Do I use a sine model or a cosine model?

For the amplitude, period and vertical shift the choice makes no difference — a cosine curve is just a shifted sine curve, so both fit the same oscillation. In 2.5 the model is usually handed to you already in the form $y = a\sin(bx) + d$ or $y = a\cos(bx) + d$; your job is to read the parameters, predict values and solve. When you do get to choose, cosine is tidiest if the cycle starts at a maximum or minimum, and sine if it starts on the principal axis and is rising.

My period or my substituted value is wrong — what did I miss?

By far the most common cause is calculator mode. If the model uses degrees — for instance $y = 4\sin(30t) + 7$, where the period is $\frac{360^\circ}{30} = 12$ — your GDC must be in degree mode. In radian mode the same keystrokes give nonsense. Always check the mode before you substitute, and read whether the question is written with $\frac{360^\circ}{b}$ (degrees) or $\frac{2\pi}{b}$ (radians).

How do I find the FIRST time something happens, not just any time?

Set the model equal to the target value and solve on the GDC, but remember a sinusoidal equation has infinitely many solutions — one per cycle. Graph the model together with the horizontal target line and read the intersection points, then choose the smallest one that lies in the interval the question gives (often $t \ge 0$). For a plain maximum or minimum you can shortcut this: the first maximum of a rising sine model is a quarter of a period after the start.

Why is the amplitude positive when the graph goes below the axis?

Amplitude measures the size of the swing, not its direction, so it is defined as the positive distance $|a| = \frac{\text{max} - \text{min}}{2}$. The curve naturally dips $|a|$ below the principal axis as well as rising $|a|$ above it. A negative coefficient in the equation only flips whether the cycle starts by rising or falling; it never makes the amplitude negative.

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Modelling with sinusoidal functions

From the rise and fall of the tide to the swing of a Ferris-wheel car and the slow march of daylight hours through the year, the world is full of quantities that go up, come back down and repeat. A sinusoidal model captures that rhythm in a single equation built from a sine or cosine curve. This lesson covers the full 2.5 picture: what the amplitude, period and vertical shift mean in the model $y = a\sin(bx) + d$ ; how to read them off a graph, from a stated maximum and minimum, or from a worded context; and how to use the finished model with a GDC to predict a value or find when a value first occurs. Because 2.5 is examined on Paper 2, the calculator handles the arithmetic — so the skill being marked is setting up the right model and writing down the method the examiner needs. As in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Interpret the amplitude $|a|$ , the period $\frac{360^\circ}{b}$ and the vertical shift (principal axis) $d$ in the model $y = a\sin(bx) + d$ or $y = a\cos(bx) + d$ .
2Determine these parameters from a graph, from a given maximum and minimum, or from a real-world context such as tides, temperature, a Ferris wheel or daylight hours.
3Use a sinusoidal model to predict a value at a given input and to find when a stated value first occurs.
4Use a GDC to solve equations arising from a sinusoidal model, and present working so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Riding the Waves of Data

Lots of things in the real world go up and down and then repeat: the tide, the temperature, the seat on a Ferris wheel. A sinusoidal function — a stretched, shifted sine or cosine curve — captures that rhythm in one equation. Pin down three numbers, the height of the swing (amplitude), how long one cycle lasts (period) and the height of the middle line (vertical shift), and you can predict the value at any moment.

THINK OF IT LIKE…

Picture yourself on a Ferris wheel. Your height above the ground never settles — it climbs to a top, sinks to a bottom, and repeats every turn. The centre of the wheel is the principal axis, the line $y = d$ that the curve wobbles around. The radius is the amplitude $|a|$ , how far above and below that line you swing. The time for one full turn is the period. Once you know those three numbers you know your height at every instant, without ever getting back on the wheel.

1
Find the middle line, $d$ . Average the maximum and minimum values: $d = \frac{\text{max} + \text{min}}{2}$ . This is the principal axis, the horizontal line the curve oscillates about.
2
Find the amplitude, $a$ . Take half the gap between the maximum and minimum: $a = \frac{\text{max} - \text{min}}{2}$ . It is the height of the swing above (and below) the middle line.
3
Find $b$ from the period. Measure the length of one complete cycle, then use $\text{period} = \frac{360^\circ}{b}$ , so $b = \frac{360^\circ}{\text{period}}$ when working in degrees.
4
Use the model. Substitute a value to predict an output, or set the model equal to a target value and let the GDC find the input — reading the answer that fits the interval the question asks for.

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Step 1

Find the middle line, $d$ . Average the maximum and minimum values: $d = \frac{\text{max} + \text{min}}{2}$ . This is the principal axis, the horizontal line the curve oscillates about.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$y = a\sin(bx) + d \qquad \text{or} \qquad y = a\cos(bx) + d$

Full topic notes

Formal explanation with the rigour you need for the exam.

The anatomy of a sinusoidal model

Almost every periodic quantity in this course can be written in one of two closely related forms. The parameters $a$ , $b$ and $d$ do all the work: $a$ sets the size of the swing, $b$ sets how quickly the cycle repeats, and $d$ sets the height of the line the curve oscillates about. Read those three numbers correctly and the whole model is under your control.

$y = a\sin(bx) + d \qquad \text{or} \qquad y = a\cos(bx) + d$

A sign warning: if the coefficient in front of the sine or cosine is negative — say $y = -3\cos(bx) + d$ — the curve is reflected in the principal axis, so it begins at a minimum instead of a maximum. The amplitude is still $|-3| = 3$ . The sign changes the starting shape, never the size of the swing.

Amplitude $|a|$ : the distance from the principal axis up to a maximum (or down to a minimum). From data, $|a| = \frac{\text{max} - \text{min}}{2}$ . It is always positive — it measures the size of the swing, not its direction.
Period and $b$ : the period is the length of one full cycle. In degrees, $\text{period} = \frac{360^\circ}{b}$ , so $b = \frac{360^\circ}{\text{period}}$ ; in radians replace $360^\circ$ with $2\pi$ .
Vertical shift / principal axis $d$ : the curve oscillates about the horizontal line $y = d$ , where $d = \frac{\text{max} + \text{min}}{2}$ . This is the 'middle' of the wave.
Maximum and minimum: with $a > 0$ the model reaches a maximum of $d + a$ and a minimum of $d - a$ . Knowing $d$ and $a$ hands you the top and bottom instantly.

Reading the parameters: from a graph, from max/min, or from a context

Whichever way a question presents the oscillation, the same three questions unlock it. Where is the middle line? That is $d = \frac{\text{max} + \text{min}}{2}$ . How big is the swing? That is $a = \frac{\text{max} - \text{min}}{2}$ . How long is one cycle? That gives $b = \frac{360^\circ}{\text{period}}$ . From a graph, read the highest and lowest values for $a$ and $d$ , and the distance between consecutive peaks for the period. From a worded context — a tide with a stated high and low, a temperature that ranges between two values, a Ferris wheel of known diameter — translate the words into a maximum, a minimum and a cycle length, then do exactly the same arithmetic.

Using the model: predicting and solving with the GDC

Once the parameters are known, the model earns its keep. To predict a value, substitute the input and evaluate on the GDC — checking first that the calculator mode matches the model. To find WHEN a quantity reaches a target, set the model equal to that target and solve: graph the model together with the horizontal target line and read the intersections, or use the equation solver, then choose the solution that lies in the interval the question asks about. A sinusoidal equation repeats every cycle, so there are always several answers — reading the right one is part of the skill.

Common mistakes examiners penalise

Calling $d$ the amplitude — the amplitude is the coefficient $|a|$ that multiplies the sine or cosine, not the vertical shift $d$ . In $y = 4\sin(30x) + 7$ the amplitude is $4$ , not $7$ .
Getting the period the wrong way up — the period is $\frac{360^\circ}{b}$ (degrees) or $\frac{2\pi}{b}$ (radians), not $360^\circ \times b$ and not simply $b$ . For $y = 6\cos(20x) + 5$ the period is $\frac{360^\circ}{20} = 18^\circ$ , not $20$ .
Working in the wrong calculator mode — a model written with $\frac{360^\circ}{b}$ or an angle like $30t$ is in degrees, so the GDC must be in degree mode. Radian mode gives a plausible-looking but wrong answer.
Forgetting the maximum is $d + a$ , not $a$ — with $a>0$ the maximum is principal axis plus amplitude, $d + a$ , and the minimum is $d - a$ . Quoting the amplitude as the maximum loses the mark.
Putting the principal axis at $y = 0$ — the curve oscillates about $y = d$ , not the $x$ -axis. State the principal axis as $y = d$ using the vertical-shift value.
Taking amplitude as negative when the graph dips below the axis — amplitude is the positive distance $\frac{\text{max}-\text{min}}{2}$ . A negative coefficient only reflects the curve; the amplitude is still $|a|$ .
Reporting every solution instead of the first — a sinusoidal equation repeats each cycle, so pick the solution inside the interval the question specifies (often the smallest $t > 0$ ), not just any intersection the GDC shows.

Model answer — marked the way our engine marks it

On Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and an accuracy mark depends on the method mark it follows. Follow-through (FT) means an earlier slip need not cost you the marks that depend on it, provided the later step is done correctly on your own figure, and a GDC answer is expected rather than hand-computed values. Study how each mark below is earned by a specific line.

Where this leads

Sinusoidal modelling is where the trigonometry of angles meets the language of functions. The amplitude, period and vertical shift you read here are the same transformations of graphs you met earlier, now attached to a repeating curve; the GDC solving you practise is exactly the technique you will reuse whenever a model has to be set equal to a target. Every context that repeats — tides, temperatures, daylight, rotating wheels, alternating signals — reduces to the same three questions: how big is the swing, how long is the cycle, and where is the middle line. Answer those and read the model with a calculator in the right mode, and periodic phenomena stop being mysterious and become just another function you can predict from.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

The height above the ground, $H$ metres, of a seat on a Ferris wheel is modelled by $H(t) = 30\sin(4t) + 35$ , where $t$ is the time in seconds and the angle is in degrees. State the amplitude and the principal axis, find the period, and hence write down the maximum and minimum heights of the seat. [5]

1
Read the parameters straight from the model $H(t) = 30\sin(4t) + 35$ .\n\nAmplitude. The coefficient of the sine term is $30$ , so the amplitude is $|a| = 30$ m. [A1]\n\nPrincipal axis. The vertical shift is $d = 35$ , so the seat oscillates about the line $H = 35$ m. [A1]\n\nPeriod. Working in degrees, $\text{period} = \dfrac{360^\circ}{b} = \dfrac{360^\circ}{4} = 90$ seconds. [M1 for $\frac{360}{b}$ ] [A1]\n\nMaximum and minimum heights. With $a > 0$ , maximum $= d + a = 35 + 30 = 65$ m and minimum $= d - a = 35 - 30 = 5$ m. [A1]\n\nSo the wheel carries the seat between $5$ m and $65$ m above the ground, completing a full turn every $90$ seconds.

✓ FULLY WORKED

EXAMPLE 2

The depth of water in a harbour is modelled by $d(t) = 4\sin(30t) + 7$ , where $d$ is the depth in metres, $t$ is the time in hours after midnight, and the angle is measured in degrees. State the amplitude and the period, and find the maximum depth and the first time it occurs. [5]

1
Amplitude. The coefficient of the sine term is $4$ , so the amplitude is $|a| = 4$ m. [A1]\n\nPeriod. In degrees, $\text{period} = \dfrac{360^\circ}{b} = \dfrac{360^\circ}{30} = 12$ hours. [M1 for $\frac{360}{b}$ ] [A1]\n\nMaximum depth. The principal axis is $d = 7$ and the amplitude is $4$ , so the greatest depth is $d + a = 7 + 4 = 11$ m. [A1]\n\nFirst time it occurs. A sine model that starts on its principal axis and rises reaches its first maximum a quarter of a period after $t = 0$ . Here a quarter period is $\frac{1}{4} \times 12 = 3$ hours, so the depth first reaches $11$ m at $t = 3$ (i.e. 03:00). A GDC graph of $d(t) = 4\sin(30t) + 7$ confirms the first peak at $t = 3$ . [A1]\n\nAnswers: amplitude $4$ m, period $12$ h, maximum depth $11$ m first reached at $t = 3$ h.

✓ FULLY WORKED

EXAMPLE 3

The number of hours of daylight in a northern town is modelled by $L(t) = 4\cos(30t) + 12$ , where $L$ is measured in hours, $t$ is the number of months after the longest day, and the angle is in degrees. \n (a) State the maximum and minimum hours of daylight the model predicts. \n (b) Use your GDC to find the first value of $t$ (for $t > 0$ ) at which the town has exactly $12$ hours of daylight. [5]

1
(a) Maximum and minimum. The amplitude is $|a| = 4$ and the principal axis is $d = 12$ , so maximum $= d + a = 12 + 4 = 16$ hours and minimum $= d - a = 12 - 4 = 8$ hours. [A1 max] [A1 min]\n\n**(b) When is $L = 12$ ?** Set the model equal to $12$ : $4\cos(30t) + 12 = 12$ , so $4\cos(30t) = 0$ . [M1 setting model $=$ target]\nGraph $Y_1 = 4\cos(30t) + 12$ and $Y_2 = 12$ on the GDC (degree mode) and read the first intersection for $t > 0$ . [M1 GDC method]\nThe first intersection is at $t = 3$ months. [A1]\n\n(The cosine model starts at its maximum of $16$ hours at $t = 0$ , the longest day, and first crosses the principal axis of $12$ hours a quarter of a cycle later — a quarter of the $\frac{360^\circ}{30} = 12$ -month period, i.e. $t = 3$ .)

✓ FULLY WORKED

EXAMPLE 4

The depth of water is modelled by $d(t) = 4\sin(30t) + 7$ (metres, $t$ hours, degrees). State the amplitude and period, and find the maximum depth and the first time it occurs. [5]

1
Model answer — full working.\n\nThe model is $d(t) = 4\sin(30t) + 7$ , with the angle in degrees, so set the GDC to degree mode.\n\nAmplitude. The coefficient of $\sin(30t)$ is $4$ , so the amplitude is $|a| = 4$ m.\n\nPeriod. Using $\text{period} = \dfrac{360^\circ}{b}$ with $b = 30$ : $\dfrac{360^\circ}{30} = 12$ hours.\n\nMaximum depth. The principal axis is $d = 7$ and the amplitude is $4$ , so the maximum depth is $7 + 4 = 11$ m.\n\nFirst time it occurs. The sine model starts on the principal axis at $t = 0$ and rises, so its first maximum is a quarter of a period later: $\frac{1}{4} \times 12 = 3$ hours. A GDC graph of $d(t)$ confirms the first peak at $t = 3$ .\n\nAnswers: amplitude $4$ m, period $12$ h, maximum depth $11$ m, first reached at $t = 3$ h.\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — amplitude. Awarded for stating the amplitude is $4$ , read as the coefficient of the sine term. This accuracy mark stands on correctly identifying $|a|$ , not $d$ .\n- M1 — period method. A method mark for using $\text{period} = \dfrac{360^\circ}{b}$ with $b = 30$ , i.e. forming $\frac{360}{30}$ . It is the approach that is rewarded, so it survives an arithmetic slip in the division.\n- A1 — period. Awarded for $\text{period} = 12$ hours. This A-mark depends on the M1 above: it is earned once $\frac{360^\circ}{b}$ is set up and evaluated correctly. FT applies — a candidate who mis-copied $b$ but divided $\frac{360^\circ}{b}$ correctly still earns it on their own figure.\n- A1 — maximum depth. Awarded for maximum $= d + a = 7 + 4 = 11$ m, combining the principal axis and amplitude. FT on the candidate's own $a$ and $d$ .\n- A1 — first time. Awarded for $t = 3$ h, identified as a quarter of the period (or read from the GDC graph). This is FT on the candidate's own period: a student whose period differed but who correctly took a quarter of THEIR period keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the period written as $\frac{360}{30}$ or $12$ or $12.0$ , the maximum as $7+4$ or $11$ , and the time as $t = 3$ or '03:00'. Once a correct value appears, subsequent restatements do not lose marks (ISW).\n\nBottom line: of the 5 marks, the period method mark survives an arithmetic slip and three accuracy marks are shielded by follow-through. A student who writes only 'max $= 11$ , $t = 3$ ' with no amplitude, no $\frac{360^\circ}{b}$ and no reasoning risks losing 3 marks; a student who shows the amplitude, the period formula, the sum $d + a$ and the quarter-period keeps the method regardless of a slip in one number.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAModelling with sinusoidal functions

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 11

What is a sinusoidal model?

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ANSWER

A model of a repeating (periodic) quantity built from a sine or cosine curve, usually written $y = a\sin(bx) + d$ or $y = a\cos(bx) + d$ . The three numbers $a$ , $b$ and $d$ fix the size, speed and height of the oscillation.

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1 / 11 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Amplitude $|a|$ : the distance from the principal axis up to a maximum (or down to a minimum). From data, $|a| = \frac{\text{max} - \text{min}}{2}$ . It is always positive — it measures the size of the swing, not its direction.
✓
Period and $b$ : the period is the length of one full cycle. In degrees, $\text{period} = \frac{360^\circ}{b}$ , so $b = \frac{360^\circ}{\text{period}}$ ; in radians replace $360^\circ$ with $2\pi$ .
✓
Vertical shift / principal axis $d$ : the curve oscillates about the horizontal line $y = d$ , where $d = \frac{\text{max} + \text{min}}{2}$ . This is the 'middle' of the wave.
✓
Maximum and minimum: with $a > 0$ the model reaches a maximum of $d + a$ and a minimum of $d - a$ . Knowing $d$ and $a$ hands you the top and bottom instantly.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

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Get a Paper 2 sinusoidal model marked: state the amplitude and period, then predict and solve with full working

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Checkpoint

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Modelling with sinusoidal functions

BY THE END, YOU CAN…

1Interpret the amplitude $|a|$ , the period $\frac{360^\circ}{b}$ and the vertical shift (principal axis) $d$ in the model $y = a\sin(bx) + d$ or $y = a\cos(bx) + d$ .
2Determine these parameters from a graph, from a given maximum and minimum, or from a real-world context such as tides, temperature, a Ferris wheel or daylight hours.
3Use a sinusoidal model to predict a value at a given input and to find when a stated value first occurs.
4Use a GDC to solve equations arising from a sinusoidal model, and present working so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Riding the Waves of Data

THINK OF IT LIKE…

1
Find the middle line, $d$ . Average the maximum and minimum values: $d = \frac{\text{max} + \text{min}}{2}$ . This is the principal axis, the horizontal line the curve oscillates about.
2
Find the amplitude, $a$ . Take half the gap between the maximum and minimum: $a = \frac{\text{max} - \text{min}}{2}$ . It is the height of the swing above (and below) the middle line.
3
Find $b$ from the period. Measure the length of one complete cycle, then use $\text{period} = \frac{360^\circ}{b}$ , so $b = \frac{360^\circ}{\text{period}}$ when working in degrees.
4
Use the model. Substitute a value to predict an output, or set the model equal to a target value and let the GDC find the input — reading the answer that fits the interval the question asks for.

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

GEOGEBRASTEP 1 / 4 · Step 1

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GeoGebra Sinusoidal Model Explorer

Use “GeoGebra Sinusoidal Model Explorer” — shift curves or drag points to see equilibrium and slopes change.

If the embed does not load, use Open on GeoGebra.

GeoGebra · GeoGebra Terms of Service

Step 1

Find the middle line, $d$ . Average the maximum and minimum values: $d = \frac{\text{max} + \text{min}}{2}$ . This is the principal axis, the horizontal line the curve oscillates about.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$y = a\sin(bx) + d \qquad \text{or} \qquad y = a\cos(bx) + d$

Full topic notes

Formal explanation with the rigour you need for the exam.

The anatomy of a sinusoidal model

$y = a\sin(bx) + d \qquad \text{or} \qquad y = a\cos(bx) + d$

Amplitude $|a|$ : the distance from the principal axis up to a maximum (or down to a minimum). From data, $|a| = \frac{\text{max} - \text{min}}{2}$ . It is always positive — it measures the size of the swing, not its direction.
Period and $b$ : the period is the length of one full cycle. In degrees, $\text{period} = \frac{360^\circ}{b}$ , so $b = \frac{360^\circ}{\text{period}}$ ; in radians replace $360^\circ$ with $2\pi$ .
Vertical shift / principal axis $d$ : the curve oscillates about the horizontal line $y = d$ , where $d = \frac{\text{max} + \text{min}}{2}$ . This is the 'middle' of the wave.
Maximum and minimum: with $a > 0$ the model reaches a maximum of $d + a$ and a minimum of $d - a$ . Knowing $d$ and $a$ hands you the top and bottom instantly.

Reading the parameters: from a graph, from max/min, or from a context

Using the model: predicting and solving with the GDC

Common mistakes examiners penalise

Calling $d$ the amplitude — the amplitude is the coefficient $|a|$ that multiplies the sine or cosine, not the vertical shift $d$ . In $y = 4\sin(30x) + 7$ the amplitude is $4$ , not $7$ .
Getting the period the wrong way up — the period is $\frac{360^\circ}{b}$ (degrees) or $\frac{2\pi}{b}$ (radians), not $360^\circ \times b$ and not simply $b$ . For $y = 6\cos(20x) + 5$ the period is $\frac{360^\circ}{20} = 18^\circ$ , not $20$ .
Working in the wrong calculator mode — a model written with $\frac{360^\circ}{b}$ or an angle like $30t$ is in degrees, so the GDC must be in degree mode. Radian mode gives a plausible-looking but wrong answer.
Forgetting the maximum is $d + a$ , not $a$ — with $a>0$ the maximum is principal axis plus amplitude, $d + a$ , and the minimum is $d - a$ . Quoting the amplitude as the maximum loses the mark.
Putting the principal axis at $y = 0$ — the curve oscillates about $y = d$ , not the $x$ -axis. State the principal axis as $y = d$ using the vertical-shift value.
Taking amplitude as negative when the graph dips below the axis — amplitude is the positive distance $\frac{\text{max}-\text{min}}{2}$ . A negative coefficient only reflects the curve; the amplitude is still $|a|$ .
Reporting every solution instead of the first — a sinusoidal equation repeats each cycle, so pick the solution inside the interval the question specifies (often the smallest $t > 0$ ), not just any intersection the GDC shows.

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

1
Read the parameters straight from the model $H(t) = 30\sin(4t) + 35$ .\n\nAmplitude. The coefficient of the sine term is $30$ , so the amplitude is $|a| = 30$ m. [A1]\n\nPrincipal axis. The vertical shift is $d = 35$ , so the seat oscillates about the line $H = 35$ m. [A1]\n\nPeriod. Working in degrees, $\text{period} = \dfrac{360^\circ}{b} = \dfrac{360^\circ}{4} = 90$ seconds. [M1 for $\frac{360}{b}$ ] [A1]\n\nMaximum and minimum heights. With $a > 0$ , maximum $= d + a = 35 + 30 = 65$ m and minimum $= d - a = 35 - 30 = 5$ m. [A1]\n\nSo the wheel carries the seat between $5$ m and $65$ m above the ground, completing a full turn every $90$ seconds.

✓ FULLY WORKED

EXAMPLE 2

1
Amplitude. The coefficient of the sine term is $4$ , so the amplitude is $|a| = 4$ m. [A1]\n\nPeriod. In degrees, $\text{period} = \dfrac{360^\circ}{b} = \dfrac{360^\circ}{30} = 12$ hours. [M1 for $\frac{360}{b}$ ] [A1]\n\nMaximum depth. The principal axis is $d = 7$ and the amplitude is $4$ , so the greatest depth is $d + a = 7 + 4 = 11$ m. [A1]\n\nFirst time it occurs. A sine model that starts on its principal axis and rises reaches its first maximum a quarter of a period after $t = 0$ . Here a quarter period is $\frac{1}{4} \times 12 = 3$ hours, so the depth first reaches $11$ m at $t = 3$ (i.e. 03:00). A GDC graph of $d(t) = 4\sin(30t) + 7$ confirms the first peak at $t = 3$ . [A1]\n\nAnswers: amplitude $4$ m, period $12$ h, maximum depth $11$ m first reached at $t = 3$ h.

✓ FULLY WORKED

EXAMPLE 3

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(a) Maximum and minimum. The amplitude is $|a| = 4$ and the principal axis is $d = 12$ , so maximum $= d + a = 12 + 4 = 16$ hours and minimum $= d - a = 12 - 4 = 8$ hours. [A1 max] [A1 min]\n\n**(b) When is $L = 12$ ?** Set the model equal to $12$ : $4\cos(30t) + 12 = 12$ , so $4\cos(30t) = 0$ . [M1 setting model $=$ target]\nGraph $Y_1 = 4\cos(30t) + 12$ and $Y_2 = 12$ on the GDC (degree mode) and read the first intersection for $t > 0$ . [M1 GDC method]\nThe first intersection is at $t = 3$ months. [A1]\n\n(The cosine model starts at its maximum of $16$ hours at $t = 0$ , the longest day, and first crosses the principal axis of $12$ hours a quarter of a cycle later — a quarter of the $\frac{360^\circ}{30} = 12$ -month period, i.e. $t = 3$ .)

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EXAMPLE 4

The depth of water is modelled by $d(t) = 4\sin(30t) + 7$ (metres, $t$ hours, degrees). State the amplitude and period, and find the maximum depth and the first time it occurs. [5]

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Model answer — full working.\n\nThe model is $d(t) = 4\sin(30t) + 7$ , with the angle in degrees, so set the GDC to degree mode.\n\nAmplitude. The coefficient of $\sin(30t)$ is $4$ , so the amplitude is $|a| = 4$ m.\n\nPeriod. Using $\text{period} = \dfrac{360^\circ}{b}$ with $b = 30$ : $\dfrac{360^\circ}{30} = 12$ hours.\n\nMaximum depth. The principal axis is $d = 7$ and the amplitude is $4$ , so the maximum depth is $7 + 4 = 11$ m.\n\nFirst time it occurs. The sine model starts on the principal axis at $t = 0$ and rises, so its first maximum is a quarter of a period later: $\frac{1}{4} \times 12 = 3$ hours. A GDC graph of $d(t)$ confirms the first peak at $t = 3$ .\n\nAnswers: amplitude $4$ m, period $12$ h, maximum depth $11$ m, first reached at $t = 3$ h.\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — amplitude. Awarded for stating the amplitude is $4$ , read as the coefficient of the sine term. This accuracy mark stands on correctly identifying $|a|$ , not $d$ .\n- M1 — period method. A method mark for using $\text{period} = \dfrac{360^\circ}{b}$ with $b = 30$ , i.e. forming $\frac{360}{30}$ . It is the approach that is rewarded, so it survives an arithmetic slip in the division.\n- A1 — period. Awarded for $\text{period} = 12$ hours. This A-mark depends on the M1 above: it is earned once $\frac{360^\circ}{b}$ is set up and evaluated correctly. FT applies — a candidate who mis-copied $b$ but divided $\frac{360^\circ}{b}$ correctly still earns it on their own figure.\n- A1 — maximum depth. Awarded for maximum $= d + a = 7 + 4 = 11$ m, combining the principal axis and amplitude. FT on the candidate's own $a$ and $d$ .\n- A1 — first time. Awarded for $t = 3$ h, identified as a quarter of the period (or read from the GDC graph). This is FT on the candidate's own period: a student whose period differed but who correctly took a quarter of THEIR period keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the period written as $\frac{360}{30}$ or $12$ or $12.0$ , the maximum as $7+4$ or $11$ , and the time as $t = 3$ or '03:00'. Once a correct value appears, subsequent restatements do not lose marks (ISW).\n\nBottom line: of the 5 marks, the period method mark survives an arithmetic slip and three accuracy marks are shielded by follow-through. A student who writes only 'max $= 11$ , $t = 3$ ' with no amplitude, no $\frac{360^\circ}{b}$ and no reasoning risks losing 3 marks; a student who shows the amplitude, the period formula, the sum $d + a$ and the quarter-period keeps the method regardless of a slip in one number.

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How it all connects

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MAIN IDEAModelling with sinusoidal functions

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What is a sinusoidal model?

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Key takeaways

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Amplitude $|a|$ : the distance from the principal axis up to a maximum (or down to a minimum). From data, $|a| = \frac{\text{max} - \text{min}}{2}$ . It is always positive — it measures the size of the swing, not its direction.
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Period and $b$ : the period is the length of one full cycle. In degrees, $\text{period} = \frac{360^\circ}{b}$ , so $b = \frac{360^\circ}{\text{period}}$ ; in radians replace $360^\circ$ with $2\pi$ .
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Vertical shift / principal axis $d$ : the curve oscillates about the horizontal line $y = d$ , where $d = \frac{\text{max} + \text{min}}{2}$ . This is the 'middle' of the wave.
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Maximum and minimum: with $a > 0$ the model reaches a maximum of $d + a$ and a minimum of $d - a$ . Knowing $d$ and $a$ hands you the top and bottom instantly.

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Revise Modelling with sinusoidal functions for IB Mathematics: Applications and Interpretation (2.5)

In simple terms

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Key formulas

Full topic notes

The anatomy of a sinusoidal model

Reading the parameters: from a graph, from max/min, or from a context

Using the model: predicting and solving with the GDC

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

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How it all connects

Glossary

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Key takeaways

Practice — then mark it

Get a Paper 2 sinusoidal model marked: state the amplitude and period, then predict and solve with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

The anatomy of a sinusoidal model

Reading the parameters: from a graph, from max/min, or from a context

Using the model: predicting and solving with the GDC

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 sinusoidal model marked: state the amplitude and period, then predict and solve with full working

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Discuss Modelling with sinusoidal functions