HSG.SRT.C.7: Sine and Cosine of Complementary Angles
In plain English: HSG.SRT.C.7 is the Common Core geometry standard that asks students to explain and use the fact that the sine of an acute angle equals the cosine of its complement: sin θ = cos(90° - θ). The reason is that the two acute angles of a right triangle add to 90°, and the leg opposite one is adjacent to the other. It is usually taught in Geometry.
Explain and use the relationship between the sine and cosine of complementary angles.
Common Core State Standards for Mathematics · Domain: Similarity, Right Triangles, and Trigonometry (SRT) · Cluster: Define trigonometric ratios and solve problems involving right triangles Also written as HSG-SRT.C.7 or G-SRT.7 · Official standard
Students explain why the sine of an acute angle equals the cosine of its complement and use that fact to rewrite, evaluate and solve. The explanation comes from a single right triangle: its two acute angles add to 90°, and the leg that is opposite one of them is the leg next to the other, so the two ratios are the same fraction of the same hypotenuse.
Students then use the relationship sin θ = cos(90° - θ) in three ways: rewriting a sine as a cosine and back, finding a value from a known value of the complement, and solving equations such as sin(3x + 4)° = cos(2x + 6)° by setting the angle sum equal to 90°. Every solution is checked to make sure both angles are acute.
Learning Objectives
By the end of this lesson, students will be able to:
Explain, using a right triangle, why sin θ = cos(90° - θ) and cos θ = sin(90° - θ) for an acute angle θ
Rewrite the sine of an acute angle as the cosine of its complement, and the reverse
Find the sine or cosine of an angle from a known value of its complement without a calculator
Solve equations of the form sin(expression)° = cos(expression)° and check that both angles are acute
Prior Knowledge Required
Students should already be comfortable with:
Complementary angles and the triangle angle sum 7.G.B.5
Definitions of sine, cosine and tangent for acute angles HSG.SRT.C.6
Naming opposite and adjacent sides relative to an angle
Solving linear equations in one variable HSA.REI.B.3
Draw right triangle ABC with the right angle at C, BC = 3, AC = 4 and AB = 5. Label the two acute angles A and B.
Warm-Up Prompt
"Write sin A, cos A, sin B and cos B as fractions. Which values match? If m∠A is about 37°, what is m∠B, and why?"
Students find sin A = 3/5, cos A = 4/5, sin B = 4/5 and cos B = 3/5, so sin A = cos B and cos A = sin B. Because the angles of a triangle add to 180° and ∠C is 90°, m∠B is about 53°. Ask students whether the match is a coincidence of the 3-4-5 triangle or something that happens in every right triangle. Leave the question open for Direct Instruction.
Direct Instruction20 minutes
Part 1: Explaining the relationship. Use Diagram 1. In any right triangle, the two acute angles are complementary: if one measures θ, the other measures 90° - θ. Build the explanation in steps:
Name the sides: in right △ABC with right angle C, let a = BC, b = AC and c = AB.
Stand at A: side a is opposite ∠A, so sin A = a/c.
Stand at B: the same side a is now the leg next to ∠B, so cos B = a/c.
Compare: sin A and cos B are the same fraction, a/c. In the same way, cos A = b/c = sin B.
Use the angle sum: m∠B = 90° - m∠A, so sin θ = cos(90° - θ) and cos θ = sin(90° - θ) for every acute angle θ.
Tell students that the "co" in cosine refers to the complement: the cosine of an angle is the sine of its complement. Use Diagram 2 to show the same fact with calculator values: the cosine curve is the sine curve read from the other end.
Part 2: Using the relationship. Work through the examples below.
Explaining with one triangle
Right △PQR has its right angle at R, QR = 9, PR = 40 and PQ = 41. Compare sin P and cos Q.
Equation: sin P = QR/PQ = 9/41 and cos Q = QR/PQ = 9/41, and m∠P + m∠Q = 90°
Rewriting with the complement
Write sin 41° as a cosine and cos 14° as a sine.
Equation: sin 41° = cos 49° and cos 14° = sin 76°
Using a known value
A table gives sin 23° ≈ 0.3907. Find cos 67° without a calculator.
Equation: (3x + 4) + (2x + 6) = 90, so 5x + 10 = 90 and x = 16; the angles are 52° and 38°
Finding the angle
Find the acute angle θ with cos θ = sin 34°.
Equation: θ = 90° - 34° = 56°
For the "Solving for x" example, stress the last step: substitute x back in and check that both angles are acute and add to 90°. The relationship sin x° = cos y° gives x + y = 90 when both angles are acute, which is the setting of this standard.
Guided Practice10 minutes
Pairs use calculators in degree mode to fill in the table, rounding to four decimal places, and then write one sentence describing the pattern.
Sine and cosine of complementary angle pairs
Angle pair
sin of the smaller angle
cos of the larger angle
cos of the smaller angle
sin of the larger angle
10° and 80°
0.1736
0.1736
0.9848
0.9848
25° and 65°
0.4226
0.4226
0.9063
0.9063
40° and 50°
0.6428
0.6428
0.7660
0.7660
Ask: "Where in Diagram 1 do you see the reason for this pattern?" Students should point to the shared side and the shared hypotenuse. Watch for students who pair angles that add to 180° instead of 90°.
Independent Practice15 minutes
Students work alone:
Rewrite sin 81° as a cosine and cos 26° as a sine.
For acute angles, solve sin(x + 20)° = cos(2x + 10)° and check both angles.
For an acute angle A, cos A = 0.28. Without a calculator, find sin(90° - A), and explain your answer in a sentence.
Sketch a right triangle and use it to explain why cos 29° = sin 61°.
Answers: cos 9° and sin 64°; x = 20, so the angles are 40° and 50°; sin(90° - A) = cos A = 0.28.
Closure5 minutes
Exit ticket: (1) Fill in the blank: sin 57° = cos ___. (Answer: 33°.) (2) For acute angles, solve cos(4x)° = sin(x + 15)°. (Answer: 5x + 15 = 90, so x = 15, and the angles are 60° and 30°.) (3) In one sentence, explain why the sine of an acute angle equals the cosine of its complement.
Differentiation Strategies
For Struggling Students
Color the leg shared by the two ratios in Diagram 1 and have students trace it from each acute angle
Give a two-column frame: "Angle | Its complement" before any rewriting, so students subtract from 90 first
For equations, provide the first line "(first angle) + (second angle) = 90" and have students fill it in
For Advanced Students
Ask students to explain why tan θ · tan(90° - θ) = 1 for an acute angle θ, labeled as going beyond the standard
Ask for an equation of the form sin(expression)° = cos(expression)° whose solution makes one angle 0° and the other 90°, and discuss why it does not describe a right triangle
Have students use the relationship and Diagram 2 to explain why sin θ > cos θ exactly when θ is between 45° and 90°
Assessment Guidance
What to Look For
The standard has two parts, so check both. For explain, look for an argument that names the shared side and the shared hypotenuse, not only the formula sin θ = cos(90° - θ). For use, look for correct complements (subtracting from 90, not 180), and for equations, a check that both angles are acute and add to 90°. A student who writes sin 25° = cos 25° has confused "complementary" with "equal".
02
Classroom Activities
3 Activities
1
Cofunction Match
15 minPairs
Pairs receive 16 cards, each showing a sine or a cosine of an angle. They pair every card with the one card that has the same value, without using a calculator, and justify each match.
Card Set
sin 12° and cos 78°
sin 48° and cos 42°
cos 5° and sin 85°
cos 61° and sin 29°
sin 8° and cos 82°
sin 70° and cos 20°
cos 24° and sin 66°
sin 3° and cos 87°
Shuffle the 16 cards before handing them out.
Procedure
Pairs lay out all cards face up and find the 8 matching pairs
For each pair, one partner says the reason aloud: "12 and 78 add to 90, so the sine of one is the cosine of the other"
Pairs check two of their matches with a calculator
Discussion Questions
Why does sin 12° never match sin 78°?
Could a sine card ever match another sine card? When?
Modification for Distance Learning
Use a shared slide with draggable cards. Pairs record their matches and reasons in a two-column table on the slide.
2
Explain It Three Ways
20 minGroups of 3
Each group member explains the same fact, sin A = cos B in a right triangle, in a different way: with numbers, with a calculator table, and with letters. The group then combines the three into one poster. This targets the "explain" half of the standard.
Roles
Numbers: right △ABC with right angle C, BC = 44, AC = 117, AB = 125. Compute sin A, cos A, sin B and cos B (44/125, 117/125, 117/125, 44/125)
Calculator: choose three angle pairs that add to 90° (not the ones used in class) and record sine and cosine values to four decimal places
Letters: label the legs a and b and the hypotenuse c, write all four ratios, and use the angle sum to write the general statement
Procedure
Each member works for 8 minutes on their role
Members take turns explaining their work to the group in 1 minute each
The group writes one paragraph that uses all three pieces of evidence, then posts it
Discussion Questions
Which of the three explanations proves the fact for every right triangle? Why are the other two only evidence?
What role does the angle sum of a triangle play in the letters explanation?
3
Solve and Check Relay
15 minGroups of 3-4
Groups solve six equation cards in which a sine equals a cosine. For each card, one member sets up the angle-sum equation, the next solves it, and the next checks that both angles are acute and add to 90°.
Equation Cards
sin(2x)° = cos(x + 36)°: x = 18, angles 36° and 54°
cos(5x - 4)° = sin(3x + 6)°: x = 11, angles 51° and 39°
sin(x + 14)° = cos(4x - 9)°: x = 17, angles 31° and 59°
sin(6x)° = cos(3x)°: x = 10, angles 60° and 30°
sin(x + 50)° = cos(x + 60)°: x = -10, angles 40° and 50°
sin(3x + 40)° = cos(x + 30)°: x = 5, angles 55° and 35°
Procedure
Roles rotate after each card
The checker substitutes x into both expressions and confirms with a calculator that the sine and the cosine match
Groups flag any card with a surprising answer and discuss it with the class
Discussion Questions
On the fifth card, x is negative. Is that a problem? What matters: x or the angles?
Why is it wrong to set the two expressions equal to each other?
Challenge Variation
Going beyond the standard: in a right triangle, compare tan A and tan B. Show that tan A · tan B = 1, and explain why the tangent does not have the same complement relationship as sine and cosine.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: One Side, Two Angles
A right triangle drawn to scale with θ = 35° at A, so the angle at B is 55°. Side a is opposite θ and is also the leg next to the angle 90° - θ, and both ratios use the same hypotenuse c. That is why sin θ = cos(90° - θ).
Diagram 2: Sine and Cosine from 0° to 90°
Graphs of y = sin x and y = cos x for 0° ≤ x ≤ 90°, drawn to scale from calculator values. The cosine curve is the sine curve read from the other end: sin 20° and cos 70° are at the same height, about 0.342.
04
Homework Assignment
~30 min
HSG.SRT.C.7 Homework: Complementary Angles
Directions: Show all work. Do not use a calculator unless a problem says you may. For every equation you solve, check that both angles are acute and add to 90°.
Part 1: Explaining the Relationship (Problems 1-2)
In right △XYZ, the right angle is at Y, XY = 15, YZ = 112 and XZ = 113. Find sin X and cos Z. Explain, using the sides of the triangle, why the two values are equal.
Two students make errors. Mia writes cos 40° = sin 40°. Leo writes sin 40° = cos 140° "because the angles are supplementary." Explain each error and write a correct statement that uses the complement of 40°.
Part 2: Rewriting and Evaluating (Problems 3-4)
Rewrite each expression using the complementary angle: (a) sin 16° (b) cos 73° (c) sin 45° (d) cos 89°
A table gives cos 52° ≈ 0.6157 and sin 52° ≈ 0.7880. Without a calculator, find sin 38° and cos 38°, and explain how you know.
Part 3: Solving and Applying (Problems 5-6)
For acute angles, solve sin(4x - 2)° = cos(2x + 14)°. Find both angles and check that their sine and cosine match.
A 16-foot ladder leans against a vertical wall and makes a 68° angle with the level ground. (a) What angle does the ladder make with the wall? (b) Show that (height reached on the wall)/(ladder length) equals both sin 68° and the cosine of your angle from part (a). (c) You may use a calculator: how high up the wall does the ladder reach, to the nearest tenth of a foot?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Explanation
Names the shared side and hypotenuse and uses the 90° angle sum
States the rule without a reason from the triangle
No explanation
Complements
All complements correct (subtracting from 90°)
One complement wrong
Uses 180° or equal angles
Equations
Correct x, both angles found and checked
Correct x without a check
Sets the expressions equal or no solution
Application
Correct wall angle, both ratios shown, height correct
Two of the three parts correct
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. 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 expression is equal to sin 28°?
Answer: C
28° and 62° are complementary (28 + 62 = 90), so sin 28° = cos 62°. Choice A uses the complement but keeps the sine, which gives a different value. Choice D uses the supplement, 180° - 28°, instead of the complement.
Question 2 of 20 · Multiple Choice
Which expression is equal to cos 81°?
Answer: A
The complement of 81° is 9°, and cos θ = sin(90° - θ), so cos 81° = sin 9°. Choice B assumes sine and cosine of the same angle are equal. Choice D uses the complement but keeps the cosine.
Question 3 of 20 · Multiple Choice
In right △ABC with the right angle at C, which statement is always true?
Answer: D
The leg opposite ∠A is the leg next to ∠B, so sin A and cos B are the same ratio. Choices A, B and C are true only when A and B are both 45°.
Question 4 of 20 · Multiple Choice
Which reason best explains why sin A = cos B in right △ABC with the right angle at C?
Answer: B
sin A = BC/AB and cos B = BC/AB: the same side over the same hypotenuse. Choice A is true but does not explain the equality. Choices C and D are false in general: in a 3-4-5 triangle the acute angles differ and sin A ≠ cos A.
Question 5 of 20 · Multiple Choice
A table gives sin 58° ≈ 0.8480. What is cos 32°?
Answer: A
32° = 90° - 58°, so cos 32° = sin 58° ≈ 0.8480. Choice B subtracts the value from 1, which has no meaning here. Choice C is cos 58°, and choice D is the reciprocal 1/0.8480.
Question 6 of 20 · Multiple Choice
For acute angles, solve sin(x + 22)° = cos(3x)°.
Answer: C
The angles must be complementary: (x + 22) + 3x = 90, so 4x = 68 and x = 17. The angles are 39° and 51°, which add to 90°. Choice A sets the two angles equal. Choice B moves 22 to the wrong side (4x = 112). Choice D uses 180 instead of 90.
(2x + 5) + (x - 5) = 90, so 3x = 90 and x = 30. The angles are 65° and 25°. Choice A uses 180 instead of 90. Choice C sets the two expressions equal. Choice D gives one of the angles instead of x.
Question 8 of 20 · Multiple Choice
For an acute angle θ, sin θ = cos 47°. What is θ?
Answer: D
The sine of an angle equals the cosine of its complement, so θ = 90° - 47° = 43°. Choice A assumes the angles are equal. Choices B and C are not acute.
Question 9 of 20 · Multiple Choice
Which equation is false?
Answer: A
35° and 65° add to 100°, not 90°, so they are not complementary and sin 35° ≈ 0.574 while cos 65° ≈ 0.423. The other three pairs add to 90°. Choice C is true because 45° is its own complement.
Question 10 of 20 · Multiple Choice
For an acute angle A in a right triangle, sin A = cos A. What kind of right triangle is it?
Answer: C
sin A = cos A means opposite/hypotenuse = adjacent/hypotenuse, so the two legs are equal and the triangle is an isosceles right triangle. Choice B fails: in a 30-60-90 triangle, sin 30° = 1/2 and cos 30° = √3/2. Choice A is not a right triangle.
Question 11 of 20 · Multiple Choice
In right △ABC with the right angle at C, BC = 65, AC = 72 and AB = 97. What is cos B?
Answer: D
The leg next to ∠B is BC = 65 and the hypotenuse is 97, so cos B = 65/97. This is also sin A, as the complement relationship predicts. Choice A is sin B. Choice B is tan A.
Question 12 of 20 · Multiple Choice
For an acute angle θ, which expression equals cos(90° - θ)?
Answer: B
The cosine of the complement of θ is the sine of θ: cos(90° - θ) = sin θ. Choice A would mean an angle and its complement always have the same cosine. Choice D subtracts values instead of angles.
Question 13 of 20 · Multiple Choice
A student writes sin 18° = cos 162°. What is the error?
Answer: C
The student used the supplement (180° - 18° = 162°) instead of the complement (90° - 18° = 72°). Choice B keeps the sine on both sides, which is false. Choice D still uses 180°.
Question 14 of 20 · Multiple Choice
For an acute angle A, cos A = 0.36. What is sin(90° - A)?
Answer: A
sin(90° - A) = cos A = 0.36. Choice B computes 1 - 0.36. Choice C is sin A, found from sin²A + cos²A = 1, which is not what was asked. Choice D is 1/0.36.
Question 15 of 20 · Short Answer
Use a labeled right triangle to explain why sin 54° = cos 36°.
Draw right △ABC with the right angle at C and m∠A = 54°. Then m∠B = 36° because the acute angles of a right triangle add to 90°. Side BC is opposite ∠A and is the leg next to ∠B, so sin 54° = BC/AB and cos 36° = BC/AB. Same side over the same hypotenuse, so the values are equal.
Question 16 of 20 · Short Answer
For acute angles, solve sin(5x + 1)° = cos(2x + 26)°. Find both angles and check them.
(5x + 1) + (2x + 26) = 90, so 7x + 27 = 90, 7x = 63 and x = 9. The angles are 46° and 44°, both acute, and 46 + 44 = 90. A calculator confirms sin 46° ≈ cos 44° ≈ 0.719.
Question 17 of 20 · Short Answer
A right triangle has legs a and b and hypotenuse c, with ∠A opposite a and ∠B opposite b. Use these letters to show that sin A = cos B and cos A = sin B, and explain why this means sin θ = cos(90° - θ).
sin A = a/c and cos B = a/c (a is the leg next to ∠B), so sin A = cos B. Also cos A = b/c and sin B = b/c, so cos A = sin B. Because the angle sum is 180° and one angle is 90°, m∠B = 90° - m∠A. Replacing B gives sin A = cos(90° - A) for any acute angle A.
Question 18 of 20 · Short Answer
A table lists sin 17° ≈ 0.2924 and cos 17° ≈ 0.9563. Without a calculator, find cos 73° and sin 73°. Explain.
73° is the complement of 17°. So cos 73° = sin 17° ≈ 0.2924 and sin 73° = cos 17° ≈ 0.9563. The sine of an angle is the cosine of its complement, and the reverse.
Question 19 of 20 · Short Answer
A vertical flagpole stands on level ground, and the sun's rays make a 64° angle with the ground. What angle do the rays make with the flagpole? Explain why the sine of one of these angles equals the cosine of the other.
The pole, its shadow and a ray from the top of the pole form a right triangle with the right angle at the base of the pole. Its acute angles are 64° (at the tip of the shadow) and 26° (at the top of the pole), which add to 90°. The pole's height is opposite the 64° angle and next to the 26° angle, so sin 64° = cos 26° (both ≈ 0.899).
Question 20 of 20 · Short Answer
Solve sin(3x)° = cos(x - 30)° by setting the angle sum equal to 90. Can the resulting angles be the two acute angles of a right triangle? Explain.
3x + (x - 30) = 90, so 4x = 120 and x = 30. The angles are 3(30) = 90° and 30 - 30 = 0°. The numbers do satisfy the equation (sin 90° = 1 = cos 0°), but 0° and 90° are not acute angles, so they cannot be the acute angles of a right triangle. This is why the check step matters.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.SRT.C.7 mean?
HSG.SRT.C.7 means students can explain and use the rule that the sine of an acute angle equals the cosine of its complement: sin θ = cos(90° - θ), and cos θ = sin(90° - θ). The explanation uses one right triangle, whose two acute angles add to 90°, and the use includes rewriting, evaluating and solving equations.
Is HSG.SRT.C.7 in Geometry or Algebra 2?
It is usually taught in Geometry, right after sine, cosine and tangent are defined (HSG.SRT.C.6). In Algebra 2 or Precalculus, the same idea returns as the identity sin(π/2 - x) = cos x for all real x.
Why is it called cosine?
The name is short for "complement's sine." The cosine of an angle is the sine of its complement, which is exactly the relationship in this standard. Sharing this with students helps them remember which way the rule goes.
Does sin x = cos y always mean x + y = 90?
Yes, when x and y are both acute angles, which is the setting of this standard. For angles outside 0° to 90°, there are other solutions, so the rule x + y = 90 is not the whole story there. That is why every solution in this lesson ends with a check that both angles are acute.
How do you solve an equation like sin(2x + 10)° = cos(3x + 5)°?
Set the two angle expressions to add to 90: (2x + 10) + (3x + 5) = 90, so 5x + 15 = 90 and x = 15. Then check: the angles are 40° and 50°, both acute, and they add to 90°. Setting the two expressions equal to each other is a frequent error.
What mistakes do students make with complementary angles?
Common errors include subtracting from 180 instead of 90, writing sin 25° = cos 25° (equal angles instead of complementary ones), keeping the same function on both sides (sin 20° = sin 70°), and forgetting to check that the solutions of an equation give acute angles.
Does tangent have a complementary-angle rule too?
Yes, but it is different and goes beyond this standard. In a right triangle, tan A = a/b and tan B = b/a, so tan A · tan B = 1. The tangent of an angle is the reciprocal of the tangent of its complement, not equal to it.
How is this standard tested?
Typical items ask students to pick the expression equal to a given sine, to find a cosine from a given sine of the complement, or to solve for x in an equation like sin(x + 12)° = cos(2x)°. Explanation items ask why sin A = cos B in a labeled right triangle. The digital SAT covers right-triangle trigonometry in its Geometry and Trigonometry domain.
Why do students need to explain it and not only use it?
The standard says "explain and use." The explanation is short but important: the leg opposite one acute angle is the leg next to the other, and the hypotenuse is shared. Students who understand this can rebuild the rule instead of memorizing which way it goes.
How can parents help with this topic at home?
Ask your student to draw any right triangle, label the two acute angles, and explain which side is "opposite" from each angle. If they can show that one side plays both roles, they understand the idea. A calculator check of a pair such as sin 30° and cos 60° also helps.
07
Related Standards
5 standards
These standards connect to HSG.SRT.C.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.G.B.5Prerequisite
Use supplementary, complementary, vertical and adjacent angles to find unknown angles
