Look for this. Bounded motion between two heights suggests sine or cosine. A slope ratio y/x becomes undefined on vertical rays, giving tangent its separated branches.
For a sinusoid, average the extremes for its midline, halve their difference for amplitude, and use b = 2π/period in radian mode. For y = tan θ, mark the zeros at kπ and the vertical asymptotes at π/2 + kπ for integer k. Each branch increases through all real heights.
Worked example
A Ferris wheel seat starts at its lowest height of 3 m, reaches 29 m at the top and returns to the bottom in 52 s. Write a cosine model and sketch one revolution.
The midline is 16 m, amplitude 13 m and b = 2π/52 = π/26. Starting at the minimum gives h(t) = 16 − 13 cos(πt/26). Plot (0, 3), (13, 16), (26, 29), (39, 16) and (52, 3), then connect with a smooth wave.
Try it yourself
A ray at angle θ intersects the vertical line x = 1 when it points to the right. On −π/2 < θ < π/2, sketch its intersection height y = tan θ. Mark the angles where y = −1, 0 and 1, and describe how the full tangent graph repeats.
Check your reasoning
Plot (−π/4, −1), (0, 0) and (π/4, 1). Between the vertical asymptotes θ = ±π/2 the curve rises from arbitrarily negative to arbitrarily positive heights. The full graph repeats every π. Its domain excludes π/2 + kπ for integer k and its range is all real numbers.
Worked notes and diagrams
1
The parts of a wave
The idea: every sinusoidal question names one of these four numbers, or asks you to read one off a graph.
The four parameters are four separate facts about the picture: how tall, how often, how far along, and how high the centre sits.
Word
What it means
Sinusoidal function
Any function of the form y = a sin[b(x − c)] + d, or the same with cosine.
Cycle
One complete repetition of the pattern.
Period
How far along the x-axis one cycle takes: 2π ÷ |b| in radians.
Amplitude
Half the distance from maximum to minimum, |a|. Never negative.
Midline
The horizontal line halfway between maximum and minimum, y = d.
Phase shift
The horizontal shift c relative to the chosen sine or cosine curve. A known peak, trough or directed midline crossing locates it in time.
Maximum and minimum
d + |a| and d − |a|, the top and bottom of the curve.
Range
Every y-value the curve reaches: d − |a| ≤ y ≤ d + |a|.
Frequency
How many cycles fit into one unit — the reciprocal of the period.
2
Reading the parameters
The idea: in y = a sin[b(x − c)] + d or y = a cos[b(x − c)] + d, the extremes determine amplitude and midline, and the period determines |b|. To locate the phase shift, you also need the time of a known peak, trough or directed midline crossing.
Parameter
From a graph or description
Amplitude |a|
(max − min) ÷ 2
Midline d
(max + min) ÷ 2
Period
the time or distance for one full cycle; then b = 2π ÷ period
Horizontal shift c
for positive a and b in a cosine model, the time of a peak. For positive sine, the time of an upward midline crossing.
Max 9 and min 1 give amplitude 4 and midline y = 5. A period of 12 gives b = 2π/12 = π/6.
Try it: match the curve
Start a match challenge and use the sliders to recreate the dashed target. Reading a graph and turning it into a, b, c and d is exactly what modelling questions ask you to do.
Watch out: in y = 2 sin(3x − π) + 1 the shift is not π. Factor it: 2 sin[3(x − π/3)] + 1, so the shift is π/3 right.
3
Writing an equation
The idea: pick the function that starts where your situation starts. Cosine starts at a maximum; negative cosine starts at a minimum; sine starts at the midline going up.
Worked example
A graph has a maximum at (2, 7) and the next minimum at (8, 1). Write an equation.
Max to min is half a cycle: 8 − 2 = 6, so the period is 12 and b = 2π/12 = π/6.
There is a maximum at x = 2, and cosine starts at a maximum, so use cosine shifted 2 right.
y = 3 cos[π/6 (x − 2)] + 4
Sine and cosine are the same wave
Sine is just cosine shifted right by a quarter period: sin x = cos(x − π/2). So there are always several correct equations for the same graph — any one that matches is right.
4
Ferris wheels, tides and daylight
The idea: each real situation maps onto the parameters the same way every time.
h(t) = −10 cos(πt/20) + 12 for a wheel with radius 10 m, centre 12 m up, one rotation every 40 s, boarding at the bottom.
Situation
Amplitude
Midline
Period
Ferris wheel
radius
height of the centre
time for one rotation
Tides
half the gap between high and low tide
average water level
time from high tide to high tide
Daylight hours
half the gap between longest and shortest day
average day length
365 days
A wheel with radius 15 m, centre 17 m, 60-second rotation and boarding at the bottom: a = 15, d = 17, b = 2π/60 = π/30, and a negative cosine to start at the minimum. h(t) = −15 cos(πt/30) + 17.
Watch out: using the diameter as the amplitude. The rider only goes one radius above and below the centre.
5
Using a model
The idea: once you have the equation, questions ask you to plug in a time, or to find the times that give a certain value.
Plugging in
For h(t) = −10 cos(πt/20) + 12 at t = 0: cos 0 = 1, so h = −10 + 12 = 2 m.
Finding when
When is the rider at the top? The top means −10 cos(πt/20) is as big as possible, so cos(πt/20) = −1. That first happens when πt/20 = π, so t = 20 s.
When is the tide h(t) = 2 sin(πt/6) + 5 exactly at 5 m, for 0 ≤ t < 12? That is the midline, so sin(πt/6) = 0: t = 0 and t = 6 hours.
Using a graphing calculator
For values that are not at a maximum, minimum or midline, graph the model and the horizontal line y = value, then find the intersections. Always set the window to match the context — time cannot be negative.
Watch out: giving only the first answer. A repeating model usually hits the same height more than once per cycle — check the whole interval the question asks about.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Reading the parameters · 4 questions
Practice 1
A sinusoidal graph has a maximum of 9 and a minimum of 1. What is its amplitude?
4
8
9
5
Answer and explanation
4
Amplitude is half the distance from minimum to maximum: (9 − 1) ÷ 2 = 4. 8 is the whole height.
Practice 2
A sinusoidal graph has a maximum of 9 and a minimum of 1. What is the equation of its midline?
y = 0
y = 8
y = 5
y = 4
Answer and explanation
y = 5
The midline sits halfway between the extremes: (9 + 1) ÷ 2 = 5.
Practice 3
Where does y = cos x reach its maximum, for 0 ≤ x < 2π?
x = π/2
x = 0
x = π
x = 3π/2
Answer and explanation
x = 0
cos 0 = 1, the largest cosine can be. That is why cosine models are handy when something starts at its highest point.
Practice 4
What is the horizontal shift of y = 2 sin(3x − π) + 1?
π to the right
π/3 to the right
3π to the right
π/3 to the left
Answer and explanation
π/3 to the right
Factor the 3 out first: 3(x − π/3). The shift is π/3 right. Reading π straight from the bracket ignores the factoring.
Writing equations · 5 questions
Practice 1
A sinusoidal function has a period of 12. What is the value of b, in radians?
π/12
12
24π
π/6
Answer and explanation
π/6
Period = 2π ÷ b, so b = 2π ÷ period = 2π ÷ 12 = π/6. π/12 would give a period of 24.
Practice 2
A sinusoidal graph has a maximum at (2, 7) and the next minimum at (8, 1). What is its period?
3
8
12
6
Answer and explanation
12
From a maximum to the next minimum is half a cycle: 8 − 2 = 6. A full period is twice that, 12.
Practice 3
Which equation has an amplitude of 3, a period of π and a midline of y = −2?
y = 2 sin 3x − 3
y = 3 sin(½x) − 2
y = 3 sin 2x − 2
y = −2 sin 3x + π
Answer and explanation
y = 3 sin 2x − 2
Amplitude 3 means a = 3. Period π means b = 2π ÷ π = 2. Midline −2 means d = −2. b = ½ would give a period of 4π.
Practice 4
A sinusoidal graph has a maximum at (2, 7) and the next minimum at (8, 1). Which equation fits?
y = 3 cos[π/12 (x − 2)] + 4
y = 3 cos[π/6 (x + 2)] + 4
y = 6 cos[π/6 (x − 2)] + 4
y = 3 cos[π/6 (x − 2)] + 4
Answer and explanation
y = 3 cos[π/6 (x − 2)] + 4
Amplitude (7 − 1) ÷ 2 = 3, midline 4, period 12 so b = π/6. Cosine starts at a maximum, and the maximum is at x = 2, so the shift is 2 right: (x − 2).
Practice 5
Which equation produces exactly the same graph as y = sin x?
y = cos(x − π/2)
y = cos 2x
y = cos(x + π/2)
y = −cos x
Answer and explanation
y = cos(x − π/2)
Sine is cosine shifted right by a quarter period. Moving cos x's peak from 0 to π/2 lands it on sin x's peak.
Real-world models · 5 questions
Practice 1
In a sinusoidal model of a rider's height on a Ferris wheel, what does the amplitude represent?
the radius of the wheel
the height of the wheel's centre
the time for one rotation
the diameter of the wheel
Answer and explanation
the radius of the wheel
The rider goes one radius above and one radius below the centre. The centre's height is the midline, and one rotation is the period.
Practice 2
In a sinusoidal model of a Ferris wheel, the period represents:
the time for one full rotation
the height of the centre
the radius of the wheel
the highest point a rider reaches
Answer and explanation
the time for one full rotation
One period is one complete cycle — for a wheel, one trip all the way around.
Practice 3
A Ferris wheel has a diameter of 20 m and its centre is 12 m above the ground. What are a rider's maximum and minimum heights?
22 m and 2 m
32 m and 12 m
12 m and 2 m
20 m and 0 m
Answer and explanation
22 m and 2 m
The radius is 10 m, so the rider goes from 12 − 10 = 2 m up to 12 + 10 = 22 m. Using the diameter instead of the radius is the usual slip.
Practice 4
Daylight in a city varies sinusoidally over a year, from a maximum of 17 hours to a minimum of 7 hours. What are the amplitude and period?
12 hours and 365 days
5 hours and 182.5 days
5 hours and 365 days
10 hours and 365 days
Answer and explanation
5 hours and 365 days
Amplitude = (17 − 7) ÷ 2 = 5 hours, and the pattern repeats once a year. 12 hours is the midline, not the amplitude.
Practice 5
A Ferris wheel has a radius of 15 m, its centre is 17 m up, one rotation takes 60 s, and the rider boards at the bottom at t = 0. Which model is correct?
h(t) = −30 cos(πt/30) + 15
h(t) = −15 cos(πt/60) + 17
h(t) = 15 cos(πt/30) + 17
h(t) = −15 cos(πt/30) + 17
Answer and explanation
h(t) = −15 cos(πt/30) + 17
a = 15 (radius), d = 17 (centre), b = 2π ÷ 60 = π/30. Starting at the bottom means starting at a minimum, which a negative cosine does.
Interpreting models · 4 questions
Practice 1
The tide is modelled by h(t) = 2 sin(πt/6) + 5, in metres. How high is high tide?
2 m
12 m
5 m
7 m
Answer and explanation
7 m
High tide is the midline plus the amplitude: 5 + 2 = 7 m. Low tide would be 5 − 2 = 3 m.
Practice 2
A rider's height is h(t) = −10 cos(πt/20) + 12, in metres after t seconds. How high is the rider at t = 0?
2 m
−10 m
12 m
22 m
Answer and explanation
2 m
cos 0 = 1, so h(0) = −10 + 12 = 2 m. The negative cosine starts at its minimum — the rider boards at the bottom.
Practice 3
A rider's height is h(t) = −10 cos(πt/20) + 12, in metres after t seconds. When does the rider first reach the top?
40 seconds
20 seconds
12 seconds
10 seconds
Answer and explanation
20 seconds
The top is where −10 cos(πt/20) is largest, so cos(πt/20) = −1, which first happens when πt/20 = π. That gives t = 20, halfway through a 40-second rotation.
Practice 4
The tide is h(t) = 2 sin(πt/6) + 5, in metres, t hours after midnight. When during 0 ≤ t < 12 is the tide exactly 5 m?
t = 3 and t = 9 hours
t = 6 hours only
t = 3 hours only
t = 0 and t = 6 hours
Answer and explanation
t = 0 and t = 6 hours
5 m is the midline, so sin(πt/6) = 0. That happens when πt/6 = 0 or π, giving t = 0 and 6. At t = 3 and 9 the tide is at its highest and lowest.
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