quadrant reads the signs. The other symmetry of the circle is the full turn, in Periodicity and the periodicity identities.

On the graphs, \cos(-t) = \cos t makes the cosine curve symmetric about the vertical axis and \sin(-t) = -\sin t makes the sine curve symmetric about the origin (Graphing sine and cosine). The even and odd identities are tools in Trigonometric identities and Verifying identities, and Sine and tangent of a sum or difference uses them to get a difference formula from a sum formula. Polar circles and symmetry tests runs the same three symmetry tests on polar equations.

Exercises

Mirror images of a terminal point

The terminal point of t is \displaystyle \left(-\frac{33}{65}, \frac{56}{65}\right). Find the terminal points of -t, \pi - t and \pi + t.
The terminal point of t is \displaystyle \left(-\frac{44}{125}, \frac{117}{125}\right). Find \cos(\pi - t), \sin(\pi + t) and \tan(-t).
Given \cos 25^\circ \approx 0.9063 and \sin 25^\circ \approx 0.4226, find the cosine and sine of -25^\circ, 155^\circ and 205^\circ.
Given \sin 1.2 \approx 0.9320 and \cos 1.2 \approx 0.3624, find \sin(\pi - 1.2), \cos(\pi + 1.2), \sin(-1.2) and \cos(-1.2).

Even and odd identities

Find (a) \csc(-t) if \displaystyle \csc t = -\frac{17}{8}; (b) \sec(-t) if \sec t = 3.4; (c) \cot(-t) if \cot t = 0.75.
An angle t has \displaystyle \cos t = \frac{20}{29} and \displaystyle \sin t = -\frac{21}{29}. Find all six trig functions of -t.
Simplify \dfrac{\sin(-t)}{\cos(-t)} + \tan t.
Simplify \cos(-t)\tan(-t).
Simplify \sin(\pi - t) - \sin(\pi + t).
Simplify \cos(t - \pi).

Other angles with the same value

Let t = 2. Between 0 and 2\pi, find the other angle with the same sine, the other with the same cosine, and the other with the same tangent, to three decimal places.
Between 0^\circ and 360^\circ, find the other angle with the same sine as 230^\circ, the other with the same cosine, and the other with the same tangent.

The weathervane

Branimir’s weathervane arrow is 1.7 feet long, and at angle t its tip is at (-0.8, 1.5), with axes as in Example 5. Where is the tip at -t, at \pi - t and at \pi + t?
Ljiljana’s arrow is 2 feet long. At angle t, \sin t = 0.28 and the tip is west of the pivot. Find the tip at t, at \pi - t and at -t.

Even, odd or neither

Is f(t) = t\cos t even, odd or neither?
Is g(t) = \sin t\tan t even, odd or neither?
Is h(t) = 1 + \sin t even, odd or neither?

Reasoning

Explain, using arc lengths, why the terminal point of \pi - t is the mirror image of the terminal point of t across the y-axis.
Write \pi + t as \pi - (-t) and use the rules for -t and \pi - t to derive \cos(\pi + t) = -\cos t and \sin(\pi + t) = -\sin t. What does this say about doing the two reflections one after the other?
Prove that secant is even and cosecant is odd, and say for which t each identity holds.
Svetozar says \cos(\pi - t) = \cos(\pi + t) and \sin(\pi - t) = \sin(\pi + t) for every t. Which claim is right? Explain with the figure.
Ljiljana notes that the unit circle is also symmetric about the line y = x. Explain why reflecting across that line does not give a sign rule like the three in this lesson.

Answers

-t: \displaystyle \left(-\frac{33}{65}, -\frac{56}{65}\right); \pi - t: \displaystyle \left(\frac{33}{65}, \frac{56}{65}\right); \pi + t: \displaystyle \left(\frac{33}{65}, -\frac{56}{65}\right). Change the sign of y, of x, and of both.
\displaystyle \cos(\pi - t) = -\cos t = \frac{44}{125}; \displaystyle \sin(\pi + t) = -\sin t = -\frac{117}{125}; \displaystyle \tan t = \frac{117/125}{-44/125} = -\frac{117}{44}, so \displaystyle \tan(-t) = \frac{117}{44}.
155^\circ = 180^\circ - 25^\circ and 205^\circ = 180^\circ + 25^\circ. \cos(-25^\circ) \approx 0.9063, \sin(-25^\circ) \approx -0.4226; \cos 155^\circ \approx -0.9063, \sin 155^\circ \approx 0.4226; \cos 205^\circ \approx -0.9063, \sin 205^\circ \approx -0.4226.
\sin(\pi - 1.2) \approx 0.9320; \cos(\pi + 1.2) \approx -0.3624; \sin(-1.2) \approx -0.9320; \cos(-1.2) \approx 0.3624.
(a) Cosecant is odd: \displaystyle \csc(-t) = \frac{17}{8}. (b) Secant is even: \sec(-t) = 3.4. (c) Cotangent is odd: \cot(-t) = -0.75.
\displaystyle \cos(-t) = \frac{20}{29}, \displaystyle \sin(-t) = \frac{21}{29}, \displaystyle \tan(-t) = \frac{21}{20}, \displaystyle \sec(-t) = \frac{29}{20}, \displaystyle \csc(-t) = \frac{29}{21}, \displaystyle \cot(-t) = \frac{20}{21}. The first two from even and odd, the rest from their definitions.
0. The quotient is \tan(-t) = -\tan t, and -\tan t + \tan t = 0, wherever \cos t \ne 0.
-\sin t. It is \displaystyle \cos t \cdot(-\tan t) = -\cos t \cdot \frac{\sin t}{\cos t}, wherever \cos t \ne 0.
2\sin t. It is \sin t - (-\sin t).
-\cos t. Cosine is even, so \cos(t - \pi) = \cos(\pi - t) = -\cos t.
t = 2 is in Quadrant II. Same sine: \pi - 2 \approx 1.142; same cosine: 2\pi - 2 \approx 4.283; same tangent: \pi + 2 \approx 5.142.
Same sine: 180^\circ - 230^\circ = -50^\circ, coterminal with 310^\circ. Same cosine: -230^\circ, coterminal with 130^\circ. Same tangent: 180^\circ + 230^\circ = 410^\circ, coterminal with 50^\circ.
-t: (-0.8, -1.5); \pi - t: (0.8, 1.5); \pi + t: (0.8, -1.5). The tip is (1.7\cos t, 1.7\sin t) and the signs change as on the unit circle.
\displaystyle \cos t = -\sqrt{1 - 0.28^2} = -0.96, negative because the tip is west. Tip at t: (-1.92, 0.56); at \pi - t: (1.92, 0.56); at -t: (-1.92, -0.56).
Odd: f(-t) = (-t)\cos(-t) = -t\cos t = -f(t).
Even: the domain \cos t \ne 0 is symmetric, and g(-t) = (-\sin t)(-\tan t) = g(t).
Neither: \displaystyle h\left(\frac{\pi}{2}\right) = 2 and \displaystyle h\left(-\frac{\pi}{2}\right) = 0, which is neither 2 nor -2.
The arc from (1, 0) counterclockwise to P has length t. Reflecting across the y-axis carries (1, 0) to (-1, 0) and that arc to an arc of length t running clockwise from (-1, 0) to the image. The point (-1, 0) is \pi along the circle, so the image is \pi - t along it: the terminal point of \pi - t.
\cos(\pi - (-t)) = -\cos(-t) = -\cos t and \sin(\pi - (-t)) = \sin(-t) = -\sin t. Reflecting across the x-axis and then the y-axis sends (x, y) to (x, -y) and then to (-x, -y): the two reflections together are the half turn.
\displaystyle \sec(-t) = \frac{1}{\cos(-t)} = \frac{1}{\cos t} = \sec t for every t with \cos t \ne 0; \displaystyle \csc(-t) = \frac{1}{\sin(-t)} = -\frac{1}{\sin t} = -\csc t for every t with \sin t \ne 0. In each case t and -t are in the domain together.
The cosine claim is right: both equal -\cos t, since both images lie left of the y-axis on one vertical line. The sine claim is wrong: \sin(\pi - t) = \sin t but \sin(\pi + t) = -\sin t, since the two images are mirror images across the x-axis. At \displaystyle t = \frac{\pi}{2} they are 1 and -1.
Reflecting across y = x sends (x, y) to (y, x): it swaps the coordinates instead of changing a sign, landing on the terminal point of \displaystyle \frac{\pi}{2} - t. It gives \displaystyle \cos\left(\frac{\pi}{2} - t\right) = \sin t, a cofunction identity, not a sign rule.