HSF.TF.B.6: Restricting Domains to Build Inverse Trigonometric Functions
In plain English: HSF.TF.B.6 is an advanced (+) Common Core functions standard about why trigonometric functions need a restricted domain before they have inverses. Sine, cosine and tangent repeat, so each output comes from many inputs. On an interval where the function is always increasing or always decreasing, each output comes from one input, so the inverse exists. It is usually taught in Precalculus.
(+) Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed.
Common Core State Standards for Mathematics · Domain: Trigonometric Functions (TF) · Cluster: Model periodic phenomena with trigonometric functions Also written as HSF-TF.B.6 or F-TF.6 · Official standard
Students learn why sine, cosine and tangent do not have inverse functions on their whole domains, and how restricting each one to an interval where it is always increasing or always decreasing makes an inverse possible. The central idea is one-to-one: a function has an inverse exactly when each output comes from only one input, and a function that always increases (or always decreases) on an interval never repeats an output there.
Students find the intervals where each function is increasing or decreasing, see why the standard choices are [-π/2, π/2] for sine, [0, π] for cosine and (-π/2, π/2) for tangent, and build the graphs of the inverses by reflecting the restricted graphs across y = x. They use the restricted domains to evaluate inverse values and to explain why sin⁻¹(sin x) is not always x.
Learning Objectives
By the end of this lesson, students will be able to:
Explain why a periodic function cannot have an inverse on its whole domain
Explain why a function that is always increasing or always decreasing on an interval is one-to-one there and so has an inverse on that interval
Identify the standard restricted domains of sine, cosine and tangent and the domain and range of each inverse
Evaluate inverse trigonometric values by choosing the one angle in the restricted domain
Explain why sin⁻¹(sin x) = x only for x in the restricted domain
Prior Knowledge Required
Students should already be comfortable with:
Inverse functions and restricting a domain to make a function invertible, such as x² for x ≥ 0 HSF.BF.B.4
Sine, cosine and tangent as unit circle coordinates and their exact values HSF.TF.A.2
Periodicity and symmetry of the trigonometric functions HSF.TF.A.4
Intervals where a function is increasing or decreasing, read from a graph HSF.IF.B.4
Start with a function students already know, then move to sine:
Warm-Up Prompt
"(1) Why does f(x) = x² have no inverse function on all real numbers, but g(x) = x² for x ≥ 0 does? (2) Find three different values of x with sin x = 1/2. (3) If a calculator key called sin⁻¹ is pressed with 1/2, which of your three values should it give, and why?"
For (1), students should say that f(3) = f(-3) = 9, so "undo 9" has two answers, and that restricting to x ≥ 0 keeps only one. For (2), accept any three of π/6, 5π/6, 13π/6, -7π/6 and so on. Leave (3) open: the lesson answers it. Point out that (1) and (3) are the same problem, and that the fix for x², keeping a piece where the function only goes up, is the fix for sine too.
Direct Instruction20 minutes
Use Diagram 1 to show the problem and Diagram 2 to show the fix. Build the argument in steps:
An inverse needs one-to-one: f⁻¹(y) must name the one input with output y. If two inputs share an output, f⁻¹ is not a function. On a graph, every horizontal line must meet the graph at most once.
Trigonometric functions fail: sine and cosine have period 2π and tangent has period π, so every output repeats infinitely often. In Diagram 1, the line y = 1/2 meets y = sin x at π/6, 5π/6, -7π/6 and many more points.
Always increasing means one-to-one: if f is always increasing on an interval, then a < b gives f(a) < f(b), so two different inputs can never give the same output. The same is true for always decreasing, with f(a) > f(b).
Choose the interval: sine is always increasing on [-π/2, π/2] and takes every value from -1 to 1 there. Cosine is always decreasing on [0, π] and takes every value from -1 to 1. Tangent is always increasing on (-π/2, π/2) and takes every real value.
Build the inverse: swap inputs and outputs. sin⁻¹ has domain [-1, 1] and range [-π/2, π/2]; cos⁻¹ has domain [-1, 1] and range [0, π]; tan⁻¹ has domain all real numbers and range (-π/2, π/2). The graph of each inverse is the reflection of the restricted graph across y = x (Diagram 2).
Discuss the choice: sine is also always decreasing on [π/2, 3π/2], and that interval would give a different inverse. The standard intervals are chosen because they are connected, contain the first-quadrant angles from 0 to π/2, and give every output of the function exactly once. Then work these examples:
Why a restriction is needed
Both π/6 and 5π/6 have sine 1/2. Which one is sin⁻¹(1/2)?
Equation: Only π/6 lies in [-π/2, π/2], so sin⁻¹(1/2) = π/6
Inverse sine of a negative value
Find sin⁻¹(-√2/2). The angles 5π/4 and 7π/4 also have this sine.
Equation: sin⁻¹(-√2/2) = -π/4, the only angle in [-π/2, π/2] with this sine
Inverse cosine of a negative value
Find cos⁻¹(-1/2). The angle -2π/3 also has this cosine.
Equation: cos⁻¹(-1/2) = 2π/3, the only angle in [0, π] with this cosine
Inverse tangent
Find tan⁻¹(-1). The angle 3π/4 also has tangent -1.
Equation: tan⁻¹(-1) = -π/4, the only angle in (-π/2, π/2) with tangent -1
Undoing outside the restricted domain
Find sin⁻¹(sin(5π/6)). The input 5π/6 is outside [-π/2, π/2].
Equation: sin(5π/6) = 1/2, and sin⁻¹(1/2) = π/6, so sin⁻¹(sin(5π/6)) = π/6, not 5π/6
Close the segment with the answer to warm-up question (3): the calculator gives π/6 (about 0.524 in radian mode) because sin⁻¹ is defined as the inverse of the restricted sine. Stress the notation too: sin⁻¹x means the inverse function, not 1/sin x.
Guided Practice15 minutes
Pairs work through three tasks and justify each answer with an interval. (1) Is cosine one-to-one on [-π/2, π/2]? (No: cos(-π/3) = cos(π/3) = 1/2, because cosine increases and then decreases on that interval.) (2) Evaluate sin⁻¹(-√3/2) = -π/3, cos⁻¹(√2/2) = π/4 and tan⁻¹(√3) = π/3, and for each one name another angle with the same function value that the inverse does not give. (3) Evaluate tan⁻¹(tan(3π/4)). (Answer: tan(3π/4) = -1 and tan⁻¹(-1) = -π/4.) Listen for answers outside the range, such as cos⁻¹(√2/2) = -π/4, and for students who read sin⁻¹ as 1/sin.
Independent Practice15 minutes
Students work alone. (1) Complete a table with the restricted domain, the range, and the domain and range of the inverse, for sine, cosine and tangent. (2) Evaluate sin⁻¹(√2/2) = π/4, cos⁻¹(√3/2) = π/6 and tan⁻¹(1) = π/4. (3) Evaluate sin⁻¹(sin(4π/3)). (Answer: sin(4π/3) = -√3/2, so the value is -π/3.) (4) Explain in two sentences why [-π/2, π/2] would not work as the restricted domain for cosine. (Cosine increases on [-π/2, 0] and decreases on [0, π/2], so it repeats values, and it never takes negative values there.)
Closure5-10 minutes
Exit ticket: (1) Why is the range of sin⁻¹ the interval [-π/2, π/2]? (It is the restricted domain of sine.) (2) Find cos⁻¹(0). (Answer: π/2.) (3) Challenge: find sin⁻¹(sin 3) exactly. (Answer: π - 3, about 0.142, because π - 3 is in [-π/2, π/2] and sin(π - 3) = sin 3.)
Differentiation Strategies
For Struggling Students
Give printed graphs and have students run a ruler across them as a horizontal line test before and after shading the restricted piece
Provide a reference card with each restricted domain drawn as an arc on the unit circle: the right half for sine and tangent, the upper half for cosine
Start inverse evaluations with the question "Which angles have this value?" and then "Which one is in the interval?"
For Advanced Students
Ask students to build the inverse of sine restricted to [π/2, 3π/2] and to express it in terms of sin⁻¹
Ask students to explain why sin(sin⁻¹x) = x for every x in [-1, 1], while sin⁻¹(sin x) = x only on [-π/2, π/2]
Ask students to sketch y = sin⁻¹(sin x) on [-2π, 2π] and describe its shape
Assessment Guidance
What to Look For
Students should connect three ideas in their own words: trigonometric functions repeat outputs, so they are not one-to-one; a function that is always increasing or always decreasing on an interval is one-to-one there; and the inverse is defined from that restricted piece, so its range is the restricted domain. When evaluating, look for answers checked against the range, such as rejecting 5π/4 for sin⁻¹(-√2/2). A student who answers sin⁻¹(sin x) = x without checking the interval has not yet understood the restriction.
02
Classroom Activities
3 Activities
1
Find the One-to-One Pieces
15 minPairs
Pairs highlight every interval on which sine, cosine and tangent are always increasing or always decreasing, then decide which piece they would choose to build each inverse.
Procedure
On printed graphs over [-2π, 2π], highlight in one color the intervals where the function is always increasing and in another color where it is always decreasing
For sine, students should find increasing pieces such as [-π/2, π/2] and [3π/2, 2π] and decreasing pieces such as [π/2, 3π/2]; for cosine, decreasing on [0, π] and increasing on [π, 2π]; for tangent, increasing on each piece between asymptotes
Run a ruler across each highlighted piece as a horizontal line test
For each function, list two pieces that give every output exactly once, then choose one and explain the choice
Discussion Questions
Why does a piece such as [0, π/2] for sine pass the horizontal line test but still make a poor choice?
Why can the chosen piece for cosine not be the same interval as the one for sine?
Why do the tangent pieces have open endpoints?
Modification for Distance Learning
Students graph each function in a free online graphing calculator, add a domain restriction such as {-π/2 ≤ x ≤ π/2} to the equation, and drag a horizontal line to test the restricted graph.
2
Allowed or Not? Card Sort
20 minGroups of 3-4
Groups sort 10 statements about inverse trigonometric functions into "true" and "false". For every false card, they write the correct value and name the restricted domain that rules the statement out.
The 10 Cards
sin⁻¹(1) = π/2 (true)
sin⁻¹(0) = π (false: 0)
cos⁻¹(1) = 0 (true)
cos⁻¹(-1) = -π (false: π)
tan⁻¹(-√3/3) = 5π/6 (false: -π/6)
cos⁻¹(√2/2) = -π/4 (false: π/4)
sin⁻¹(2) is undefined (true: 2 is not in [-1, 1])
tan⁻¹(1000) is a little less than π/2 (true: about 1.5698)
cos⁻¹(cos(3π/2)) = 3π/2 (false: π/2)
sin(sin⁻¹(0.3)) = 0.3 (true)
Procedure
Each student draws a card in turn, reads it aloud and makes a claim; the group must agree before the card is placed
For each false card, the group writes the correct value and sketches the restricted graph with the point marked
Groups check two of their decimal answers on a graphing calculator in radian mode
Challenge Variation
Groups write three new cards that look true but are false, trade them with another group, and explain each error in terms of the restricted domain.
3
Build a Different Inverse
20 minGroups of 3-4
Groups restrict sine to [π/2, 3π/2], where it is always decreasing, and build the inverse of that piece. Comparing it with the usual sin⁻¹ shows that the restriction, not the sine function alone, decides the inverse.
Procedure
Call the restricted function S(x) = sin x for π/2 ≤ x ≤ 3π/2. Check on the graph that it is always decreasing and takes every value in [-1, 1]
Make a table for S: (π/2, 1), (5π/6, 1/2), (π, 0), (7π/6, -1/2), (3π/2, -1)
Swap the columns to get a table for the inverse S⁻¹, and graph both S and S⁻¹ on tracing paper by reflecting across y = x
Compare S⁻¹(1/2) = 5π/6 with sin⁻¹(1/2) = π/6, and find the rule that connects them: S⁻¹(y) = π - sin⁻¹(y)
Discussion Questions
Is S⁻¹ a correct inverse? What makes it one?
Why do calculators and textbooks use [-π/2, π/2] instead of [π/2, 3π/2]?
What would go wrong if you restricted sine to [0, π] instead?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Sine Fails the Horizontal Line Test Until It Is Restricted
The graph of y = sin x on [-2π, 2π], drawn to scale. The line y = 1/2 meets the full graph again and again, so sine has no inverse on all real numbers. On [-π/2, π/2] (thick line), sine is always increasing and meets the line once, at x = π/6.
Diagram 2: The Restricted Sine and Its Inverse
The restricted sine (solid) and y = sin⁻¹x (dashed) are reflections of each other across y = x, drawn to scale with equal units on both axes. The domain of one is the range of the other.
04
Homework Assignment
~30 min
HSF.TF.B.6 Homework: Restricted Domains and Inverse Trigonometric Functions
Directions: Give exact values in radians unless a problem asks for a decimal. For every inverse value, name the interval that the answer must lie in. When a problem asks you to explain, refer to where the function is increasing or decreasing.
Part 1: Why Restrict? (Problems 1-2)
For each interval, decide whether y = sin x is one-to-one there, and explain using increasing or decreasing: (a) [π, 3π/2] (b) [-π, 0] (c) [-3π/2, -π/2] (d) [3π/2, 5π/2]. Which of these intervals give every value from -1 to 1?
Evaluate exactly and name one other angle with the same function value that the inverse does not return: (a) sin⁻¹(-1/2) (b) cos⁻¹(-√3/2) (c) tan⁻¹(√3/3)
Part 2: Compositions (Problems 3-4)
Evaluate each expression exactly, and explain why the answer is not the original angle: (a) sin⁻¹(sin(3π/4)) (b) cos⁻¹(cos(4π/3)) (c) tan⁻¹(tan(4π/5))
Let g(x) = cos x on the interval [-π, 0]. Explain why g has an inverse. Give the domain and range of g⁻¹, and find g⁻¹(0), g⁻¹(1/2) and g⁻¹(-1).
Part 3: Graphs and Technology (Problems 5-6)
Make a table of at least four points on y = cos x for 0 ≤ x ≤ π, including (1, 0.540), rounded to 3 decimal places. Use it to graph the restricted cosine and y = cos⁻¹x on the same axes, with the line y = x. State the domain and range of each graph.
A calculator gives sin⁻¹(0.8) ≈ 0.927. Find every x in [0, 2π] with sin x = 0.8, to 3 decimal places, and explain why the calculator shows only one of them.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
One-to-One Reasoning
Uses increasing or decreasing to justify each decision
Correct decisions without reasons
Incorrect decisions
Restricted Domains and Ranges
All domains and ranges correct
One or two errors
Missing or mostly incorrect
Inverse Values
Exact values in the correct range
Correct angles, some outside the range
Most values incorrect
Graphs and Explanations
Accurate reflection across y = x and clear explanations
Graph or explanation incomplete
Missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Choose an answer, then read the explanation. Give angles in radians. The score counts 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
Why does y = sin x have no inverse function on all real numbers?
Answer: A
Sine repeats every 2π, so a horizontal line such as y = 0.3 meets the graph infinitely often and "undo 0.3" has no single answer. Choice B is true, but a limited range only limits the domain of the inverse. Choice C is not a reason: y = x³ is odd and has an inverse. Choice D is false: sine is continuous.
Question 2 of 20 · Multiple Choice
On which interval is sine restricted to define sin⁻¹?
Answer: B
Sine is always increasing on [-π/2, π/2] and takes every value from -1 to 1 there, including the endpoints, so the interval is closed. Choice A is the interval used for cosine; sine rises and then falls on it. Choice D is the open interval used for tangent, and it would leave out the outputs -1 and 1.
Question 3 of 20 · Multiple Choice
On which interval is cosine restricted to define cos⁻¹?
Answer: C
Cosine is always decreasing on [0, π], from 1 to -1, so each output appears once. Choice A fails: cosine increases and then decreases there, and it never takes negative values. Choices B and D contain a full period, so outputs repeat.
Question 4 of 20 · Multiple Choice
Find sin⁻¹(√3/2).
Answer: C
sin(π/3) = √3/2 and π/3 is in [-π/2, π/2], so sin⁻¹(√3/2) = π/3. Choice B also has sine √3/2 but lies outside the restricted domain. Choice A mixes up the special angles: sin(π/6) = 1/2.
Question 5 of 20 · Multiple Choice
Find cos⁻¹(-√2/2).
Answer: A
cos(3π/4) = -√2/2 and 3π/4 is in [0, π]. Choice B uses the range of sin⁻¹ instead of cos⁻¹, and cos(-π/4) is positive anyway. Choice C has the right cosine but lies outside [0, π]. Choice D drops the negative sign.
Question 6 of 20 · Multiple Choice
Find tan⁻¹(-√3).
Answer: D
tan(-π/3) = -√3 and -π/3 is in (-π/2, π/2). Choice A also has tangent -√3 but is outside the restricted domain; a common error is to use a second-quadrant angle as for cosine. Choice B is tan⁻¹(-√3/3).
Question 7 of 20 · Multiple Choice
What is the range of y = cos⁻¹x?
Answer: C
The range of the inverse is the restricted domain of cosine, [0, π]. Choice A is the domain of cos⁻¹, the outputs of cosine. Choice B is the range of sin⁻¹.
Question 8 of 20 · Multiple Choice
Why is [0, π] not used as the restricted domain of sine?
Answer: B
On [0, π] sine rises to 1 and falls back, so it is not always increasing or always decreasing and fails the horizontal line test. It also never takes negative values there. Choice C gets the idea backward: a full period guarantees repeated outputs. Choice D is false: sine is nonnegative on [0, π].
Question 9 of 20 · Multiple Choice
Find sin⁻¹(sin(4π/5)).
Answer: B
sin(4π/5) = sin(π - 4π/5) = sin(π/5), and π/5 is in [-π/2, π/2], so sin⁻¹(sin(4π/5)) = π/5. Choice A assumes sin⁻¹ always undoes sin, which is true only inside [-π/2, π/2]. Choice C has the wrong sign: sin(-π/5) is negative. Choice D is sin(4π/5) ≈ 0.588, the inside step, not the final answer.
Question 10 of 20 · Multiple Choice
Find cos⁻¹(cos(-π/4)).
Answer: D
cos(-π/4) = √2/2, and cos⁻¹(√2/2) = π/4, because the answer must be in [0, π]. Choice A is outside the range of cos⁻¹. Choice B has a negative cosine, the wrong sign.
Question 11 of 20 · Multiple Choice
On which interval does cosine also have an inverse, because it is always increasing there?
Answer: B
On [3π, 4π] cosine rises from -1 to 1 without turning back, so it is one-to-one and takes every value in [-1, 1]. Choice A is a full period. Choices C and D each contain a turning point (at 0 and at π), so cosine increases on part of the interval and decreases on the rest.
Question 12 of 20 · Multiple Choice
What is the domain of y = sin⁻¹x?
Answer: A
The inputs of sin⁻¹ are the outputs of the restricted sine, which are the numbers from -1 to 1. Choice B is the range of sin⁻¹. Choice C is the domain of tan⁻¹; sin⁻¹(2) is undefined because no angle has sine 2.
Question 13 of 20 · Multiple Choice
Why is the restricted domain of tangent the open interval (-π/2, π/2) rather than a closed one?
Answer: C
tan x = sin x / cos x, and cos(±π/2) = 0, so the endpoints are not in the domain. On the open interval tangent is always increasing and takes every real value. Choice B is false: tan 0 = 0, and tangent is undefined at ±π/2. Choice D is false: tangent has period π.
Question 14 of 20 · Multiple Choice
Find sin⁻¹(-1).
Answer: D
sin(-π/2) = -1 and -π/2 is in [-π/2, π/2]. Choice A has sine -1 but lies outside the restricted domain. Choices B and C have sine 0, not -1.
Question 15 of 20 · Short Answer
Explain why restricting sine to [-π/2, π/2] allows its inverse to be constructed.
On [-π/2, π/2], sine is always increasing, so if a < b then sin a < sin b: no two inputs share an output, and the restricted sine is one-to-one. It also takes every value from -1 to 1 there. So each y in [-1, 1] comes from exactly one angle in the interval, and sin⁻¹(y) is defined as that angle.
Question 16 of 20 · Short Answer
Find sin⁻¹(sin(-2π/3)) exactly.
sin(-2π/3) = -√3/2. The angle in [-π/2, π/2] with sine -√3/2 is -π/3, so sin⁻¹(sin(-2π/3)) = -π/3. The answer is not -2π/3 because -2π/3 is outside the restricted domain.
Question 17 of 20 · Short Answer
A student says cos⁻¹(0.5) could be 5π/3, because cos(5π/3) = 0.5. Explain the error and give the correct value.
cos(5π/3) is 0.5, but cos⁻¹ is the inverse of cosine restricted to [0, π], so its output must be in [0, π]. 5π/3 is not. The only angle in [0, π] with cosine 0.5 is π/3, so cos⁻¹(0.5) = π/3.
Question 18 of 20 · Short Answer
Tangent is also always increasing on (-3π/2, -π/2). Suppose a different inverse T⁻¹ is built from that piece. What are its domain and range, and what is T⁻¹(1)?
On (-3π/2, -π/2) tangent takes every real value once, so T⁻¹ has domain all real numbers and range (-3π/2, -π/2). The angle in that interval with tangent 1 is -3π/4, so T⁻¹(1) = -3π/4, while the usual tan⁻¹(1) is π/4.
Question 19 of 20 · Short Answer
The restricted sine passes through (-π/2, -1), (0, 0), (0.5, 0.479) and (π/2, 1), with values rounded to 3 decimal places. Give the matching points on y = sin⁻¹x, and explain how you found them.
Swap the coordinates: (-1, -π/2), (0, 0), (0.479, 0.5) and (1, π/2). The inverse sends each output of the restricted sine back to its input, so its graph is the reflection of the restricted graph across the line y = x. For example, sin⁻¹(0.479) ≈ 0.5.
Question 20 of 20 · Short Answer
Tangent is always increasing on every interval between two consecutive asymptotes. Give two such intervals, and explain why (-π/2, π/2) is the one used for tan⁻¹.
For example, (-π/2, π/2) and (π/2, 3π/2); any interval (-π/2 + kπ, π/2 + kπ) works. Each gives every real output once, so each would give an inverse. The interval (-π/2, π/2) is used because it contains 0 and the first-quadrant angles, so tan⁻¹ of a positive number is an angle between 0 and π/2, and tan⁻¹(0) = 0.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.TF.B.6 mean?
HSF.TF.B.6 means understanding why a trigonometric function must be restricted before it has an inverse. Sine, cosine and tangent repeat, so an output such as 1/2 comes from infinitely many angles. On an interval where the function only increases or only decreases, each output comes from one angle, and the inverse function returns that angle. The "(+)" marks it as an advanced standard.
Is HSF.TF.B.6 taught in Precalculus?
Yes, usually. Because it is a (+) standard, it is typically part of Precalculus or a trigonometry course rather than Algebra II, though some Algebra II courses introduce sin⁻¹ on the calculator. It is often taught right before HSF.TF.B.7, where the inverses are used to solve equations.
Why do trig functions need a restricted domain to have an inverse?
Because an inverse function can give only one output for each input. Sine, cosine and tangent are periodic, so each of their outputs comes from many inputs. Restricting to an interval where the function is always increasing or always decreasing removes the repeats, and the inverse can then return a single angle.
Why is sine restricted to [-π/2, π/2] and cosine to [0, π]?
Each is an interval where the function is always increasing or always decreasing and takes every value from -1 to 1 exactly once. Sine increases on [-π/2, π/2]; cosine decreases on [0, π]. Cosine cannot use [-π/2, π/2], because it increases and then decreases there and never takes negative values. Both intervals contain the first-quadrant angles from 0 to π/2.
What is the difference between sin⁻¹x and 1/sin x?
They are different: sin⁻¹x is the inverse sine (also written arcsin x), which returns an angle, while 1/sin x is the cosecant, csc x. For example, sin⁻¹(1/2) = π/6, but 1/sin(1/2) ≈ 2.09. The -1 in sin⁻¹ is inverse-function notation, not an exponent, even though sin²x does mean (sin x)².
Why doesn't sin⁻¹(sin x) always equal x?
Because sin⁻¹ always returns an angle in [-π/2, π/2]. If x is in that interval, sin⁻¹(sin x) = x. If not, the answer is the angle in the interval with the same sine: sin⁻¹(sin(5π/6)) = π/6. In the other order, sin(sin⁻¹x) = x for every x in [-1, 1].
What are the domain and range of the inverse trig functions?
sin⁻¹: domain [-1, 1], range [-π/2, π/2]. cos⁻¹: domain [-1, 1], range [0, π]. tan⁻¹: domain all real numbers, range (-π/2, π/2). In every case the range of the inverse is the restricted domain of the original function, and the domain of the inverse is its range.
Could a different restricted domain be used?
Yes. Any interval where the function is always increasing or always decreasing and takes every value in its range once would give an inverse. Sine restricted to [π/2, 3π/2] gives an inverse whose value at 1/2 is 5π/6. The standard intervals are a shared convention, used by calculators and textbooks, so everyone gets the same answer.
What mistakes do students make with inverse trig functions?
A frequent one is giving an angle outside the range, such as cos⁻¹(-1/2) = -2π/3 or tan⁻¹(-1) = 3π/4. Others are reading sin⁻¹ as 1/sin, assuming sin⁻¹(sin x) = x for every x, and leaving the calculator in degree mode when radians are expected. Asking "is my answer in the range?" catches the first three.
How does HSF.TF.B.6 connect to other standards?
It applies the general idea from HSF.BF.B.4, making a function invertible by restricting its domain, to the trigonometric functions. It prepares students for HSF.TF.B.7, where they use inverse functions to solve trigonometric equations in modeling problems and must find the solutions that the calculator does not show. Inverse trigonometric functions also appear later in calculus.
07
Related Standards
5 standards
These standards connect to HSF.TF.B.6: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSF.BF.B.4Prerequisite
Find inverse functions, including by restricting a domain to make one invertible
