HSF.TF.C.8: Proving and Using the Pythagorean Identity
In plain English: HSF.TF.C.8 is the Common Core functions standard that asks students to prove the Pythagorean identity sin²(θ) + cos²(θ) = 1 from the unit circle and to use it to find sin(θ), cos(θ) or tan(θ) when one of them and the quadrant of θ are known. The quadrant decides the sign. It is usually taught in Algebra II or Precalculus.
Prove the Pythagorean identity sin²(θ) + cos²(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle.
Common Core State Standards for Mathematics · Domain: Trigonometric Functions (TF) · Cluster: Prove and apply trigonometric identities Also written as HSF-TF.C.8 or F-TF.8 · Official standard
Students prove that sin²(θ) + cos²(θ) = 1 for every angle θ, starting from the definition of sine and cosine as the coordinates of a point on the unit circle and the fact that this point is 1 unit from the origin. The proof is short, and the lesson spends time on why it holds in all four quadrants.
Students then use the identity as a tool: given one of sin θ, cos θ or tan θ and the quadrant of θ, they find the other two values. The identity gives the size of the missing value, and the quadrant gives its sign. Students work with fractions, radicals and decimals, and check each answer by substituting back into the identity.
Learning Objectives
By the end of this lesson, students will be able to:
Prove sin²(θ) + cos²(θ) = 1 from the unit circle definition of sine and cosine and the distance formula
Explain why the identity holds for angles in every quadrant
Given sin θ or cos θ and the quadrant, find the other two trigonometric values with correct signs
Given tan θ and the quadrant, use the identity (divided by cos²θ) to find sin θ and cos θ
Prior Knowledge Required
Students should already be comfortable with:
The Pythagorean theorem and the distance formula 8.G.B.7
Sine, cosine and tangent as right triangle ratios HSG.SRT.C.6
The unit circle definition of sine and cosine for all real numbers HSF.TF.A.2
Simplifying square roots and rationalizing denominators, for example 1/√5 = √5/5
Draw a circle of radius 1 centered at the origin on graph paper. Ask students to test two points:
Warm-Up Prompt
"Is the point (0.6, 0.8) on the unit circle? Is (0.5, 0.8)? How can you decide without drawing?"
Students use the distance from the origin: 0.6² + 0.8² = 0.36 + 0.64 = 1, so (0.6, 0.8) is on the circle, while 0.5² + 0.8² = 0.89, so (0.5, 0.8) is inside it. Ask what the coordinates of a point on the unit circle mean in trigonometry. Students should recall that the point where the terminal side of θ meets the unit circle is (cos θ, sin θ). Write the question for the lesson: what does x² + y² = 1 say about cos θ and sin θ?
Direct Instruction20-25 minutes
Part 1: The proof. Use Diagram 1. Present the proof as a short chain of reasons that works for every angle, not only acute ones:
Definition: for any angle θ in standard position, the terminal side meets the unit circle at P = (cos θ, sin θ).
Distance: P is 1 unit from the origin O, so by the distance formula (the Pythagorean theorem on the legs |cos θ| and |sin θ|), √(cos²θ + sin²θ) = 1.
Square both sides: cos²θ + sin²θ = 1. Squaring removes the absolute values, so the result holds in every quadrant and on the axes.
Rewrite as needed: sin²θ = 1 - cos²θ and cos²θ = 1 - sin²θ. Dividing every term by cos²θ (when cos θ ≠ 0) gives tan²θ + 1 = 1/cos²θ.
Stress that sin²θ means (sin θ)², and that the identity is about squares, so it gives the size of the missing value but not its sign. Part 2: Using the identity. The routine: substitute the known value, solve for the square of the unknown, take the square root, and choose the sign from the quadrant (Diagram 2). Then find the third ratio from tan θ = sin θ / cos θ.
Proof check at one angle
For θ = 5π/6 (Quadrant II), the unit circle point is (-√3/2, 1/2).
Equation: (-√3/2)² + (1/2)² = 3/4 + 1/4 = 1
Given sine, Quadrant II
sin θ = 5/13 and θ is in Quadrant II. Find cos θ and tan θ.
Equation: cos²θ = 1 - 25/169 = 144/169, cosine is negative in QII, so cos θ = -12/13 and tan θ = -5/12
Given cosine, Quadrant IV
cos θ = 2/7 and θ is in Quadrant IV. Find sin θ and tan θ.
Equation: sin²θ = 1 - 4/49 = 45/49, so sin θ = -3√5/7 and tan θ = -3√5/2
Given tangent, Quadrant III
tan θ = 2 and θ is in Quadrant III. Find cos θ and sin θ.
Equation: 1/cos²θ = 1 + 4 = 5, so cos θ = -√5/5 and sin θ = tan θ · cos θ = -2√5/5
Decimal value, Quadrant III
sin θ = -0.28 and θ is in Quadrant III. Find cos θ and tan θ.
Equation: cos²θ = 1 - 0.0784 = 0.9216, so cos θ = -0.96 and tan θ = 0.28/0.96 = 7/24 ≈ 0.2917
Guided Practice15 minutes
Pairs work three problems on whiteboards and hold them up after each. (a) cos θ = -8/17, θ in Quadrant III: sin θ = -15/17 and tan θ = 15/8. (b) sin θ = 1/3, θ in Quadrant II: cos θ = -2√2/3 and tan θ = -√2/4. (c) tan θ = -2/5, θ in Quadrant II: 1/cos²θ = 29/25, so cos θ = -5√29/29 and sin θ = 2√29/29. Before each square root, ask pairs to name the sign from the quadrant. Listen for these errors: writing sin θ = 1 - cos θ without squares, taking the square root of only one term, and keeping the positive root in every quadrant.
Independent Practice15 minutes
Students work alone on four tasks. (1) sin θ = -20/29, θ in Quadrant IV: find cos θ and tan θ (21/29 and -20/21). (2) cos θ = 1/4, θ in Quadrant IV: find sin θ and tan θ (-√15/4 and -√15). (3) tan θ = 1/2, θ in Quadrant III: find sin θ and cos θ (-√5/5 and -2√5/5). (4) Write the proof of the identity for an angle θ in Quadrant III, and explain in one sentence why the negative coordinates do not change the result. Students check tasks 1-3 by confirming that their sin²θ + cos²θ equals 1.
Closure5 minutes
Exit ticket: (1) sin θ = 0.6 and θ is in Quadrant II. Find cos θ and tan θ. (cos θ = -0.8, tan θ = -0.75.) (2) A classmate says, "If sin θ = 0.6, then cos θ = 0.4 because they add to 1." Write two sentences explaining the error.
Differentiation Strategies
For Struggling Students
Draw a reference triangle for every problem: label the given side and the hypotenuse, find the third side with the Pythagorean theorem, then place the triangle in the quadrant
Give a sign card with the four quadrants and the signs of sin, cos and tan to keep on the desk
Start with Pythagorean triples such as 5-12-13 and 8-15-17 before moving to radicals and decimals
For Advanced Students
Ask students to derive 1 + 1/tan²θ = 1/sin²θ from the identity and state when it is defined
Given sin θ = k with θ in Quadrant II, ask students to write cos θ and tan θ in terms of k and state the allowed values of k
Ask for a second proof of the identity that uses a right triangle with hypotenuse r and legs x and y, and explain why it covers only acute angles unless it is extended
Assessment Guidance
What to Look For
In proofs, look for the unit circle definition P = (cos θ, sin θ) and a reason for each step, including why the identity holds when a coordinate is negative. In computations, check that students square the given value, take the square root of the difference and not of each term, and choose the sign from the quadrant before simplifying. A complete answer gives all requested values in simplest radical form or as decimals and checks that sin²θ + cos²θ = 1.
02
Classroom Activities
3 Activities
1
One Identity, Three Arguments
20 minGroups of 3
Each group member builds a different argument for sin²θ + cos²θ = 1, then the group decides which argument covers every angle. This makes the difference between checking examples and proving a statement explicit.
Roles
Student A (numeric check): uses a calculator to evaluate sin²θ + cos²θ for θ = 200°, θ = 2.5 and θ = -1 and records the results
Student B (right triangle): draws a right triangle with legs x and y and hypotenuse r, writes x² + y² = r², and divides by r² to get (x/r)² + (y/r)² = 1
Student C (unit circle): draws an angle in Quadrant III, marks P = (cos θ, sin θ), and applies the distance formula from the origin to P
Procedure
Each student presents for two minutes while the others write one question
The group writes a final proof on a poster, using Student C's argument, with Student B's argument as the acute-angle case
The group adds one sentence explaining why Student A's work supports the identity but does not prove it
Discussion Questions
Why does the right triangle argument need extra work for θ = 200°?
Where in the unit circle proof do negative coordinates stop mattering?
What happens to the identity at θ = 90°, where the right triangle has no second acute angle?
2
Quadrant Detective Cards
20-25 minPairs
Pairs receive 8 cards. Each card gives one trigonometric value and a quadrant. Pairs find the other two values, then match each card to a sign pattern on a quadrant mat.
The 8 Cards (with answers)
sin θ = 12/37, QII: cos θ = -35/37, tan θ = -12/35
cos θ = 9/41, QIV: sin θ = -40/41, tan θ = -40/9
tan θ = 60/11, QIII: sin θ = -60/61, cos θ = -11/61
sin θ = -1/2, QIII: cos θ = -√3/2, tan θ = √3/3
cos θ = -1/3, QII: sin θ = 2√2/3, tan θ = -2√2
tan θ = -1, QIV: sin θ = -√2/2, cos θ = √2/2
sin θ = 0.96, QII: cos θ = -0.28, tan θ = -24/7
cos θ = √5/3, QIV: sin θ = -2/3, tan θ = -2√5/5
Procedure
Partners alternate: one solves while the other checks that sin²θ + cos²θ = 1 and that the signs match the quadrant
Pairs place each finished card on the quadrant mat
Each pair picks the card they found hardest and explains their steps to another pair
Challenge Variation
Pairs write two cards of their own that have the same given value but different quadrants, and explain how the answers differ.
3
Error Analysis Gallery
15 minGroups of 3-4
Four posters around the room show student work with one error each. Groups rotate, name the error, and write the corrected answer.
Posters
Poster 1: "cos θ = -5/6 in QIII, so sin θ = 1 - 25/36 = 11/36." (No square root and no sign; correct: sin θ = -√11/6.)
Poster 2: "sin θ = 2/9 in QII, so cos θ = √(1 - 2/9) = √7/3." (The given value was not squared; correct: cos θ = -√77/9.)
Poster 3: "tan θ = 3/4, so sin θ = 3 and cos θ = 4." (Sine and cosine are between -1 and 1; in QI, sin θ = 3/5 and cos θ = 4/5.)
Poster 4: "sin²θ + cos²θ = 1 is true because sin 30° = 1/2 and cos 30° = √3/2 work." (One example is a check, not a proof.)
Procedure
Groups spend 3 minutes at each poster and write the error and the correction on a sticky note
After the rotation, each group reads the notes at its last poster and presents the best explanation
Modification for Distance Learning
Put the four posters on shared slides. Groups add comments in breakout rooms, then the class reviews the comments together.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Unit Circle Proof of sin²θ + cos²θ = 1
The terminal side of θ meets the unit circle at P = (cos θ, sin θ), drawn to scale here for sin θ = 5/13 in Quadrant II. The legs of the right triangle have lengths |cos θ| = 12/13 and sin θ = 5/13, and the hypotenuse is the radius 1, so (12/13)² + (5/13)² = 1.
Diagram 2: Signs of Sine, Cosine and Tangent by Quadrant
The identity gives the size of a missing value; this chart gives its sign. Sine follows the sign of y, cosine the sign of x, and tangent is positive where x and y have the same sign (Quadrants I and III).
04
Homework Assignment
~30 min
HSF.TF.C.8 Homework: The Pythagorean Identity
Directions: Show every step. Give exact answers in simplest radical form with rationalized denominators unless the problem uses decimals. For each problem in Parts 2 and 3, state which sign the quadrant requires before you take a square root, and check your answers by confirming that sin²θ + cos²θ = 1.
Part 1: Proving the Identity (Problems 1-2)
Write a proof of sin²(θ) + cos²(θ) = 1 using the unit circle. Draw θ in Quadrant IV, label the point P, and give a reason for every step. Then verify the identity for θ = 5π/3 using the exact coordinates of its unit circle point.
Start from sin²θ + cos²θ = 1 and divide every term by sin²θ. Write the new identity using only tan θ and sin θ. State the values of θ for which the new identity is not defined, and explain why.
Part 2: Finding Values from the Quadrant (Problems 3-4)
(a) sin θ = 2/5 and θ is in Quadrant II. Find cos θ and tan θ. (b) cos θ = -0.35 and θ is in Quadrant III. Find sin θ and tan θ, rounded to four decimal places.
tan θ = -7/3 and θ is in Quadrant II. Divide sin²θ + cos²θ = 1 by cos²θ and use the resulting identity to find cos θ, then find sin θ.
Part 3: Reasoning with the Identity (Problems 5-6)
A student is told that sin θ = -3/5 and θ is in Quadrant III. The student writes cos θ = 4/5 and tan θ = -3/4. Find and explain each error, then give the correct values.
Suppose sin θ = k, where 0 < k < 1 and θ is in Quadrant II. Write cos θ and tan θ in terms of k. Then check your formulas with k = 5/13.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Proof
Unit circle definition, distance formula and squaring step, each with a reason, for any quadrant
Correct steps with a missing reason or only acute angles
Examples instead of a proof, or no proof
Using the Identity
Squares and square roots handled correctly in every problem
One algebra or arithmetic error
Identity misused, for example sin θ + cos θ = 1
Signs and Quadrants
Every sign matches the quadrant and is justified
One sign error
Signs ignored
Exact Form and Checking
Simplest radical form and a check with the identity
Correct values but not simplified or not checked
Values missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order and give exact answers where you can. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
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
Which equation is the Pythagorean identity?
Answer: B
The identity is about squares: the point (cos θ, sin θ) is 1 unit from the origin, so cos²θ + sin²θ = 1. Choice A drops the squares; for θ = 45° it would give √2/2 + √2/2 = √2, not 1.
Question 2 of 20 · Multiple Choice
In the unit circle proof, why is (cos θ)² + (sin θ)² equal to 1?
Answer: C
The point where the terminal side meets the unit circle is (cos θ, sin θ), and every point on that circle is at distance 1 from the origin. Choice B describes the right triangle version, which only covers acute angles. Choice A is false: both can be negative.
Question 3 of 20 · Multiple Choice
sin θ = 13/85 and θ is in Quadrant II. What is cos θ?
Answer: A
cos²θ = 1 - 169/7225 = 7056/7225, so |cos θ| = 84/85, and cosine is negative in Quadrant II. Choice B has the wrong sign. Choice C comes from 1 - 13/85 = 72/85 without squaring.
Question 4 of 20 · Multiple Choice
cos θ = 33/65 and θ is in Quadrant IV. What is sin θ?
Answer: C
sin²θ = 1 - 1089/4225 = 3136/4225, so |sin θ| = 56/65, and sine is negative in Quadrant IV. Choice A has the wrong sign. Choice B subtracts 33/65 from 1 without squaring.
Question 5 of 20 · Multiple Choice
sin θ = -4/9 and θ is in Quadrant III. What is tan θ?
Answer: B
cos²θ = 1 - 16/81 = 65/81 and cosine is negative in Quadrant III, so cos θ = -√65/9. Then tan θ = (-4/9)/(-√65/9) = 4/√65 = 4√65/65, positive in Quadrant III. Choice A has the wrong sign. Choice C divides cosine by sine, the reciprocal of tan θ. Choice D is cos θ, not tan θ.
Question 6 of 20 · Multiple Choice
In which quadrant is sin θ positive and cos θ negative?
Answer: D
Sine is the y-coordinate and cosine is the x-coordinate, so y > 0 and x < 0 means Quadrant II. In Quadrant IV (choice C) the signs are reversed.
Question 7 of 20 · Multiple Choice
cos θ = -3/8 and θ is in Quadrant III. What is sin θ?
Answer: A
sin²θ = 1 - 9/64 = 55/64, so |sin θ| = √55/8, and sine is negative in Quadrant III. Choice B has the wrong sign. Choice C subtracts 3/8 from 1 without squaring. Choice D is cos²θ, the square of the given value, not sin θ.
Question 8 of 20 · Multiple Choice
tan θ = -45/28 and θ is in Quadrant II. What is cos θ?
Answer: B
1/cos²θ = 1 + 2025/784 = 2809/784, so cos²θ = 784/2809 and |cos θ| = 28/53. Cosine is negative in Quadrant II. Choice D is sin θ, which is positive in Quadrant II.
Question 9 of 20 · Multiple Choice
A student writes: "cos θ = 0.8, so sin θ = 1 - 0.8 = 0.2." Which statement describes the error?
Answer: C
The identity relates the squares: sin²θ = 1 - 0.64 = 0.36, so sin θ = ±0.6. The student's version has no squares. Choice B names a real issue in general, but the main error here is dropping the squares.
Question 10 of 20 · Multiple Choice
Why does sin²θ + cos²θ = 1 hold for θ = 250°, even though cos 250° and sin 250° are both negative?
Answer: B
The distance formula uses x² + y², and (-a)² = a², so the signs of the coordinates do not affect the sum. Choice C is false: the unit circle proof works for every angle. Choice D is true but does not explain the identity.
Question 11 of 20 · Multiple Choice
cos θ = √7/4 and sin θ < 0. What are the quadrant of θ and the value of sin θ?
Answer: A
Cosine positive and sine negative means Quadrant IV. sin²θ = 1 - 7/16 = 9/16, so sin θ = -3/4. Choice C gives sin²θ instead of sin θ. Choice D has the wrong quadrant: cosine is negative in Quadrant III.
Question 12 of 20 · Multiple Choice
tan θ = 4 and θ is in Quadrant III. Dividing sin²θ + cos²θ = 1 by cos²θ gives an identity relating tan θ and cos θ. Use it to find cos θ.
Answer: B
Dividing by cos²θ gives tan²θ + 1 = 1/cos²θ, so 1/cos²θ = 16 + 1 = 17 and cos²θ = 1/17. Cosine is negative in Quadrant III, so cos θ = -1/√17 = -√17/17. Choice A has the wrong sign. Choice C uses tan²θ - 1 instead of tan²θ + 1. Choice D forgets to square the tangent: 4 + 1 = 5.
Question 13 of 20 · Multiple Choice
cos θ = -0.42 and θ is in Quadrant II. To four decimal places, what is sin θ?
Answer: D
sin²θ = 1 - 0.1764 = 0.8236, so sin θ = √0.8236 ≈ 0.9075, positive in Quadrant II. Choice B is sin²θ, before the square root. Choice A is 1 - 0.42. Choice C has the wrong sign.
Question 14 of 20 · Multiple Choice
sin θ = 1/4 and the quadrant of θ is not given. How many values of cos θ are possible?
Answer: A
cos²θ = 1 - 1/16 = 15/16, so cos θ = ±√15/4: positive in Quadrant I and negative in Quadrant II. This is why the standard includes the quadrant. Choice C forgets to square 1/4.
Question 15 of 20 · Short Answer
Prove that sin²(θ) + cos²(θ) = 1 for an angle θ whose terminal side is in Quadrant III. Give a reason for each step.
Let P be the point where the terminal side of θ meets the unit circle. By definition, P = (cos θ, sin θ), where both coordinates are negative in Quadrant III. P is on the unit circle, so its distance from O(0, 0) is 1. By the distance formula, √((cos θ - 0)² + (sin θ - 0)²) = 1. Squaring both sides gives cos²θ + sin²θ = 1. The negative signs do not matter because each coordinate is squared.
Question 16 of 20 · Short Answer
sin θ = -5/8 and θ is in Quadrant IV. Find cos θ and tan θ in simplest form.
cos²θ = 1 - 25/64 = 39/64, and cosine is positive in Quadrant IV, so cos θ = √39/8. Then tan θ = (-5/8)/(√39/8) = -5/√39 = -5√39/39.
Question 17 of 20 · Short Answer
tan θ = 3/2 and θ is in Quadrant III. Find cos θ and sin θ.
1/cos²θ = 1 + 9/4 = 13/4, so cos²θ = 4/13. Cosine is negative in Quadrant III: cos θ = -2/√13 = -2√13/13. Then sin θ = tan θ · cos θ = (3/2)(-2√13/13) = -3√13/13. Check: 4/13 + 9/13 = 1.
Question 18 of 20 · Short Answer
cos θ = 3/10 and θ is in Quadrant IV. Find sin θ and tan θ.
sin²θ = 1 - 9/100 = 91/100, and sine is negative in Quadrant IV, so sin θ = -√91/10. Then tan θ = (-√91/10)/(3/10) = -√91/3.
Question 19 of 20 · Short Answer
Is there an angle θ with sin θ = 0.6 and cos θ = 0.6? Use the identity to explain.
No. For any angle, sin²θ + cos²θ must equal 1, but 0.6² + 0.6² = 0.36 + 0.36 = 0.72. If sin θ = 0.6, then cos θ must be ±0.8.
Question 20 of 20 · Short Answer
Explain why the standard gives the quadrant of θ when you are asked to find cos θ from sin θ = 3/7. Include both possible values in your answer.
The identity gives cos²θ = 1 - 9/49 = 40/49, which has two square roots, 2√10/7 and -2√10/7. Both are possible: an angle in Quadrant I and an angle in Quadrant II both have sine 3/7. The quadrant tells you which one: 2√10/7 in Quadrant I, -2√10/7 in Quadrant II.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.TF.C.8 mean?
HSF.TF.C.8 asks students to prove that sin²(θ) + cos²(θ) = 1 and to use that identity to find one trigonometric value from another when the quadrant is known. For example, if sin θ = 5/13 and θ is in Quadrant II, the identity gives cos θ = -12/13 and then tan θ = -5/12.
Is HSF.TF.C.8 Algebra 2 or Precalculus?
It is usually taught in Algebra II, after the unit circle, and reviewed in Precalculus. Unlike several other trigonometry standards, it is not marked (+), so Common Core expects all students to learn it.
Why is it called the Pythagorean identity?
Because it is the Pythagorean theorem applied to the unit circle. The point (cos θ, sin θ) and the origin form a right triangle with legs |cos θ| and |sin θ| and hypotenuse 1, so the squares of the legs add up to 1. The distance formula says the same thing for every angle.
What counts as a proof of the Pythagorean identity?
A proof must work for every angle, not only the ones you test. The standard proof uses the unit circle definition P = (cos θ, sin θ) and the distance from the origin to P. Checking θ = 30° or θ = 45° is useful but is not a proof. A right triangle argument proves it for acute angles, and the unit circle extends it to all angles.
How do you know whether the answer is positive or negative?
From the quadrant. The identity gives the square of the missing value, which has a positive and a negative square root. Sine follows the sign of y and cosine the sign of x, and tangent is positive in Quadrants I and III. Choose the sign before simplifying.
How do you find sin θ and cos θ if you only know tan θ?
Divide the identity by cos²θ to get tan²θ + 1 = 1/cos²θ. Substitute tan θ, solve for cos²θ, and take the square root with the sign from the quadrant. Then sin θ = tan θ · cos θ. For tan θ = 2 in Quadrant III, cos θ = -√5/5 and sin θ = -2√5/5.
What are common mistakes with the Pythagorean identity?
Common errors include writing sin θ + cos θ = 1 without the squares, forgetting to square the given value, taking the square root of each term separately, and always keeping the positive root. Many students also stop at cos²θ and forget the square root.
Can students just use a right triangle instead of the identity?
Yes, a reference triangle is a good way to find the size of the missing value when the given value is a fraction, and many teachers teach both. The identity is faster with decimals and radicals, and it is the form used later to simplify expressions. With either method, the quadrant still decides the sign.
Does sin²θ mean sin(θ²)?
No. sin²θ means (sin θ)², the square of the value of sine. For θ = π/6, sin²θ = (1/2)² = 1/4. The notation is a convention, and students should write (sin θ)² when they are unsure.
Where is the Pythagorean identity used later?
It is used to simplify trigonometric expressions, to derive the double-angle formula cos 2θ = 1 - 2sin²θ, to solve trigonometric equations, and in calculus for integrals and derivatives of trigonometric functions. The related form tan²θ + 1 = 1/cos²θ comes from dividing it by cos²θ, and dividing by sin²θ gives another.
07
Related Standards
6 standards
These standards connect to HSF.TF.C.8: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.B.7Prerequisite
Apply the Pythagorean Theorem to find unknown side lengths in right triangles
