Pose and solve real-world sinusoidal problems

Ontario Grade 11 Mathematics

Trigonometric Functions

MCR3U · Unit D3 · Expectation D3.5

Tides rise and fall. Temperatures climb in summer and drop in winter. A Ferris wheel carries passengers up and then back down in a smooth, repeating arc. All of these situations share the same mathematical shape: a smooth wave that repeats at regular intervals. That shape is modelled by a sinusoidal function — one built from sine or cosine. In this lesson you will read a real-world scenario, extract the key numbers that define its wave, build an equation, and then use that equation to answer specific questions about the situation. Every step will be explained in plain language before any symbols appear.

What you will learn

Identify the key features (amplitude, period, vertical shift, phase shift) of a sinusoidal situation described in words or a table of values.
Write a sinusoidal equation of the form y = a sin(kx + d) + c or y = a cos(kx + d) + c that models a real-world context.
Substitute given values into a sinusoidal model and solve for an unknown quantity using algebraic reasoning and inverse sine or inverse cosine.
Interpret the solution in the context of the original problem, including checking whether answers make sense.

Prerequisite Bridge: The Four Features of a Sinusoidal Function

Before posing any real-world problem, you need to be comfortable naming the four key features of a sinusoidal function and locating them on a graph or in a description. This is a quick review — you met these ideas earlier in Unit D.

The amplitude, written

a

, measures how far the wave travels above or below its middle line. It is always a positive number. If the highest value of the function is

M

and the lowest value is

m

, then

a = \frac{M - m}{2}

.

The vertical shift, written

c

, is the value of the horizontal middle line (also called the equation of the axis or the sinusoidal axis). It sits exactly halfway between the maximum and the minimum:

c = \frac{M + m}{2}

.

The period, written

T

, is the length of one complete cycle — the horizontal distance before the wave repeats exactly. The value

k

in the equation controls the period:

k = \frac{360^{\circ}}{T}

.

The phase shift, written

d

, moves the entire wave left or right along the horizontal axis. It is found by identifying where a reference point (the first maximum for cosine, or the first upward crossing of the axis for sine) occurs.

a = \frac{M - m}{2}, c = \frac{M + m}{2}, k = \frac{360°}{T}

Amplitude: $a = \frac{M - m}{2}$ — always positive.
Vertical shift: $c = \frac{M + m}{2}$ — the middle value between max and min.
Period $T$ : the horizontal length of one full cycle; $k = \frac{360^{\circ}}{T}$ .
Phase shift $d$ : the horizontal shift of the wave from a standard starting position.
The general model is $y = a\sin(k(x - d)) + c$ or $y = a\cos(k(x - d)) + c$ .

Choosing Sine or Cosine and Writing the Model

Both sine and cosine can model any sinusoidal situation — they produce the same family of curves, just shifted horizontally relative to each other. The choice is a matter of convenience based on where the problem gives you information.

Choose cosine when the problem tells you the maximum (or minimum) value at a clearly stated starting point. The standard cosine curve starts at its maximum, so the phase shift calculation is usually simpler.

Choose sine when the problem tells you the value at the midline at a clearly stated starting point, because the standard sine curve begins at its midline on the way up.

Once you choose your function, plug in

a

,

k

,

d

, and

c

. Keep track of units: if the horizontal axis is time in hours, make sure the period

T

is also in hours before you calculate

k

.

After writing the equation, always do a quick sanity check: substitute the time of the maximum into your equation and confirm you get the maximum value. If not, recheck the sign of

a

or the value of

d

.

y = a\cos(k(x - d)) + c

Cosine is convenient when a maximum or minimum is given at the reference point.
Sine is convenient when the midline value is given at the reference point.
Either choice produces an equivalent model; pick the one that makes $d$ simplest.
Always verify the model by substituting a known point back in.
Keep the horizontal units consistent throughout (e.g., all in hours, or all in degrees).

Solving for an Unknown: Using the Inverse Function

Once your model equation is written, you may be asked two types of questions. The first type gives you an

x

-value (a time, angle, or position) and asks for the corresponding

y

-value (a height, temperature, or depth). For this type, simply substitute and evaluate — no inverse function is needed.

The second type gives you a

y

-value and asks when or where it occurs. This requires isolating the trigonometric expression and then applying the inverse sine (

\sin^{-1}

) or inverse cosine (

\cos^{-1}

) to both sides.

Remember that inverse sine and inverse cosine each give only one angle in a restricted range. Because a sinusoidal wave repeats and is symmetric, there are usually two solutions within each cycle. You find the second solution using the symmetry of the sine or cosine graph: for cosine, if

\cos^{-1}

gives angle

\theta

, then

-\theta

(or equivalently

360^{\circ} - \theta

in the next position) is also a solution. For sine, if

\sin^{-1}

gives

\theta

, then

180^{\circ} - \theta

is the second solution in the same cycle.

After finding your angles, reverse every algebraic step you used to isolate the trigonometric expression to recover the original variable (usually time or position). Finally, check every answer against the real-world context — for example, a time cannot be negative if the problem starts at

t = 0

, and you may need to add or subtract full periods to find all solutions in a stated interval.

\cos^{-1} (\frac{y - c}{a}) = k(x - d)

Type 1 (find $y$ ): substitute the given $x$ -value and evaluate directly.
Type 2 (find $x$ ): isolate the trig expression, apply inverse sin or cos, then solve for $x$ .
Inverse sin/cos gives one principal angle; use graph symmetry to find the second angle per cycle.
Undo all algebraic steps in reverse order to recover the original variable.
Verify each answer against the real-world context and the stated time interval.

Setting Up the Problem from a Word Description

Real-world sinusoidal problems are described in words, so your first job is to extract the four features from the language of the problem. Look for the words 'maximum' and 'minimum' — these give you

M

and

m

, and therefore

a

and

c

. Look for the word 'period', 'cycle', 'repeats every', or 'takes _ to complete one full rotation' — this gives you

T

.

Next, identify the starting condition. The problem might say 'at

t = 0

, the height is at its maximum' (use cosine with

d = 0

) or 'at

t = 2

hours, the depth reaches its lowest point' (use cosine with a negative amplitude or a phase shift of

d = 2

).

It helps to sketch a rough wave on a number line before writing any equation. Mark the maximum, minimum, and midline. Mark one or two known points. This sketch does not need to be precise — its purpose is to prevent sign errors when you set up the equation.

Label every quantity with its unit as you write it down. Writing

T = 12 \text{ h}

and

M = 4.8 \text{ m}

from the start prevents confusion later. Only strip the units when substituting pure numbers into your calculator.

Scan the problem for max, min, and period language first.
Identify the starting condition to determine the phase shift $d$ .
Sketch a rough wave to visualise the situation before writing the equation.
Label every extracted value with its unit.
Write the equation only after all four features are identified.

Interpreting and Communicating Solutions

A numerical answer is not a complete solution in a real-world problem. You must state what the number means in context. For example, writing

t \approx 3.2

is incomplete; writing 'the water depth first reaches 2.0 m at approximately 3.2 hours after midnight' is a complete answer.

Pay attention to whether the problem asks for the first time, all times within an interval, or the duration of time spent above or below a threshold. Each phrasing requires a slightly different strategy: finding all times in an interval means adding full period lengths to your initial solutions until you exceed the interval boundary.

Always re-read the question after you have an answer to confirm you answered what was actually asked. It is common to solve for

x

when the question asked for a duration, or to give one answer when two were needed.

Round only at the final step. Carry extra decimal places through intermediate calculations to avoid rounding errors accumulating.

State every answer in a full sentence that includes units and real-world meaning.
Find additional solutions in an interval by adding whole multiples of the period.
Re-read the question to confirm you answered exactly what was asked.
Round only at the final step; keep extra precision in intermediate steps.

Extracting the Four Sinusoidal Features from a Word Problem

Feature	Symbol	What to Look For in the Problem	Formula
Amplitude	$a$	Half the distance between max and min values	$a = \frac{M - m}{2}$
Vertical shift	$c$	The average of the max and min; the middle value	$c = \frac{M + m}{2}$
Period	$T$	'Repeats every …', 'one full cycle takes …', 'rotates once in …'	Given directly
Angular frequency	$k$	Calculated from the period; controls how fast the wave oscillates	$k = \frac{360^{\circ}}{T}$
Phase shift	$d$	Where the maximum (cosine) or midline crossing (sine) occurs on the horizontal axis	Read from the starting condition

Worked example

Ferris Wheel Height Over Time

A Ferris wheel has a diameter of 20 m. The centre of the wheel is 12 m above the ground. The wheel completes one full rotation every 40 seconds. A passenger starts at the lowest point of the wheel at

t = 0

seconds. (a) Write a sinusoidal equation for the passenger's height

h

, in metres, as a function of time

t

, in seconds. (b) Find the passenger's height at

t = 25

seconds. (c) Find the first time the passenger is exactly 19 m above the ground.

Identify maximum and minimum heights
The centre of the wheel is 12 m above the ground and the radius is half the diameter, so the radius is $20 \div 2 = 10$ m. The highest point is $12 + 10 = 22$ m and the lowest point is $12 - 10 = 2$ m. So $M = 22$ and $m = 2$ .
$M = 22 \text{ m}, m = 2 \text{ m}$
Calculate amplitude and vertical shift
Using the formulas from the prerequisite section, the amplitude is $a = \frac{M - m}{2} = \frac{22 - 2}{2} = 10$ and the vertical shift is $c = \frac{M + m}{2} = \frac{22 + 2}{2} = 12$ .
$a = 10, c = 12$
Find k from the period
The period is $T = 40$ seconds. The value $k$ controls the period through the relationship $k = \frac{360^{\circ}}{T}$ , giving $k = \frac{360^{\circ}}{40} = 9^{\circ}$ per second.
$k = \frac{360°}{40} = 9°\text{/s}$
Choose cosine with a reflection to match the starting condition
At $t = 0$ , the passenger is at the lowest point, height 2 m. A standard cosine with positive amplitude starts at the maximum. To start at the minimum instead, use a negative amplitude: $a = -10$ . This way, when $t = 0$ , the cosine equals 1 and $h = -10(1) + 12 = 2$ , which matches. The phase shift is $d = 0$ because the starting point is exactly at $t = 0$ .
$h = -10\cos(9° t) + 12$
Part (b): Find height at t = 25 s
Substitute $t = 25$ into the model. First compute the angle: $9^{\circ} \times 25 = 225^{\circ}$ . Then $\cos(225^{\circ}) = -\frac{\sqrt{2}}{2} \approx -0.7071$ . So $h = -10(-0.7071) + 12 = 7.071 + 12 \approx 19.1$ m.
$h = -10\cos(225°) + 12 \approx 19.1 \text{ m}$
Part (c): Set h = 19 and isolate cosine
Set the equation equal to 19 and solve step by step. Subtract 12 from both sides: $-10\cos(9^{\circ} t) = 7$ . Divide both sides by $-10$ : $\cos(9^{\circ} t) = -0.7$ .
$\cos(9° t) = -0.7$
Apply inverse cosine and find the principal angle
Take $\cos^{-1}(-0.7)$ to get the principal angle. Since the cosine is negative, the angle is in the second quadrant. $\cos^{-1}(0.7) \approx 45.57^{\circ}$ , so the principal angle is $180^{\circ} - 45.57^{\circ} = 134.43^{\circ}$ . Therefore $9^{\circ} t = 134.43^{\circ}$ , giving $t = \frac{134.43^{\circ}}{9^{\circ}} \approx 14.9$ s.
$t = \frac{\cos^{-1}(-0.7)}{9°} \approx \frac{134.43°}{9°} \approx 14.9 \text{ s}$
Find the second solution using cosine symmetry and select the first
Cosine is symmetric about $0^{\circ}$ (and about $360^{\circ}$ ), so the second angle in one cycle is $360^{\circ} - 134.43^{\circ} = 225.57^{\circ}$ . This gives $t = \frac{225.57^{\circ}}{9^{\circ}} \approx 25.1$ s. Both 14.9 s and 25.1 s are valid times. The question asks for the first time, which is $t \approx 14.9$ s. This also makes sense: part (b) showed $h \approx 19.1$ m at $t = 25$ s, which is close to 19 m, confirming the second solution near 25.1 s is reasonable.
$t_1 \approx 14.9 \text{ s}, t_2 \approx 25.1 \text{ s}$

Answer: (a)

h = -10\cos(9^{\circ} t) + 12

. (b) The passenger is approximately 19.1 m above the ground at

t = 25

s. (c) The passenger first reaches 19 m above the ground at approximately 14.9 seconds after the ride begins.

Check: Substitute

t \approx 14.9

s back:

9^{\circ} \times 14.9 = 134.1^{\circ}

;

\cos(134.1^{\circ}) \approx -0.6967

;

h = -10(-0.6967) + 12 = 6.967 + 12 = 18.97 \approx 19

m. ✓ Substitute

t = 0

:

h = -10\cos(0^{\circ}) + 12 = -10(1) + 12 = 2

m. ✓ This matches the lowest point.

Worked example

Monthly Average Temperature in a Canadian City

Environment Canada records show that for a city in Ontario, the average monthly temperature (in °C) reaches a maximum of

26^{\circ}

C in July (month 7) and a minimum of

-8^{\circ}

C in January (month 1). Assume the temperature varies sinusoidally throughout the year. (a) Write a cosine equation for the average temperature

T

, in °C, as a function of month number

m

, where

m = 1

is January. (b) Predict the average temperature in October (month 10). (c) Determine during which months the average temperature is above

20^{\circ}

C.

Extract maximum, minimum, amplitude, and vertical shift
The maximum temperature is $26^{\circ}$ C and the minimum is $-8^{\circ}$ C. Amplitude: $a = \frac{26 - (-8)}{2} = \frac{34}{2} = 17$ . Vertical shift: $c = \frac{26 + (-8)}{2} = \frac{18}{2} = 9$ .
$a = 17, c = 9$
Determine the period and k
Temperature repeats every 12 months (one full year), so $T = 12$ months. Then $k = \frac{360^{\circ}}{12} = 30^{\circ}$ per month.
$k = \frac{360°}{12} = 30°\text{/month}$
Identify the phase shift using the maximum
Cosine is the convenient choice here because the maximum is given at a specific month. The standard cosine $\cos(k(m - d))$ reaches its maximum when the argument equals $0^{\circ}$ , meaning $k(m - d) = 0^{\circ}$ . The maximum occurs at month 7, so $30^{\circ}(7 - d) = 0^{\circ}$ , which gives $d = 7$ .
$d = 7$
Write the full equation
Assembling all four values into the cosine model gives the equation below. The positive amplitude is used because cosine starts at its maximum when the argument is $0^{\circ}$ , and the maximum occurs at $m = 7$ as required.
$T = 17\cos(30°(m - 7)) + 9$
Part (b): Evaluate at m = 10 (October)
Substitute $m = 10$ : the argument is $30^{\circ}(10 - 7) = 30^{\circ}(3) = 90^{\circ}$ . Then $\cos(90^{\circ}) = 0$ , so $T = 17(0) + 9 = 9^{\circ}$ C.
$T = 17\cos(90°) + 9 = 9°\text{C}$
Part (c): Set T > 20 and isolate cosine
To find when the temperature is above $20^{\circ}$ C, set $T = 20$ to find the boundary months first. Subtract 9 from both sides: $17\cos(30^{\circ}(m - 7)) = 11$ . Divide both sides by 17: $\cos(30^{\circ}(m - 7)) = \frac{11}{17} \approx 0.6471$ .
$\cos(30°(m - 7)) = \frac{11}{17}$
Apply inverse cosine to find the principal angle
Take $\cos^{-1}(0.6471) \approx 49.67^{\circ}$ . This gives two equations: $30^{\circ}(m - 7) = 49.67^{\circ}$ and $30^{\circ}(m - 7) = -49.67^{\circ}$ (the symmetric solution, since cosine is symmetric about $0^{\circ}$ ).
$30°(m - 7) = ± 49.67°$
Solve each equation for m
From $30^{\circ}(m - 7) = 49.67^{\circ}$ : divide by $30^{\circ}$ to get $m - 7 = 1.656$ , so $m \approx 8.66$ (late August, partway through month 8–9). From $30^{\circ}(m - 7) = -49.67^{\circ}$ : divide by $30^{\circ}$ to get $m - 7 = -1.656$ , so $m \approx 5.34$ (partway through month 5). Since cosine is above 0.6471 between $-49.67^{\circ}$ and $49.67^{\circ}$ — that is, between $m \approx 5.34$ and $m \approx 8.66$ — the temperature is above $20^{\circ}$ C during this interval. m \approx 5.34 and m \approx 8.66
Interpret in the context of whole months
Month 5 is May and month 8 is August. Since $m \approx 5.34$ means part of May has already passed before the temperature crosses $20^{\circ}$ C, and $m \approx 8.66$ means the temperature drops below $20^{\circ}$ C partway through August, only the whole months entirely inside this range qualify. Months 6 (June) and 7 (July) are fully inside the interval. Month 8 (August) is partially inside. To be precise about the question 'during which months', full months above $20^{\circ}$ C on average are June and July; August is partially above. The temperature is above $20^{\circ}$ C for portions of May, all of June and July, and portions of August.
$5.34 < m < 8.66$

Answer: (a)

T = 17\cos(30^{\circ}(m - 7)) + 9

. (b) The predicted average temperature in October is

9^{\circ}

C. (c) The average temperature is above

20^{\circ}

C from approximately mid-May (month 5.3) to late August (month 8.7), meaning the complete months of June and July are fully above

20^{\circ}

C, with portions of May and August also exceeding that threshold.

Check: Check at

m = 7

(July):

T = 17\cos(0^{\circ}) + 9 = 17 + 9 = 26^{\circ}

C. ✓ Check at

m = 1

(January): argument

= 30^{\circ}(1 - 7) = -180^{\circ}

;

\cos(-180^{\circ}) = -1

;

T = 17(-1) + 9 = -8^{\circ}

C. ✓ Check at

m \approx 5.34

: argument

= 30^{\circ}(5.34 - 7) = 30^{\circ}(-1.66) = -49.8^{\circ}

;

\cos(-49.8^{\circ}) \approx 0.645

;

T \approx 17(0.645) + 9 \approx 10.97 + 9 = 19.97 \approx 20^{\circ}

C. ✓

Common mistakes and how to avoid them

Using the diameter of a circular path as the amplitude instead of the radius.

Correction: The amplitude equals the radius (half the diameter), because the wave travels that distance above and below the centre.

Forgetting that inverse cosine gives only one angle, then reporting only one solution when two exist in a cycle.

Correction: Always use the symmetry of the cosine (or sine) graph to find the second angle in the cycle: for cosine, the second solution is

360^{\circ}

minus the principal angle.

Setting the phase shift equal to the time of the maximum without checking which function (sine or cosine) is being used.

Correction: The phase shift

d

places the maximum of cosine (or the upward midline crossing of sine) at that time. Confirm by substituting back into the equation.

Rounding intermediate values (angles, cosine outputs) to one or two decimal places, which causes the final answer to be noticeably off.

Correction: Keep at least four decimal places until the very last step, then round the final answer to the precision the question requires.

Giving a purely numerical answer without linking it to the context, such as writing

t = 14.9

without specifying what that means.

Correction: Write a complete sentence: state the units, what the variable represents, and what the answer means in the real situation.

Lesson summary

A real-world sinusoidal model requires four features: amplitude $a$ , vertical shift $c$ , angular frequency $k$ (from the period $T$ ), and phase shift $d$ .
Extract these features by identifying the maximum, minimum, period, and starting condition from the problem description.
Use cosine when the starting condition gives a maximum or minimum; use sine when the starting condition gives a midline value.
To find $y$ from a given $x$ , substitute directly. To find $x$ from a given $y$ , isolate the trig expression and apply the inverse function, then find both solutions per cycle.
Add full period lengths to initial solutions to find all values of $x$ within a stated interval.
Always interpret numerical answers in context, include units, and verify by substituting back into the model equation.

Check your understanding

Question 1

A buoy bobs up and down in the ocean. Its highest point is 3 m above the water surface and its lowest point is 1 m below the surface. What is the amplitude of this sinusoidal motion?

1 m
2 m
3 m
4 m

Show answer and explanation

2 m

The maximum is

+3

m and the minimum is

-1

m. Amplitude

= \frac{3 - (-1)}{2} = \frac{4}{2} = 2

m. The amplitude is the distance from the midline to the maximum, not the full range.

Question 2

A sinusoidal function has the equation

h = 5\cos(18^{\circ} t) + 8

. What is the period of this function?

5 seconds
8 seconds
18 seconds
20 seconds

Show answer and explanation

20 seconds

The period is found using

T = \frac{360^{\circ}}{k}

. Here

k = 18^{\circ}

, so

T = \frac{360^{\circ}}{18^{\circ}} = 20

seconds.

Question 3

You are solving

\cos(30^{\circ}(t - 2)) = 0.5

and the inverse cosine gives a principal angle of

60^{\circ}

. What is the second angle within the same cycle that also satisfies this equation?

$120^{\circ}$
$180^{\circ}$
$270^{\circ}$
$300^{\circ}$

Show answer and explanation

300^{\circ}

Cosine is symmetric about

0^{\circ}

(and

360^{\circ}

), so if one solution is

60^{\circ}

, the other is

360^{\circ} - 60^{\circ} = 300^{\circ}

. Both angles have cosine equal to

0.5

.

Question 4

A sinusoidal temperature model gives

T = 15\cos(30^{\circ}(m - 7)) + 10

. What is the maximum temperature predicted by this model, and in which month does it occur?

Maximum of $25^{\circ}$ C in month 7
Maximum of $15^{\circ}$ C in month 7
Maximum of $25^{\circ}$ C in month 10
Maximum of $10^{\circ}$ C in month 7

Show answer and explanation

Maximum of

25^{\circ}

C in month 7

The maximum of cosine is 1, occurring when its argument equals

0^{\circ}

, i.e., when

m = 7

. At that point,

T = 15(1) + 10 = 25^{\circ}

C. The vertical shift

c = 10

is the midline, not the maximum.

Key terms

Sinusoidal function: A function whose graph has the smooth, repeating wave shape of sine or cosine. It models situations that cycle between a maximum and a minimum at regular intervals.
Amplitude: The positive distance from the midline of the wave to its maximum (or minimum) value. Calculated as half the difference between the maximum and minimum values.
Period: The horizontal length of one complete cycle of the wave — the distance (or time) before the pattern repeats exactly.
Vertical shift (sinusoidal axis): The horizontal middle line of the wave, equal to the average of the maximum and minimum values. Also called the equation of the axis.
Phase shift: The horizontal displacement of the wave from the standard starting position of sine or cosine. It shifts the wave left or right along the horizontal axis.
Angular frequency (k): The value $k$ in the equation $y = a\sin(kx)$ or $y = a\cos(kx)$ that controls the period. It equals $\frac{360^{\circ}}{T}$ , where $T$ is the period.
Inverse cosine / Inverse sine: Operations written as $\cos^{-1}$ and $\sin^{-1}$ that reverse the cosine or sine function. They return the angle whose cosine or sine equals a given value, within a restricted range.
Sinusoidal model: A specific sinusoidal equation, with numerical values of $a$ , $k$ , $d$ , and $c$ , built to represent a particular real-world situation.

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