HSF.TF.A.3: Special Triangles and Unit Circle Symmetry for Sine, Cosine and Tangent
In plain English: HSF.TF.A.3 is an advanced (+) Common Core functions standard, usually taught in Precalculus. Students use the 30-60-90 and 45-45-90 triangles to find exact sine, cosine and tangent values for π/6, π/4 and π/3, then use unit circle symmetry to write the values for π - x, π + x and 2π - x in terms of the values for x, for any real number x.
(+) Use special triangles to determine geometrically the values of sine, cosine, tangent for π/3, π/4 and π/6, and 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.
Common Core State Standards for Mathematics · Domain: Trigonometric Functions (TF) · Cluster: Extend the domain of trigonometric functions using the unit circle Also written as HSF-TF.A.3 or F-TF.3 · Official standard
This lesson has two halves that match the two halves of the standard. First, students cut an equilateral triangle and a square along a line of symmetry to get the 30-60-90 and 45-45-90 triangles, and read exact values of sine, cosine and tangent for π/6, π/4 and π/3 from the side lengths. The values are found geometrically, from the triangles, not from a memorized chart.
Second, students use the symmetry of the unit circle. If P(x) = (a, b), then reflecting across the y-axis gives P(π - x) = (-a, b), a half turn about the origin gives P(π + x) = (-a, -b), and reflecting across the x-axis gives P(2π - x) = (a, -b). Those three facts turn into nine rules, such as cos(π - x) = -cos x, that hold for every real number x, including inputs that are not special angles and inputs that are not acute.
Learning Objectives
By the end of this lesson, students will be able to:
Derive the side ratios of 30-60-90 and 45-45-90 triangles from an equilateral triangle and a square
Use those triangles to find exact values of sine, cosine and tangent for π/6, π/4 and π/3
Use reflections and a half turn of the unit circle to express sine, cosine and tangent of π - x, π + x and 2π - x in terms of their values at x
Apply the rules to special angles and to any real number x, and check them with a calculator
Prior Knowledge Required
Students should already be comfortable with:
Right-triangle trigonometric ratios HSG.SRT.C.6
The Pythagorean Theorem 8.G.B.7
Sine and cosine as coordinates on the unit circle for any real input HSF.TF.A.2
Reflections and rotations in the coordinate plane HSG.CO.A.5
Simplifying radicals and rationalizing denominators
Hand out a paper equilateral triangle and a paper square. Students fold each along a line of symmetry and answer the prompt.
Warm-Up Prompt
"An equilateral triangle has side 2. Fold it in half along an altitude. What are the angles and side lengths of each half? A square has side 1. Fold it along a diagonal. What are the angles and side lengths of each half? Give the angles in radians."
Students should find a triangle with angles π/6, π/3 and π/2 and sides 1, √3 and 2 (the altitude is √(2² - 1²) = √3), and a triangle with angles π/4, π/4 and π/2 and sides 1, 1 and √2. Record both on the board; they are the only two facts needed for the first half of the lesson.
Direct Instruction25 minutes
Part 1: Values from special triangles. Using Diagram 1, apply sin = opposite/hypotenuse, cos = adjacent/hypotenuse and tan = opposite/adjacent to each triangle. Point out that any triangle similar to these gives the same ratios, so the side lengths chosen do not matter.
π/3 from the half equilateral triangle
The angle π/3 is at a base vertex: opposite side √3, adjacent side 1, hypotenuse 2.
Part 2: Symmetry on the unit circle. Show Diagram 2 with a point P(x) = (a, b) that is not at a special angle. Explain each move with arc lengths:
π - x: go to (-1, 0), then back x units. The arc from (-1, 0) matches the arc from (1, 0) in a mirror, so P(π - x) is the reflection of P(x) across the y-axis: (-a, b). So sin(π - x) = sin x, cos(π - x) = -cos x and tan(π - x) = -tan x.
π + x: go to (-1, 0), then x more units. That is P(x) turned a half turn about the origin: (-a, -b). So sin(π + x) = -sin x, cos(π + x) = -cos x and tan(π + x) = (-b)/(-a) = tan x.
2π - x: go a full turn, then back x units, the same as going x units clockwise. That is the reflection across the x-axis: (a, -b). So sin(2π - x) = -sin x, cos(2π - x) = cos x and tan(2π - x) = -tan x.
Any real x: none of these steps needed x to be small or positive. The reflections and the half turn move every point of the circle, so the rules hold for every real x (for tangent, wherever tan x is defined).
Finish with the third rule applied to a special angle: 5π/3 = 2π - π/3, so sin(5π/3) = -sin(π/3) = -√3/2, cos(5π/3) = cos(π/3) = 1/2 and tan(5π/3) = -tan(π/3) = -√3.
Guided Practice15 minutes
Pairs name the form of each input (π - x, π + x or 2π - x), sketch the reflection or half turn, then write the value. (1) All three functions at 2π/3 = π - π/3: sin = √3/2, cos = -1/2, tan = -√3. (2) Given sin 0.4 ≈ 0.389 and cos 0.4 ≈ 0.921, find sin(π - 0.4), cos(π + 0.4) and tan(2π - 0.4): about 0.389, -0.921 and -0.423. (3) The rules also work when x is not acute: with x = 2, cos(π - 2) = -cos 2 ≈ 0.416. Pairs confirm each decimal with a calculator in radian mode. Watch for students who change the sign of sine for π - x, or who decide the sign from the rule without checking the quadrant of the new point.
Independent Practice15 minutes
Students work alone. Find exact sine, cosine and tangent at 7π/6 = π + π/6 (-1/2, -√3/2 and √3/3) and at 7π/4 = 2π - π/4 (-√2/2, √2/2 and -1). Then show that cos(π - x) + cos(2π - x) = 0 for every real x, using the rules, and test the identity with x = 1.5 on a calculator. Students write which symmetry they used next to every answer.
Closure5-10 minutes
Exit ticket: (1) Write cos(2π - x) in terms of cos x, and name the symmetry. (cos x; reflection across the x-axis.) (2) Explain with the unit circle why sin(π - x) = sin x. (Reflecting across the y-axis changes only the x-coordinate, so the y-coordinate stays the same.) (3) Suppose sin x = 0.28 and cos x = 0.96. Give sin(π + x) and cos(π - x). (-0.28 and -0.96.)
Differentiation Strategies
For Struggling Students
Keep the two folded paper triangles on the desk, labeled with side lengths, and have students point to opposite, adjacent and hypotenuse before writing any ratio
Give a unit circle with the four points (a, b), (-a, b), (-a, -b) and (a, -b) already drawn, and ask students only to match each to π - x, π + x or 2π - x
Use a sign check in every problem: first decide the quadrant of the new point, then attach the sign to the value from the special triangle
For Advanced Students
Ask students to explain why the rules for π - x, π + x and 2π - x together with the three special angles give exact values at every multiple of π/6 and π/4 between 0 and 2π
Ask students to find and prove a similar rule for π/2 - x, and to connect it with the complementary-angle relationship in HSG.SRT.C.7
Ask students to find exact values at -5π/6 and at 17π/6 by applying the rules more than once, and to check the quadrant of each point
Assessment Guidance
What to Look For
For the first half, students should derive values from side lengths, not quote them: ask "Where does the √3 come from?" and expect an answer about the altitude of an equilateral triangle. For the second half, look for students who explain each rule with a reflection or a half turn, and who can apply it when x is not a special angle, since the standard says x is any real number. A frequent error is to use the rule for π - x on an input of the form π + x; asking students to sketch the new point first prevents it.
02
Classroom Activities
3 Activities
1
Fold, Measure, Compare
20 minPairs
Pairs fold paper shapes into the two special triangles, measure the sides with a ruler, and compare their measured ratios with the exact values from the geometry. The goal is to see that the exact values come from the shapes, and that the size of the triangle does not matter.
Procedure
Fold an equilateral triangle with side 20 cm along an altitude. Measure the altitude: the exact length is 10√3 ≈ 17.3 cm
Fold a square with side 15 cm along a diagonal. Measure the diagonal: the exact length is 15√2 ≈ 21.2 cm
Compute each measured ratio to two decimals: opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent for the angles π/6, π/3 and π/4
Compare with the exact values: √3/2 ≈ 0.87, 1/2 = 0.50, √3 ≈ 1.73, √3/3 ≈ 0.58, √2/2 ≈ 0.71 and 1
Discussion Questions
Why does the 20 cm triangle give the same ratios as the side-2 triangle in the warm-up?
Which of the six ratios for π/6 and π/3 are equal to each other? Why does that happen?
Why is tan(π/4) exactly 1, whatever the size of the square?
Modification for Distance Learning
Students draw the two triangles in a free geometry app with the side lengths 2 and 1, use the measurement tools to read the altitude and the diagonal, and paste a screenshot with their three ratios for each angle.
2
Mirror Moves on Patty Paper
20 minGroups of 3
Each group traces a unit circle and marks P(x) for an input from one of 5 x-value cards. By folding the patty paper and turning it a half turn, the group finds P(π - x), P(π + x) and P(2π - x), then checks the nine rules with a calculator.
The 5 x-Value Cards
x = 0.5, x = 1.3, x = 2.4, x = 4.0 and x = -1.0
Three of the five inputs are not acute, on purpose: the rules must work for any real x
Procedure
Mark P(x) at its calculator coordinates (cos x, sin x), rounded to two decimals
Fold across the y-axis and mark the image: this is P(π - x). Unfold, fold across the x-axis and mark P(2π - x). Turn the paper a half turn about the center and mark P(π + x)
Read the coordinates of each image from the grid, then compute cos, sin and tan of π - x, π + x and 2π - x on the calculator. For example, with x = 2.4, cos(π + 2.4) ≈ 0.74 and -cos 2.4 ≈ 0.74
Record the nine results in a table next to the rules they confirm
Challenge Variation
Groups predict, before folding, the quadrant of P(π - x) when x = 4.0, and explain why it is not in Quadrant II even though π - x "looks like" a Quadrant II input.
3
Rule Match
15 minPairs
Pairs match 15 expression cards to their simpler forms and justify each match with a sketch of the symmetry. The deck mixes the nine rules with six distractor cards that state a wrong sign.
Six wrong rules: sin(π - x) = -sin x, cos(π - x) = cos x, tan(π + x) = -tan x, sin(π + x) = sin x, cos(2π - x) = -cos x, sin(2π - x) = sin x
Procedure
Sort the cards into "always true" and "not always true"
For each "not always true" card, find one x that breaks it. With x = 3, sin(π + 3) ≈ -0.141 but sin 3 ≈ 0.141
Glue the nine true rules into a 3-by-3 table with rows π - x, π + x, 2π - x and columns sin, cos, tan
Discussion Questions
How many rules in each row keep the same sign? In each column? What pattern do you notice?
Why does tangent keep its sign for π + x but not for the other two?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Two Special Triangles
Halving an equilateral triangle of side 2 gives a 30-60-90 triangle with sides 1, √3 and 2. Halving a square of side 1 along a diagonal gives a 45-45-90 triangle with sides 1, 1 and √2. Each figure is drawn to scale; the two figures use different scales.
Diagram 2: Symmetry Gives π - x, π + x and 2π - x
P(x) is drawn for x = 0.7, not a special angle, to show that the rules work for any x. The three images are exact reflections across the axes and a half turn about the origin, and each highlighted arc has length x.
04
Homework Assignment
~30 min
HSF.TF.A.3 Homework: Special Triangles and Unit Circle Symmetry
Directions: Show your work. Give exact values with simplified radicals and rationalized denominators. For every problem that uses π - x, π + x or 2π - x, sketch the unit circle and mark both P(x) and the new point.
Part 1: Special Triangles (Problems 1-2)
An equilateral triangle has side 12. (a) Draw an altitude and find its exact length. (b) Use one of the two halves to find the exact sine, cosine and tangent of π/6 and of π/3. (c) Explain why you would get the same six values from an equilateral triangle of side 50.
A 45-45-90 triangle is placed in the unit circle with its hypotenuse from the origin to P(π/4), so the hypotenuse has length 1. (a) Find the exact length of each leg. (b) Give the coordinates of P(π/4). (c) Find tan(π/4) and explain why it must equal 1.
Part 2: Using Symmetry (Problems 3-4)
Let P(x) = (a, b) be a point in Quadrant I. (a) Give the coordinates of P(π - x), P(π + x) and P(2π - x) and name the symmetry for each. (b) Write sine, cosine and tangent of each input in terms of sin x, cos x and tan x. (c) Test your three cosine rules with x = 2.7, which is not acute, using a calculator.
The standard says x can be any real number. Use a rule and a special triangle to find each exact value, then name the input and check its quadrant on the unit circle: (a) sin(π - x) for x = 4π/3 (b) cos(π + x) for x = 7π/6 (c) tan(2π - x) for x = -π/4
Part 3: Reasoning with the Rules (Problems 5-6)
The point P(x) = (-5/13, 12/13) is on the unit circle. (a) In which quadrant is P(x)? (b) Find sine, cosine and tangent of π - x, π + x and 2π - x, and the quadrant of each new point.
Jordan writes three rules: sin(π - x) = sin x, cos(π + x) = cos x and tan(2π - x) = tan x. Test each with x = 0.9 on a calculator. Explain with the unit circle which rules are wrong, and correct them.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Special Triangles
Side lengths derived and all ratios exact and simplified
Correct ratios with one error or unsimplified radicals
Values quoted without the triangle, or incorrect
Symmetry Rules
Correct rule and symmetry named for every input
Rules correct but symmetry not explained
Wrong rules or signs
Any Real x
Rules applied correctly to non-special or non-acute inputs and checked
Applied with one sign or quadrant error
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Give exact values where a question asks for them, with simplified radicals. For the decimal questions, keep the calculator in radian mode. Your score updates as you answer, and Reset quiz starts over.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
A 30-60-90 triangle has hypotenuse 10 and short leg 5. What is sin(π/6)?
Answer: A
The angle π/6 is opposite the short leg, so sin(π/6) = 5/10 = 1/2. Choice B is cos(π/6), adjacent over hypotenuse. Choice C gives a side length instead of a ratio, and choice D inverts the ratio.
Question 2 of 20 · Multiple Choice
An equilateral triangle has side 8. Its altitude is 4√3. Use one half of the triangle to find tan(π/3).
Answer: C
At the base angle π/3 the opposite side is the altitude 4√3 and the adjacent side is half the base, 4. So tan(π/3) = 4√3/4 = √3. Choice A is tan(π/6), which uses the ratio the other way up. Choice D forgets to divide by the adjacent side.
Question 3 of 20 · Multiple Choice
A square has side 6, so its diagonal is 6√2. What is cos(π/4)?
Answer: B
In the half square, cos(π/4) = adjacent/hypotenuse = 6/(6√2) = 1/√2 = √2/2. Choice A inverts the ratio (hypotenuse over leg). Choice C is tan(π/4), leg over leg.
Question 4 of 20 · Multiple Choice
What is the exact value of tan(π/6)?
Answer: D
At π/6 in the half equilateral triangle, the opposite side is 1 and the adjacent side is √3, so tan(π/6) = 1/√3 = √3/3. Choice A is tan(π/3), from reading the ratio at the wrong vertex. Choice B is sin(π/6). Choice C is hypotenuse over adjacent.
Question 5 of 20 · Multiple Choice
Which expression equals cos(π - x) for every real number x?
Answer: B
P(π - x) is the reflection of P(x) across the y-axis, which changes the sign of the x-coordinate. So cos(π - x) = -cos x. Choice A is the rule for cos(2π - x). Choice C confuses it with the complementary-angle rule for π/2 - x.
Question 6 of 20 · Multiple Choice
Which expression equals tan(π + x) wherever it is defined?
Answer: A
A half turn sends (a, b) to (-a, -b), and (-b)/(-a) = b/a, so tan(π + x) = tan x. Choice B would be right if only one coordinate changed sign, as happens for π - x and 2π - x.
Question 7 of 20 · Multiple Choice
Which expression equals sin(2π - x) for every real number x?
Answer: D
P(2π - x) is the reflection of P(x) across the x-axis, which changes the sign of the y-coordinate. So sin(2π - x) = -sin x. Choice A would come from treating 2π - x as a full turn plus x instead of a full turn minus x.
Question 8 of 20 · Multiple Choice
What is the exact value of sin(3π/4)?
Answer: C
3π/4 = π - π/4, so sin(3π/4) = sin(π/4) = √2/2. The point is in Quadrant II, where sine is positive. Choice A applies a sign change that belongs to cosine, not sine. Choice B is tan(π/4).
Question 9 of 20 · Multiple Choice
What is the exact value of cos(4π/3)?
Answer: B
4π/3 = π + π/3, so cos(4π/3) = -cos(π/3) = -1/2. The point is in Quadrant III, where cosine is negative. Choice C is sin(4π/3): the half-equilateral triangle values were swapped. Choice A forgets the sign change.
Question 10 of 20 · Multiple Choice
What is the exact value of tan(11π/6)?
Answer: D
11π/6 = 2π - π/6, so tan(11π/6) = -tan(π/6) = -√3/3. The point is in Quadrant IV, where tangent is negative. Choice A misses the sign change. Choice B uses tan(π/3) instead of tan(π/6).
Question 11 of 20 · Multiple Choice
Suppose sin x = 0.35. What is sin(π + x)?
Answer: A
A half turn changes the sign of the y-coordinate, so sin(π + x) = -sin x = -0.35. Choice B applies the π - x rule. Choice C subtracts from 1, which is not a symmetry of the unit circle.
Question 12 of 20 · Multiple Choice
Suppose tan x = 2.5. What is tan(π - x)?
Answer: C
Reflecting across the y-axis changes the sign of the x-coordinate only, so the quotient y/x changes sign: tan(π - x) = -tan x = -2.5. Choice A uses the π + x rule. Choice D takes a reciprocal, which does not come from any of the three symmetries.
Question 13 of 20 · Multiple Choice
The point P(x) on the unit circle is (a, b). Which point is P(2π - x)?
Answer: B
Going a full turn and then back x units ends at the reflection of P(x) across the x-axis: (a, -b). Choice A is P(π - x) and choice C is P(π + x). Choice D swaps the coordinates, which is the reflection for π/2 - x.
Question 14 of 20 · Multiple Choice
On the unit circle, P(π - x) is the image of P(x) under which transformation?
Answer: A
Going to (-1, 0) and then back x units mirrors the arc from (1, 0) to P(x), so the image is the reflection across the y-axis. Choice B gives P(2π - x) and choice C gives P(π + x).
Question 15 of 20 · Short Answer
Use an equilateral triangle with side 6 to explain why cos(π/3) = 1/2.
An altitude cuts the triangle into two 30-60-90 triangles. In each, the hypotenuse is a side of the original triangle, 6, and the side adjacent to the base angle π/3 is half the base, 3. So cos(π/3) = 3/6 = 1/2, and the same ratio holds for any size of equilateral triangle.
Question 16 of 20 · Short Answer
Explain, using a 45-45-90 triangle, why sin(π/4) = cos(π/4), and give the exact value.
In the half square the two legs are equal, so the side opposite π/4 and the side adjacent to it have the same length. With legs 1 and hypotenuse √2, both ratios are 1/√2, so sin(π/4) = cos(π/4) = √2/2.
Question 17 of 20 · Short Answer
Suppose cos x = -0.8 and sin x = 0.6, so P(x) is in Quadrant II. Find cos(π - x), sin(π + x) and tan(2π - x).
cos(π - x) = -cos x = 0.8. sin(π + x) = -sin x = -0.6. tan x = 0.6/(-0.8) = -0.75, so tan(2π - x) = -tan x = 0.75. The rules apply even though x is not acute.
Question 18 of 20 · Short Answer
Explain with the unit circle why tan(π + x) = tan x for every x where tan x is defined.
P(π + x) is P(x) turned a half turn about the origin, so if P(x) = (a, b) then P(π + x) = (-a, -b). Then tan(π + x) = (-b)/(-a) = b/a = tan x: both coordinates change sign, so their quotient does not.
Question 19 of 20 · Short Answer
Use a calculator in radian mode to find sin 1.1, cos 1.1 and tan 1.1 to three decimals. Then use the 2π - x rules to write sin(2π - 1.1), cos(2π - 1.1) and tan(2π - 1.1) without entering 2π - 1.1.
sin 1.1 ≈ 0.891, cos 1.1 ≈ 0.454, tan 1.1 ≈ 1.965. Reflecting across the x-axis gives sin(2π - 1.1) ≈ -0.891, cos(2π - 1.1) ≈ 0.454 and tan(2π - 1.1) ≈ -1.965. Entering 2π - 1.1 directly gives the same values.
Question 20 of 20 · Short Answer
Simplify sin(π - x) + sin(π + x) + sin(2π - x) for any real number x.
Use the rules: sin(π - x) = sin x, sin(π + x) = -sin x and sin(2π - x) = -sin x. The sum is sin x - sin x - sin x = -sin x.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.TF.A.3 mean?
It means students find sine, cosine and tangent of π/6, π/4 and π/3 from the geometry of special triangles, and use the unit circle to rewrite the values at π - x, π + x and 2π - x in terms of the values at x. For example, cos(π - x) = -cos x for every real x. It is marked (+), so it is part of the additional mathematics for students in advanced courses.
What does the (+) on HSF.TF.A.3 mean?
It marks the standard as additional mathematics that Common Core describes for students who take advanced courses such as Precalculus. It is not required of every high school student. Many Algebra II courses still teach the special-angle values, and Precalculus usually adds the symmetry rules.
Why use special triangles instead of memorizing the unit circle?
Because the standard asks students to determine the values geometrically. A student who knows that an equilateral triangle of side 2 has altitude √3, and that a square of side 1 has diagonal √2, can rebuild every value for π/6, π/4 and π/3 in under a minute. Memorized charts are easy to mix up, especially sin(π/6) and sin(π/3).
How do I remember which values are √3/2 and which are 1/2?
Use the triangle: the side opposite the smaller angle is shorter. In the half equilateral triangle, the shortest side, 1, is opposite π/6, so sin(π/6) = 1/2, and the longer leg √3 is opposite π/3, so sin(π/3) = √3/2. Cosine switches them, because the side adjacent to one acute angle is opposite the other.
Why is tan(π/6) written as √3/3?
It is the same number as 1/√3, with the denominator rationalized: 1/√3 = √3/(√3 × √3) = √3/3 ≈ 0.577. Both forms are correct. Many textbooks and tests use √3/3, so students should recognize both.
Where do the rules for π - x, π + x and 2π - x come from?
From symmetries of the unit circle. If P(x) = (a, b), then P(π - x) is its reflection across the y-axis, (-a, b); P(π + x) is its image under a half turn about the origin, (-a, -b); and P(2π - x) is its reflection across the x-axis, (a, -b). Reading off the coordinates gives the sine and cosine rules, and dividing gives the tangent rules.
Do these rules only work for special angles?
No. The standard says x is any real number. The rules work for x = 0.4, x = 2.7 or x = -5, because the reflections and the half turn move every point of the circle. Special angles are simply the case where the values at x are known exactly, so the rules give exact answers.
What is a common mistake with these rules?
Changing the sign of the wrong function. For π - x only cosine and tangent change sign; for 2π - x only sine and tangent change sign; for π + x sine and cosine both change sign and tangent does not. A quick sketch of where the new point lands prevents most sign errors.
How are HSF.TF.A.3 values used later?
The exact values are used to graph sine and cosine accurately, to solve trigonometric equations, and with the Pythagorean identity (HSF.TF.C.8) and the addition formulas (HSF.TF.C.9). The rules for π - x and π + x also explain why equations such as sin x = 1/2 have two solutions in each full turn.
Is this on the SAT?
The digital SAT includes right-triangle trigonometry and radians in its Geometry and Trigonometry domain, so special right triangles and the values they give can appear. The symmetry rules for π - x, π + x and 2π - x are more typical of Precalculus course tests than of the SAT.
07
Related Standards
6 standards
These standards connect to HSF.TF.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.SRT.C.6Prerequisite
Side ratios in right triangles define the trigonometric ratios for acute angles
