Use the unit circle to explain the values at π − x, π + x and 2π − x, odd and even symmetry, and periodicity

Learning goal: Use the unit circle to express the values of sine, cosine and tangent for π − x, π + x and 2π − x in terms of their values for x, where x is any real number; explain why cosine is an even function and sine and tangent are odd functions; and explain the periodicity of the three functions.

Scope note. F.TF.3 and F.TF.4 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

The three reflections of a point. Let x be any real number and let P(x) = (cos x, sin x) be its point on the unit circle. The unit circle has three symmetries that send P(x) to another point on the circle: reflection across the y-axis, reflection through the origin, and reflection across the x-axis. Each reflection changes the coordinates in a predictable way, and each corresponds to an angle built from x.

π − x: reflection across the y-axis. Travel x counterclockwise from (1, 0) to reach P(x). Now travel π − x counterclockwise: that is the same as traveling π (a half turn, ending at (−1, 0)) and then coming back x clockwise. The point you reach is the mirror image of P(x) across the y-axis, because it is the same distance x along the circle from (−1, 0) as P(x) is from (1, 0). Reflecting across the y-axis changes the sign of the x-coordinate only. So cos(π − x) = −cos x, sin(π − x) = sin x, and dividing, tan(π − x) = −tan x wherever tan x is defined. Example: sin(5π/6) = sin(π − π/6) = sin(π/6) = 1/2 and cos(5π/6) = −√3/2.

π + x: reflection through the origin. Traveling π + x is a half turn followed by a further x. A half turn about the origin sends every point (a, b) to (−a, −b), so P(π + x) is the point diametrically opposite P(x). Both coordinates change sign: cos(π + x) = −cos x, sin(π + x) = −sin x, and the quotient of two negatives is positive: tan(π + x) = tan x wherever tan x is defined. Example: cos(4π/3) = cos(π + π/3) = −cos(π/3) = −1/2, sin(4π/3) = −√3/2, and tan(4π/3) = tan(π/3) = √3.

2π − x: reflection across the x-axis. Traveling 2π − x is a full turn (back to (1, 0)) and then x clockwise. Traveling x clockwise from (1, 0) reaches the mirror image of P(x) across the x-axis, since the two points are the same arc length from (1, 0) on opposite sides of the axis. Reflecting across the x-axis changes the sign of the y-coordinate only: cos(2π − x) = cos x, sin(2π − x) = −sin x, tan(2π − x) = −tan x (the tangent rules hold wherever tan x is defined). Example: tan(7π/4) = tan(2π − π/4) = −tan(π/4) = −1.

A fully worked model. Find sin, cos and tan of 7π/6 exactly. Write 7π/6 = π + π/6, so this is the "π + x" case with x = π/6. From the special triangle, half of an equilateral triangle with side 2 (hypotenuse 2, short leg 1 opposite the π/6 angle, long leg √(2² − 1²) = √3), cos(π/6) = √3/2 and sin(π/6) = 1/2; the same triangle gives the values at π/3, and the isosceles right triangle with legs 1 and hypotenuse √2 gives cos(π/4) = sin(π/4) = √2/2. Reflecting through the origin, cos(7π/6) = −√3/2, sin(7π/6) = −1/2 and tan(7π/6) = (−1/2)/(−√3/2) = 1/√3 = √3/3. Check the quadrant: 7π/6 is between π and 3π/2, quadrant III, where both coordinates are negative and tangent is positive. Check with a calculator in radian mode: sin(3.6652) ≈ −0.5, cos ≈ −0.866, tan ≈ 0.577. Everything agrees.

Odd and even functions. The point P(−x) is reached by traveling x clockwise, which as we saw is the reflection of P(x) across the x-axis. That is the same point as P(2π − x), so cos(−x) = cos x and sin(−x) = −sin x. A function f with f(−x) = f(x) for every x is called even: its graph is symmetric across the vertical axis. A function with f(−x) = −f(x) is called odd: its graph is symmetric through the origin. Cosine is even, sine is odd, and tangent, being an odd function divided by an even one, is odd: tan(−x) = −tan x. Examples: cos(−π/3) = cos(π/3) = 1/2; sin(−π/4) = −sin(π/4) = −√2/2; tan(−π/3) = −√3.

Periodicity. Traveling 2π along the unit circle is exactly one full turn, so P(x + 2π) = P(x) for every real x, and therefore sin(x + 2π) = sin x and cos(x + 2π) = cos x. The same holds for any whole number of turns: sin(x + 2πk) = sin x for every integer k. A function that repeats in this way is periodic, and the smallest positive repeat distance is its period. The period of sine is 2π, not smaller: if p were a period, sin(π/2 + p) would equal sin(π/2) = 1, but sine equals 1 only at π/2 + 2πk, so p must be a whole number of full turns. The same argument with cosine's maximum, cos 0 = 1, gives period 2π for cosine. Tangent repeats sooner: P(x + π) is the reflection of P(x) through the origin, both coordinates change sign, and their quotient does not, so tan(x + π) = tan x. No smaller positive p works: tan(0 + p) would have to equal tan 0 = 0, and tangent is 0 only at whole multiples of π, so the period of tangent is π. Examples: sin(13π/6) = sin(π/6 + 2π) = 1/2; cos(9π/4) = cos(π/4 + 2π) = √2/2; tan(5π/4) = tan(π/4 + π) = 1.

Combining the rules. Any input can be reduced to a reference angle in [0, π/2] by first subtracting full turns and then applying one of the three reflections. For 25π/6: subtract 4π = 24π/6 to get π/6, so sin(25π/6) = sin(π/6) = 1/2. For −7π/6: cosine is even, so cos(−7π/6) = cos(7π/6) = −cos(π/6) = −√3/2. Each step is a symmetry of the circle, which is why the answer never depends on which order you apply them in as long as each step is correct.

The misconception. Students often guess that sin(π − x) = −sin x "because there is a minus sign." The sign inside the argument describes a position on the circle, not the sign of the output. The reliable procedure is to locate the point: π − x is in quadrant II when x is acute, where sine is positive, so the sine cannot change sign. Whenever a rule is in doubt, draw the circle, mark P(x), apply the reflection, and read the coordinates.

Worked, π − x: Find cos(3π/4) and sin(3π/4) exactly. 3π/4 = π − π/4, reflection of P(π/4) = (√2/2, √2/2) across the y-axis. So cos(3π/4) = −√2/2 ≈ −0.707 and sin(3π/4) = √2/2 ≈ 0.707. Then tan(3π/4) = −1. Quadrant II check: x negative, y positive, tangent negative.

Worked, 2π − x: Find sin(11π/6), cos(11π/6) and tan(11π/6). 11π/6 = 2π − π/6, the reflection of P(π/6) = (√3/2, 1/2) across the x-axis. So cos(11π/6) = √3/2, sin(11π/6) = −1/2 and tan(11π/6) = (−1/2)/(√3/2) = −√3/3 ≈ −0.577. Quadrant IV check: x positive, y negative, tangent negative.

Worked, odd and even with a large input: Evaluate cos(−11π/3) and sin(−11π/3). Cosine is even and sine is odd, so cos(−11π/3) = cos(11π/3) and sin(−11π/3) = −sin(11π/3). Now 11π/3 = 4π − π/3, and subtracting the two full turns leaves −π/3, so cos(11π/3) = cos(−π/3) = cos(π/3) = 1/2 and sin(11π/3) = sin(−π/3) = −√3/2. Therefore cos(−11π/3) = 1/2 and sin(−11π/3) = −(−√3/2) = √3/2. Calculator check: cos(−11.519) ≈ 0.5, sin(−11.519) ≈ 0.866.

Worked, why the period of tangent is π: Show tan(x + π) = tan x from the unit circle, then find tan(10π/3). P(x + π) is P(x) reflected through the origin, so cos(x + π) = −cos x and sin(x + π) = −sin x; dividing, tan(x + π) = (−sin x)/(−cos x) = tan x. For 10π/3, subtract 3π: 10π/3 − 9π/3 = π/3, and three periods of tangent leave the value unchanged, so tan(10π/3) = tan(π/3) = √3 ≈ 1.732. Note that sin(10π/3) is not sin(π/3): 10π/3 = 2π + 4π/3, and sin(4π/3) = −√3/2. Only tangent has period π.

Representation task: Draw a large unit circle and mark P(π/6) = (√3/2, 1/2) ≈ (0.866, 0.5). Reflect it across the y-axis, through the origin and across the x-axis, and label the three images with their angle names 5π/6, 7π/6 and 11π/6 and their coordinates (−0.866, 0.5), (−0.866, −0.5) and (0.866, −0.5). Record a table with rows for x, π − x, π + x, 2π − x and −x listing cosine, sine, tangent and the symmetry used. Then, on a separate axis from −2π to 4π, sketch y = sin x from the coordinates you read off and mark two places where the graph repeats exactly 2π apart.

Independent performance: Write a one-page explanation, with diagrams, of why cos is even, sin and tan are odd, and why sin and cos have period 2π while tan has period π. Every claim must be justified by a reflection or rotation of the unit circle, not by "the calculator says so." Then evaluate, exactly and with the symmetry named at each step, sin(17π/4), cos(−5π/4), tan(−7π/6) and cos(2π − 1) in terms of cos 1. Success criteria: each of the four evaluations shows the reduction (subtract full turns, then apply one reflection), gives the correct sign, and matches a calculator check to three decimal places; and the explanation states the definitions of even, odd and period correctly.