HSF.TF.B.7: Solving Trigonometric Equations in Modeling Contexts with Inverse Functions
In plain English: HSF.TF.B.7 is an advanced (+) Common Core functions standard, usually taught in Precalculus. Students solve trigonometric equations that come from models such as Ferris wheels, tides and daylight by applying inverse sine, cosine or tangent, evaluate the answers with a calculator, and decide what each solution means in the situation.
(+) Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context.
Common Core State Standards for Mathematics · Domain: Trigonometric Functions (TF) · Cluster: Model periodic phenomena with trigonometric functions Also written as HSF-TF.B.7 or F-TF.7 · Official standard
Students use inverse trigonometric functions to solve equations that come from periodic models: electrical voltage, a Ferris wheel, tides, air temperature and daylight. The core idea is that sin⁻¹, cos⁻¹ and tan⁻¹ each return only one angle, so students must use the symmetry and period of the model to find every solution in the time window the question asks about.
Technology does the evaluation: students compute inverse values in radian mode and confirm answers by graphing the model and a horizontal line. Every answer ends in a sentence about the context, including cases where a solution is outside the time window or where the equation has no solution because the model never reaches the value.
Learning Objectives
By the end of this lesson, students will be able to:
Rewrite a sinusoidal model equation as sin(u) = r or cos(u) = r and decide whether a solution exists
Use inverse sine, cosine and tangent to find one solution, and use symmetry and the period to find the others in a given interval
Evaluate inverse trigonometric values and check solutions with a calculator or graphing tool in radian mode
Interpret solutions in the units of the context, reject solutions outside the time window, and explain a model with no solution
Prior Knowledge Required
Students should already be comfortable with:
Radian measure and the unit circle HSF.TF.A.2
Modeling periodic situations with amplitude, frequency and midline HSF.TF.B.5
Restricting the domain of sine, cosine and tangent so that each has an inverse HSF.TF.B.6
Solving equations of the form f(x) = c using an inverse function HSF.BF.B.4
Put a graph of y = sin(x) on the board with the horizontal line y = 0.5 drawn across two full periods. Ask students to answer without a calculator:
Warm-Up Prompt
"The calculator says sin⁻¹(0.5) = π/6. How many times does the line y = 0.5 cross the curve between x = 0 and x = 4π? Why does the calculator give only one of them?"
Students should count four crossings: π/6, 5π/6, 13π/6 and 17π/6. Draw out the reason: sin⁻¹ is a function, so it returns exactly one value, the one in [-π/2, π/2]. The other solutions come from the symmetry of the graph (π - π/6) and from adding the period 2π. Tell students that every model in today's lesson works the same way: the inverse function gives one answer, and the context decides which of the others matter.
Direct Instruction20 minutes
Present a four-step routine for any model of the form y = A sin(B(t - h)) + k or y = A cos(B(t - h)) + k:
Isolate the trig function: subtract the midline k and divide by the amplitude A, so the equation reads sin(u) = r or cos(u) = r, where u is the whole argument.
Check that a solution exists: if r < -1 or r > 1, the model never reaches that value. Say so in words.
Apply the inverse function with technology: in radian mode, u₁ = sin⁻¹(r) or u₁ = cos⁻¹(r). Find the second angle in the cycle: π - u₁ for sine, 2π - u₁ for cosine.
Solve for t and interpret: undo B and h, keep the solutions inside the time window of the question, add or subtract whole periods when needed, and answer in the units of the context.
Sine model, first times in a cycle
A US wall outlet has voltage V(t) = 170 sin(120πt) volts, t in seconds. When in the first cycle is the voltage 100 V?
Equation: sin(120πt) = 100/170, so t = sin⁻¹(10/17)/(120π) ≈ 0.00167 s and t = (π - sin⁻¹(10/17))/(120π) ≈ 0.00667 s
Cosine model, interval above a height
A Ferris wheel seat has height h(t) = 33 - 30 cos(πt/10) meters, t in minutes. When during the first ride is the seat 50 m high?
Equation: cos(πt/10) = -17/30, so t ≈ 6.92 min and t ≈ 13.08 min; the seat is above 50 m for about 6.16 min
Tide model, time window
Harbor depth is d(t) = 3.5 + 1.5 cos(2πt/12.4) meters, t in hours after high tide. A boat needs 2.8 m of water.
Equation: cos(2πt/12.4) = -0.7/1.5, so t ≈ 4.06 h and t ≈ 8.34 h; the harbor is too shallow between them
Daylight model, dates
An invented model for a mid-latitude city: D(t) = 12.2 + 2.4 sin(2π(t - 80)/365) hours of daylight on day t of the year. When is there 14 hours of daylight?
Equation: sin(2π(t - 80)/365) = 0.75, so t ≈ 129.3 and t ≈ 213.2, about day 129 (May 9) and day 213 (August 1)
No solution, interpreted
Using the tide model above, when is the harbor 5.5 m deep?
Equation: cos(2πt/12.4) = 2/1.5 ≈ 1.33 > 1, so there is no solution: the maximum depth is 5 m
Use Diagram 2 to show where the second Ferris wheel time comes from, and Diagram 1 to show both times on the graph. For the voltage example, stress that the calculator must be in radian mode because the argument 120πt is in radians. In the daylight example, point out that a day number like 129.3 is rounded to day 129 and translated into a date, and that the model is invented for class use, not a forecast.
Guided Practice15 minutes
Pairs work one problem with the teacher checking each step. On a spring day, the temperature is modeled by T(t) = 58 + 12 sin(π(t - 10)/12) degrees Fahrenheit, t in hours after midnight. When is it 65°F? Students isolate sin(π(t - 10)/12) = 7/12, use technology to get sin⁻¹(7/12) ≈ 0.6228, find the second angle π - 0.6228 ≈ 2.5188, and solve t ≈ 12.38 and t ≈ 19.62. Ask pairs to translate: about 12:23 pm and 7:37 pm, and to state that it is warmer than 65°F between those times. Then have pairs graph T and the line y = 65 on a graphing calculator to confirm both intersections. Listen for these errors: dividing by 12 before subtracting 58, forgetting the second angle, and reporting 0.6228 as a time.
Independent Practice15-20 minutes
Students solve two models on their own and write one sentence of interpretation for each. (1) A buoy moves up and down with waves: y(t) = 1.2 sin(πt/3) meters from its rest level, t in seconds. When in the first period is the buoy 0.5 m above rest? (t ≈ 0.41 s and t ≈ 2.59 s.) (2) A simplified blood pressure model is P(t) = 100 + 20 sin(2.4πt) mm Hg, t in seconds. When in the first beat is the pressure 110 mm Hg? (sin(2.4πt) = 0.5, so t = 1/14.4 ≈ 0.069 s and t = 5/14.4 ≈ 0.347 s; the beat lasts 5/6 s, which is 72 beats per minute.) Students check each answer by substituting it back into the model with a calculator.
Closure5 minutes
Exit ticket: A point on a water wheel has height h(t) = 1.5 + 2 sin(πt/4) meters above the water, t in seconds. (1) Solve h(t) = 0 for 0 ≤ t < 8. (sin(πt/4) = -0.75; the calculator gives sin⁻¹(-0.75) ≈ -0.848, which is negative, so use π + 0.848 and 2π - 0.848 to get t ≈ 5.08 s and t ≈ 6.92 s.) (2) What happens to the point between those two times? (It is under water for about 1.84 s of each 8-second turn.)
Differentiation Strategies
For Struggling Students
Give a two-column organizer: the left column holds each algebra step, the right column holds the calculator keystrokes, including a mode check
Start with models that have no horizontal shift, such as y = 5 sin(t), before adding a midline and a period change
Have students sketch one period of the graph and the horizontal line first, so they know how many solutions to expect
For Advanced Students
Ask for all solutions of the Ferris wheel equation in the first hour, written as t ≈ 6.92 + 20k and t ≈ 13.08 + 20k
Ask students to find the fraction of each ride the seat spends above 50 m, and to explain why it does not depend on the ride number
Give a model with a horizontal shift and a negative amplitude, and ask students to explain why the second-angle rule still works
Assessment Guidance
What to Look For
Check that students isolate the trig function completely before applying an inverse, and that they state whether a solution exists. Look for the second solution in every cycle and for answers written in the units of the context (seconds, minutes, clock times or dates), not as bare angles. When students give a negative value from sin⁻¹ or tan⁻¹, ask how they turned it into a time inside the window. A strong answer checks at least one solution on a graph or by substitution.
02
Classroom Activities
3 Activities
1
County Fair Wheel Timeline
20 minPairs
Pairs build a timeline for one 40-second ride on a small fair wheel whose seat height is h(t) = 10 - 8 cos(πt/20) meters. Each height card leads to a cosine equation, and pairs place the solutions on a shared timeline.
Height Cards
14 m: cos(πt/20) = -1/2, so t = 40/3 ≈ 13.33 s and t = 80/3 ≈ 26.67 s
4 m: cos(πt/20) = 3/4, so t ≈ 4.60 s and t ≈ 35.40 s
17 m: cos(πt/20) = -7/8, so t ≈ 16.78 s and t ≈ 23.22 s
19 m: cos(πt/20) = -9/8, no solution, because the top of the wheel is 18 m
Procedure
For each card, one partner solves with cos⁻¹ and the second-angle rule while the other graphs h(t) and the horizontal line on a graphing calculator
Partners compare results and resolve any difference before placing the times on the timeline
Pairs shade the part of the timeline when the seat is above 14 m and state its length (about 13.33 s)
Discussion Questions
Why does every reachable height, except the top and the bottom, happen exactly twice per ride?
What does the calculator do with cos⁻¹(-9/8), and what does that tell a rider?
How would the times change on the second ride?
Modification for Distance Learning
Share a Desmos graph with a slider for the height. Pairs move the slider to each card value, read the intersection points, and compare them with their inverse-function answers in a shared document.
2
Technology Check: Radians, Degrees and Missing Solutions
20 minGroups of 3
Groups solve one tide equation three ways and compare: by hand with an inverse function, with a calculator in the wrong mode, and by graphing. The goal is to use technology to evaluate solutions and to catch the errors technology can hide.
The Model
A ferry dock has depth d(t) = 4.2 + 1.6 sin(2πt/12.4) meters, t in hours after midnight. The ferry needs 3.5 m. Solve d(t) = 3.5 for 0 ≤ t < 12.4.
Roles
Student A isolates sin(2πt/12.4) = -0.4375 and uses sin⁻¹ in radian mode: -0.4528, then π + 0.4528 and 2π - 0.4528, giving t ≈ 7.09 h and t ≈ 11.51 h
Student B repeats the calculation in degree mode and gets -25.94, then tries to turn that number into a time
Student C graphs d(t) and y = 3.5 and reads the intersections
Discussion Questions
Which result matches the graph, and why is the degree-mode number meaningless as an argument of this model?
The calculator returned a negative angle. How did Student A turn it into two times after midnight?
Between which clock times should the ferry not dock? (About 7:06 am to 11:30 am.)
3
Which Answer Makes Sense? Pedal Sort
15-20 minGroups of 3-4
Groups sort answer cards for one model and keep only the answers that make sense. A bicycle pedal on a 17 cm crank, with the crank axle 28 cm above the ground and a cadence of 72 revolutions per minute, has height p(t) = 28 + 17 sin(2.4πt) cm, t in seconds. The question: when is the pedal 40 cm above the ground during the first revolution?
Answer Cards
t ≈ 0.104 s (keep: sin⁻¹(12/17)/(2.4π))
t ≈ 0.313 s (keep: the second angle in the first revolution)
t ≈ -0.104 s (reject: negative time, and it solves sin = -12/17, not 12/17)
t ≈ 0.937 s (reject for this question: it is in the second revolution, 0.104 + 0.833)
t ≈ 44.9 (reject: a degree-mode angle, not a time)
t ≈ 0.784 (reject: the angle sin⁻¹(12/17) in radians, not yet divided by 2.4π)
Procedure
Groups sort the six cards into Keep and Reject piles and write a reason on each rejected card
Each group checks the kept answers by substituting them into p(t)
Groups write one sentence a cyclist could understand: the pedal is above 40 cm for about 0.21 s of each 0.83-second revolution
Challenge Variation
Groups write their own model for a different cadence and make six cards for another group: two correct answers and four with different, realistic errors.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Ferris Wheel Height and the Line h = 50
The graph of h(t) = 33 - 30 cos(πt/10) for one 20-minute ride, drawn to scale. The dashed line h = 50 meets the curve twice, at t ≈ 6.92 and t ≈ 13.08 minutes. The green part of the curve is the time the seat is above 50 m, about 6.16 minutes.
Diagram 2: Why cos⁻¹ Gives Only One of the Two Angles
Both angles on the unit circle with x-coordinate -17/30 solve cos(u) = -17/30. The calculator returns u₁ ≈ 2.173 because cos⁻¹ is restricted to [0, π]. The reflection across the x-axis gives u₂ = 2π - u₁ ≈ 4.110. Dividing each by π/10 gives the two Ferris wheel times.
04
Homework Assignment
~30 min
HSF.TF.B.7 Homework: Solving Trigonometric Models
Directions: Use radian mode. For each problem, isolate the trig function, apply an inverse function with a calculator, find every solution in the stated window, and round times to two decimal places unless the problem says otherwise. End each problem with a sentence that explains your answer in the context. Check one solution per problem on a graph or by substitution.
Part 1: Solving with Inverse Functions (Problems 1-2)
In many European countries, outlet voltage is modeled by V(t) = 325 sin(100πt) volts, t in seconds. Find both times in the first cycle (0 ≤ t < 0.02) when the voltage is 200 V. Give your answers in milliseconds.
A Ferris wheel seat has height h(t) = 25 - 22 cos(πt/6) meters, t in minutes. Find the times during the first ride (0 ≤ t ≤ 12) when the seat is 40 m high, and find how long each ride the seat stays above 40 m.
Part 2: Technology and Time Windows (Problems 3-4)
High tide at a marina is at 6:00 am, and the depth of the channel is d(t) = 6 + 2.5 cos(2πt/12.4) meters, t in hours after 6:00 am. A sailboat needs 5 m of water. Between which two clock times, to the nearest minute, can the sailboat not use the channel? Confirm your times by graphing d(t) and y = 5.
On a summer day, the temperature in a classroom without air conditioning is modeled by T(t) = 68 + 9 sin(π(t - 9)/12) degrees Fahrenheit, t in hours after midnight. The school turns on fans when the temperature reaches 74°F. Find the two times, to the nearest minute, when T(t) = 74, and state when the fans run.
Part 3: Interpreting Solutions (Problems 5-6)
A weight on a spring has position y(t) = 10 cos(3t) centimeters from rest, t in seconds. A student solved y(t) = -4 with the calculator in degree mode and got t = 37.86. Explain the error. Then find every solution with 0 ≤ t ≤ 2 and say what the weight is doing at those times.
An invented daylight model for a southern US city is D(t) = 12.2 + 1.9 sin(2π(t - 80)/365) hours, t the day of the year. (a) Find the two days of the year with 13.5 hours of daylight. (b) A student asks when the city has 14.5 hours of daylight. What happens when you try to solve D(t) = 14.5, and what does that tell you about the city?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Setting Up the Equation
Trig function isolated correctly and existence of a solution checked
Set up correctly with one algebra error
Equation missing or incorrect
Inverse Functions and All Solutions
Inverse value and every solution in the window found
One solution found, second missing
Inverse function not used correctly
Technology
Radian mode, accurate rounding, one solution checked on a graph or by substitution
Accurate values but no check
Degree-mode or rounding errors change the answer
Interpretation
Every answer stated in context units with a correct explanation
Units given but explanation incomplete
Bare numbers with no interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Use a calculator in radian mode. 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
A calculator in radian mode gives sin⁻¹(0.3) ≈ 0.3047. Which other angle u with 0 ≤ u < 2π also satisfies sin(u) = 0.3?
Answer: A
Sine is positive in Quadrants I and II, and the second angle is π - 0.3047 ≈ 2.8369. Choice B uses the cosine rule 2π - u, which gives an angle whose sine is -0.3. Choice C is outside the interval and has sine -0.3. Choice D, π + 0.3047, is in Quadrant III, where sine is negative.
Question 2 of 20 · Multiple Choice
Which set of values can cos⁻¹ return?
Answer: D
Cosine is restricted to [0, π], where it is always decreasing, so cos⁻¹ returns angles from 0 to π. Choice B is the range of sin⁻¹, a frequent mix-up. Choice C includes angles where cosine repeats values, so cos⁻¹ could not be a function on it.
Question 3 of 20 · Multiple Choice
A student solves 4 + 2 cos(πt/6) = 5 and gets cos⁻¹(0.5) = 60, then writes t = 60 · 6/π ≈ 114.6. What went wrong?
Answer: B
The argument πt/6 is in radians, so cos⁻¹(0.5) must be π/3 ≈ 1.047, which gives t = 2. The value 60 is the angle in degrees. Choice C is wrong because 5 - 4 = 1 and 1/2 = 0.5 were done correctly.
Question 4 of 20 · Multiple Choice
A child on a swing has horizontal position x(t) = 1.2 sin(2πt/3.2) meters, t in seconds. When does the swing first reach x = 0.6 m?
Answer: B
sin(2πt/3.2) = 0.5, so 2πt/3.2 = π/6 and t = 3.2/12 ≈ 0.27 s. Choice C is the second time in the cycle, (5π/6)(3.2/(2π)) ≈ 1.33 s, not the first. Choice D divides π/6 by 2π but forgets to multiply by the period 3.2.
Question 5 of 20 · Multiple Choice
A seat on a wheel has height h(t) = 12 - 10 cos(πt/15) meters, t in seconds. The seat first reaches 17 m at t = 10 s. When does it next reach 17 m?
Answer: A
cos(πt/15) = -1/2 gives πt/15 = 2π/3 or 4π/3, so t = 10 or t = 20 in the first 30-second ride. Choice B adds the full period 30 to the first time, which skips the descent. Choice D is the time at the top of the wheel, where h = 22.
Question 6 of 20 · Multiple Choice
For a harbor, d(t) = 3 is solved to give t ≈ 3.9 and t ≈ 8.5 hours after high tide, where d is depth in meters. The depth is below 3 m between these times. Which statement interprets this correctly?
Answer: C
The solutions are times, not depths, and the model is below 3 m between them, for 8.5 - 3.9 = 4.6 hours. Choice A reads times as depths. Choice B reverses the meaning of the interval. Choice D is wrong because low tide is halfway through the cycle, not at a crossing.
Question 7 of 20 · Multiple Choice
In a model with t ≥ 0, a student gets sin⁻¹(-0.4) ≈ -0.4115 for the argument u = 2t. What should the student do next to find the first positive times?
Answer: B
Sine is -0.4 in Quadrants III and IV, so u ≈ 3.5531 or u ≈ 5.8717, giving t ≈ 1.78 and t ≈ 2.94. Choice A is a negative time, outside the context. Choice C solves sin(u) = 0.4 instead. Choice D confuses a negative angle with a value outside [-1, 1].
Question 8 of 20 · Multiple Choice
Which values of t with 0 ≤ t < 2π solve 5 + 4 sin(t) = 6?
Answer: A
sin(t) = 0.25, so t = sin⁻¹(0.25) ≈ 0.2527 or t = π - 0.2527 ≈ 2.8889. Choice B uses 2π - t, the cosine rule. Choice C comes from cos⁻¹(0.25). Choice D misses the Quadrant II solution.
Question 9 of 20 · Multiple Choice
A model gives the water level L(t) = 5 + 3 sin(πt/6) meters. When is L(t) = 9?
Answer: D
L(t) = 9 requires sin(πt/6) = 4/3, but sine never exceeds 1, so the maximum level is 5 + 3 = 8 m and there is no solution. Choice A would give a calculator error. Choice B is when the level reaches its maximum of 8 m.
Question 10 of 20 · Multiple Choice
A rider boards at the bottom of a wheel that turns once every 12 minutes and first reaches a camera platform at t ≈ 2.1 minutes, going up. When does the rider next pass the platform going up?
Answer: C
Each rotation repeats every 12 minutes, so the same point on the way up comes at 2.1 + 12 = 14.1 minutes. Choice A, 12 - 2.1, is the time the rider passes the platform going down. Choice B doubles the time, which has no meaning for a periodic model.
Question 11 of 20 · Multiple Choice
A temperature model gives the solution t ≈ 13.6, where t is hours after midnight. What clock time is this?
Answer: B
0.6 hours is 0.6 × 60 = 36 minutes, so t ≈ 13.6 is 1:36 pm. Choice A reads the decimal .6 as 6 minutes. Choice C writes .6 as 60 minutes, which is not a clock time. Choice D has the right minutes but the wrong half of the day: 13.6 hours after midnight is in the afternoon.
Question 12 of 20 · Multiple Choice
A student graphs y = 2 + 3 cos(x) and y = 4 on a graphing calculator and finds intersections at x ≈ 0.841 and x ≈ 5.442. How can the student check these with inverse functions?
Answer: A
2 + 3 cos(x) = 4 gives cos(x) = 2/3, and cos⁻¹(2/3) ≈ 0.841 with the second angle 2π - 0.841 ≈ 5.442, so the graph and the algebra agree. Choice B skips isolating cosine. Choice D divides the amplitude 3 by the target 4 without subtracting the midline 2 first.
Question 13 of 20 · Multiple Choice
Outlet voltage is V(t) = 170 sin(120πt) volts. When in the first cycle is the voltage first equal to -85 V?
Answer: C
sin(120πt) = -1/2 happens first at 120πt = 7π/6, so t = 7/720 ≈ 0.0097 s. The second time is 11π/6, giving t = 11/720 (choice D), which is later. Choices A and B are when V = +85 V.
Question 14 of 20 · Multiple Choice
An invented daylight model is D(t) = 12 + 3 sin(2π(t - 80)/365) hours on day t of the year. On which days is there 10.5 hours of daylight?
Answer: A
sin(2π(t - 80)/365) = -1/2, so the angle is 7π/6 or 11π/6. Then t = 80 + 365(7/12) ≈ 292.9 and t = 80 + 365(11/12) ≈ 414.6, which is day 49.6 of the next year, so about day 50 (February 19) and day 293 (October 20). Choice B uses +1/2, giving the spring and summer days with 13.5 hours. Choice D forgets to subtract 365.
Question 15 of 20 · Short Answer
A Ferris wheel seat has height h(t) = 20 - 18 cos(πt/5) meters, t in minutes. Find both times in the first ride when the seat is 30 m high, and how long the seat is above 30 m each ride.
cos(πt/5) = -10/18 = -5/9. cos⁻¹(-5/9) ≈ 2.1598, so t = 5(2.1598)/π ≈ 3.44 min, and the second time is 10 - 3.44 ≈ 6.56 min. The seat is above 30 m for about 3.1 minutes of each 10-minute ride.
Question 16 of 20 · Short Answer
A weight on a spring has position y(t) = 5 cos(2t) centimeters, t in seconds. Find the first two positive times when y(t) = 3.
cos(2t) = 0.6. cos⁻¹(0.6) ≈ 0.9273, so 2t = 0.9273 or 2t = 2π - 0.9273 ≈ 5.3559. Then t ≈ 0.46 s and t ≈ 2.68 s. At the first time the weight is moving toward rest (y is decreasing); at the second it is moving away from rest.
Question 17 of 20 · Short Answer
Explain in two or three sentences why sin⁻¹ gives only one solution of 30 sin(u) = 12 when a periodic model can reach the value 12 many times.
sin⁻¹ is the inverse of sine restricted to [-π/2, π/2], where sine is one-to-one, so it returns exactly one angle: sin⁻¹(0.4) ≈ 0.4115. The model repeats every period and reaches each value between its minimum and maximum twice per cycle, so the other solutions are π - 0.4115 plus whole periods. The context tells you which of those times to keep.
Question 18 of 20 · Short Answer
A dock has depth d(t) = 3 + 1.2 sin(2πt/12.4) meters, t in hours after midnight. A boat needs at least 3.6 m. Find the time window after midnight, in the first cycle, when the boat can use the dock.
sin(2πt/12.4) = 0.6/1.2 = 0.5, so 2πt/12.4 = π/6 or 5π/6. Then t = 12.4/12 ≈ 1.03 h and t = 5(12.4)/12 ≈ 5.17 h. The depth is at least 3.6 m between about 1:02 am and 5:10 am, a window of about 4.13 hours.
Question 19 of 20 · Short Answer
Describe how to use a graphing calculator or Desmos to check your solutions of 8 - 5 sin(πt/4) = 6 for 0 ≤ t < 8, and give the solutions.
Graph y = 8 - 5 sin(πt/4) and y = 6 on the window 0 ≤ t ≤ 8 and read the intersection points. Algebra: sin(πt/4) = 0.4, sin⁻¹(0.4) ≈ 0.4115, so πt/4 ≈ 0.4115 or π - 0.4115 ≈ 2.7301, giving t ≈ 0.52 and t ≈ 3.48. The two intersection points on the graph should have these t-values.
Question 20 of 20 · Short Answer
A simplified model of blood pressure is P(t) = 95 + 25 sin(2.5πt) mm Hg, t in seconds. Find when in the first beat the pressure is 110 mm Hg, and interpret the result.
sin(2.5πt) = 15/25 = 0.6. sin⁻¹(0.6) ≈ 0.6435, so t = 0.6435/(2.5π) ≈ 0.082 s and t = (π - 0.6435)/(2.5π) ≈ 0.318 s. Each beat lasts 2/2.5 = 0.8 s (75 beats per minute), and the pressure is above 110 mm Hg for about 0.24 s of each beat.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.TF.B.7 mean?
HSF.TF.B.7 asks students to solve trigonometric equations that come from real models by using inverse functions, to evaluate the answers with technology, and to explain what the answers mean. For example, a student solves 33 - 30 cos(πt/10) = 50 to find when a Ferris wheel seat is 50 m high, uses a calculator for cos⁻¹(-17/30), and reports two times in minutes.
Is HSF.TF.B.7 taught in Algebra 2 or Precalculus?
It is usually taught in Precalculus. The standard is marked (+), which Common Core uses for additional mathematics that students need for advanced courses such as calculus. Some Algebra II courses that cover inverse trigonometric functions include a first version of it.
Why does the calculator give only one answer to a trig equation?
Because sin⁻¹, cos⁻¹ and tan⁻¹ are functions, each returns one angle from a restricted interval: [-π/2, π/2] for sine and tangent (open at the ends for tangent) and [0, π] for cosine. A periodic model reaches most values twice per cycle, so students use π - u for sine, 2π - u for cosine, and add whole periods to get the rest.
Should students use degrees or radians for these problems?
Radians, whenever the model is written with π in the argument, such as sin(πt/6). The argument is a real number of radians, and a degree-mode result like 30 instead of π/6 gives a time that is off by a factor of about 57. A quick check: substitute the answer back into the model and see whether it gives the target value.
How do you find all solutions in a time interval?
Find the two solutions in one cycle, then add or subtract multiples of the period until you leave the interval. For the Ferris wheel h(t) = 33 - 30 cos(πt/10), the first ride gives t ≈ 6.92 and t ≈ 13.08, so the first hour gives those plus 26.92, 33.08, 46.92 and 53.08 minutes.
What should students do when sin⁻¹ gives a negative answer?
A negative angle is not wrong, but it may not be in the window of the question. Add the period, or use the symmetry of sine: if sin⁻¹(r) = -0.85, the solutions in [0, 2π) are π + 0.85 and 2π - 0.85. Then convert the angles to times.
What does it mean when a trig model equation has no solution?
It means the model never reaches that value. If the isolated equation is sin(u) = 1.33, the calculator gives an error, and the right answer is a sentence such as "The tide never reaches 5.5 m; the maximum depth in this model is 5 m." Students should find and state the maximum or minimum to support it.
What are common mistakes on this standard?
Common errors include applying the inverse function before isolating the sine or cosine, dropping the second solution in a cycle, using degree mode, reporting the angle u instead of the time t, and giving times outside the window of the question. Many students also forget to convert decimal hours into minutes.
How is this standard different from HSF.TF.B.6?
HSF.TF.B.6 is about why an inverse exists: restricting a trig function to an interval where it always increases or always decreases. HSF.TF.B.7 uses those inverse functions as tools to solve equations in context. Students who understand B.6 know why the calculator's answer lies in a particular interval, which is exactly what they need to find the other solutions.
Where do trigonometric modeling equations appear after high school?
They appear in calculus, physics and engineering courses: alternating current, springs and pendulums, sound waves, and seasonal data. The same routine applies: isolate the trig function, apply an inverse, use periodicity, and interpret the result in units.
07
Related Standards
6 standards
These standards connect to HSF.TF.B.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSF.TF.B.6Prerequisite
Restrict a trig function to an increasing or decreasing domain to build its inverse
