B. Trigonometric Functions · Trigonometric Identities and Equations

Solving Problems with Trigonometric Equations

B3.4About 75 minutes

Learning goals

I can set up and solve a trigonometric equation from a sinusoidal model of a real situation.
I can interpret the solutions in context, including intervals of time when a quantity is above or below a value.
I can solve equations that cannot be solved with algebra alone by using graphing technology or a table of values.
I can check that my answers are reasonable for the situation.

Words to know

sinusoidal model
A function such as h(t) = a cos(k(t − d)) + c that describes a repeating real-world quantity.

period
The time for one full cycle; for y = a cos(kt) + c it is 2π/k.

midline and amplitude
The midline y = c is halfway between the maximum and minimum; the amplitude |a| is the distance from the midline to a maximum.

numerical solution
An approximate solution found with a graph, a table or repeated guessing, when algebra cannot isolate x (as in sin x = x/2).

Why solve trigonometric equations?

In the unit Radian Measure and Trigonometric Functions you built sinusoidal models for Ferris wheels, tides and hours of daylight, and answered some "when?" questions with inverse ratios. Now you have the full equation-solving toolkit of this unit: the five-step method (lesson 10), substitution for angles like πt/3 (lesson 11), and numerical methods for equations that algebra cannot solve. The curriculum asks you to "solve related problems", with and without graphing technology. The method is always the same:

Set the model equal to the target value.
Isolate the trigonometric ratio.
Let θ be the whole angle (such as θ = πt/3) and find its interval from the time interval.
Solve for θ, then convert back to t.
Interpret: units, which solution is which, and whether the answer makes sense.

Worked example

Example 1: A Ferris wheel

A rider's height above the ground on a Ferris wheel at a fall fair is modelled by h(t) = 12 − 10 cos(πt/3), where h is in metres and t is in minutes, for one turn 0 ≤ t ≤ 6. When is the rider exactly 17 m high, and for how long is the rider above 17 m?

Set up: 12 − 10 cos(πt/3) = 17, so −10 cos(πt/3) = 5 and cos(πt/3) = −1/2.
Let θ = πt/3. Since 0 ≤ t ≤ 6, 0 ≤ θ ≤ 2π.
cos θ = −1/2: θ = 2π/3 or 4π/3.
t = 3θ/π: t = 2 or t = 4.
The rider starts at the bottom (h(0) = 2 m), reaches the top (22 m) at t = 3, and is above 17 m between t = 2 and t = 4.

The rider is at 17 m at t = 2 min and t = 4 min, and above 17 m for 2 minutes of each turn.

The rider is above the dashed line h = 17 between t = 2 and t = 4 minutes (shaded). The solutions of the equation are where the curve crosses the line.

Worked example

Example 2: Tides in the Bay of Fundy

The Bay of Fundy, between New Brunswick and Nova Scotia, has some of the highest tides in the world. A simplified model for the depth of water at a harbour there is d(t) = 8 + 7 cos(πt/6.2), in metres, where t is hours after high tide (0 ≤ t ≤ 12.4). A fishing boat needs at least 4 m of water. When must it leave, and when can it return?

Set up: 8 + 7 cos(πt/6.2) = 4, so cos(πt/6.2) = −4/7 ≈ −0.5714.
θ = πt/6.2 runs from 0 to 2π. α = cos⁻¹(4/7) ≈ 0.9626. Cosine is negative in quadrants II and III: θ ≈ π − 0.9626 ≈ 2.1790 and θ ≈ π + 0.9626 ≈ 4.1042.
t = 6.2θ/π: t ≈ 4.30 and t ≈ 8.10.
The depth falls below 4 m at about 4.3 h after high tide and rises above 4 m again at about 8.1 h.

The boat must leave within about 4.3 hours after high tide and can return about 8.1 hours after high tide (a wait of about 3.8 hours).

Worked example

Example 3: Short winter days

A simplified model for the hours of daylight in a city in southern Ontario is D(t) = 12.2 + 3.5 sin(2π(t − 80)/365), where t is the day of the year (t = 1 is January 1). On which days does the model give 10 hours of daylight, and for about how many days is daylight shorter than that?

12.2 + 3.5 sin θ = 10 with θ = 2π(t − 80)/365, so sin θ = −2.2/3.5 ≈ −0.6286.
For 1 ≤ t ≤ 365, θ runs from about −1.36 to about 4.91. Note that this interval starts at a negative angle.
α = sin⁻¹(0.6286) ≈ 0.6797. Sine is negative in quadrants III and IV. Quadrant III: θ ≈ π + 0.6797 ≈ 3.8213. Quadrant IV: 2π − 0.6797 ≈ 5.60 is past 4.91, but one turn earlier, θ ≈ −0.6797, is inside the interval.
t = 80 + 365θ/(2π): t ≈ 80 − 39.5 ≈ 40.5 and t ≈ 80 + 222.0 ≈ 302.0.
In a non-leap year, day 40 is February 9 and day 302 is October 29. Daylight is shorter than 10 hours from day 302 to the end of the year and from the start of the year to day 40.5: about (365 − 302) + 40.5 ≈ 103.5 days.

The model gives 10 hours of daylight around February 9 or 10 (day 40.5) and October 29 (day 302), with less than 10 hours for about 103 to 104 days in late fall and winter.

When algebra cannot do it

Some equations mix a trigonometric function with a polynomial, such as sin x = x/2. No identity turns this into a single ratio, so there is no algebraic method. Instead, solve numerically:

Graphing technology: graph y = sin x and y = x/2 (radian mode) and find the intersection points.
Without technology: make a table of sin x − x/2 and look for a sign change, then narrow the interval (guess and check).

For sin x = x/2 on 0 ≤ x ≤ 2π, x = 0 is a solution by inspection. A table shows sin x − x/2 is positive at x = 1.8 (0.9738 − 0.9 > 0) and negative at x = 1.9 (0.9463 − 0.95 < 0), so another solution lies between them. Narrowing further gives x ≈ 1.8955. For x > 2, x/2 > 1 ≥ sin x, so there are no more solutions.

y = sin x and y = x/2 meet at x = 0 and x ≈ 1.8955. After x = 2 the line is above 1, so the sine curve can never catch up.

Check that the answer makes sense

Is the time inside the stated interval (one turn, one tide cycle, one year)?
Does the sign make sense? A Ferris wheel rider rising from the bottom passes 17 m first while going up (t = 2), then while going down (t = 4).
Did you convert θ back to t? Answers like 2π/3 "minutes" are a sign you forgot.
Models are simplified: real tides and daylight vary from these curves, so round sensibly.

Just for fun

A Ferris wheel is a cosine graph you can ride. Every time you go around, your height traces out one full period, and your stomach does its best to point out the maximum and minimum.

Hands-on activity

Build a Ferris wheel model

You need: a large plate or round lid, a marker, a ruler, a protractor, graph paper and a calculator.

Step 1. Stand the plate on its edge against a box, or lay it flat and imagine the table edge as the ground. Mark a dot at the bottom of the rim. Measure the radius r.

Step 2. Turn the plate in steps of 30° (π/6). After each step, measure the height of the dot above the bottom of the plate. Record angle (in radians) and height.

Step 3. Plot height against angle θ from 0 to 2π. Compare with the model h = r − r cos θ.

Step 4. Use the model to predict the two angles where the dot is at height 1.5r, then check them on your graph.

What to look for: with r = 12 cm, heights should be close to 0, 1.6, 6, 12, 18, 22.4, 24, ... cm. Height 1.5r means r − r cos θ = 1.5r, so cos θ = −1/2 and θ = 2π/3 or 4π/3 (120° and 240°). Your graph should cross 18 cm (for r = 12) near those angles.

Your turn

For the Ferris wheel h(t) = 12 − 10 cos(πt/3), 0 ≤ t ≤ 6, when is the rider 7 m high?

For the same wheel, when is the rider at the top (22 m)?

For the tide model d(t) = 8 + 7 cos(πt/6.2), 0 ≤ t ≤ 12.4, when is the depth 11.5 m?

A mass on a spring has displacement d(t) = 4 cos(2πt) cm after t seconds, 0 ≤ t ≤ 1. When is d = −2 cm?

For the daylight model D(t) = 12.2 + 3.5 sin(2π(t − 80)/365), on which days is D = 12.2 h?

Use a table or graphing technology to solve cos x = x for 0 ≤ x ≤ 2π, to three decimal places.

More practice: warm-up

For h(t) = 12 − 10 cos(πt/3), what is h(0)?

What is the period of h(t) = 12 − 10 cos(πt/3)?

What are the amplitude and midline of h(t)?

What is the period of d(t) = 8 + 7 cos(πt/6.2)?

What are the greatest and least depths in the tide model?

More practice: on your own

For how long during each turn is the Ferris wheel rider (h(t) = 12 − 10 cos(πt/3)) higher than 7 m?

In the tide model d(t) = 8 + 7 cos(πt/6.2), when is the depth 8 m?

For the spring d(t) = 4 cos(2πt), 0 ≤ t ≤ 1, for what fraction of the second is the mass more than 2 cm from its rest position?

Household electricity in Canada alternates 60 times per second. A model for the voltage is V(t) = 170 sin(120πt), t in seconds. Find the first two times after t = 0 when V = 85 V.

Solve 3 sin x = x for 0 ≤ x ≤ 2π, using a table or graphing technology. Give answers to two decimal places.

Challenge

In the daylight model D(t) = 12.2 + 3.5 sin(2π(t − 80)/365), for about how many days of the year is there more than 15 hours of daylight?

A rectangle is drawn inside a semicircle of radius 1, with its base on the diameter and its top corners on the arc. If a top corner is at angle x (0 < x < π/2) from the centre, the width is 2 cos x and the height is sin x. Find x when the area is 1/2.

Show, without technology, that sin x = x/2 has a solution between π/2 and 2, and explain why there are no solutions greater than 2.

Check your understanding

Quiz

For h(t) = 12 − 10 cos(πt/3), 0 ≤ t ≤ 6, when does the rider first reach 17 m?

What is the period of h(t) = 12 − 10 cos(πt/3)?

Why does the daylight equation D(t) = 10 have two solutions in one year?

How should you solve sin x = x/2?

In the tide model, the solution t ≈ 4.30 for d(t) = 4 means:

For V(t) = 170 sin(120πt), when is V = 85 V for the first time after t = 0?

For parents and teachers

Big idea (B3.4): trigonometric equations answer "when?" questions about repeating things: Ferris wheels, tides, daylight, springs, electricity. Set the model equal to the target, solve for the angle, convert back to time, and interpret.

How to help: Ask your teen to explain each answer in a sentence with units ("The rider is above 17 m from 2 to 4 minutes"). If a free graphing app is available, graph the model and the target line together to see the solutions.

Common difficulties:

Forgetting to convert θ back to t (answers in radians instead of minutes or days).
Using the wrong interval for θ.
Not interpreting which solution is "going up" and which is "going down".
Trying to solve equations like sin x = x/2 with algebra; these need a graph or table.

Note: the tide and daylight models are simplified teaching models, not official data.