IB Math AI HL Trigonometric Graphs — Complete Cheatsheet
Every skill for the sinusoidal model in IB Mathematics Applications & Interpretation HL — amplitude, period, principal axis and phase, reading the four constants off a graph, and solving trig equations on the GDC, with the real-world applications of tides, Ferris wheels & temperature. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Trigonometric graphs are where IB Mathematics Applications & Interpretation HL turns the sine and cosine curves into a genuine modelling tool. Anything that rises and falls in a steady repeating rhythm — the daily tides, the height of a Ferris-wheel car, the temperature through the year, a sound wave — is captured by a single sinusoidal model $y=a\sin(b(x-c))+d$. Because a calculator is always allowed, the marks live in reading the four constants from a context or a graph and in solving on the GDC — not in hand-grinding trigonometry.
This cheatsheet condenses the whole of SL 3.5 · AHL 3.7–3.8 — amplitude, period, principal axis and phase, reading $a,b,c,d$ off a graph, evaluating the model, and solving trig equations in a given interval — onto one page, and flags the traps (the missing $\div2$, the wrong GDC mode) that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Anatomy of the sinusoidal model SL 3.5
A sinusoidal model describes any quantity that oscillates steadily about a central level. Every constant carries a physical meaning, so reading them off a context — or back off a graph — is the whole game.
Sine form:$y=a\sin\big(b(x-c)\big)+d$
Cosine form:$y=a\cos\big(b(x-c)\big)+d$ — the same wave, shifted
Amplitude $a$:$a=\dfrac{\text{max}-\text{min}}{2}$ — how far the curve swings from the midline
Principal axis $d$:$d=\dfrac{\text{max}+\text{min}}{2}$ — the midline $y=d$ the wave oscillates about
Period:$\dfrac{360^\circ}{b}$ (degrees) or $\dfrac{2\pi}{b}$ (radians) — the length of one full cycle
Phase shift $c$:slides the whole wave sideways along the $x$-axis
The sinusoidal model: amplitude $a$ is the swing from the midline to a peak, the period is the crest-to-crest distance, and $d$ is the midline the wave rocks about. Sketch this first to fix what each constant means.
NoteThe four constants are independent knobs: $a$ stretches the wave vertically, $d$ raises or lowers it, $b$ squeezes it horizontally, and $c$ slides it left or right. Change one and only that feature moves.
TrickChoose sine or cosine by the starting point. A $+a\cos$ model starts at its maximum at $x=0$ (since $\cos 0=1$); a $-a\cos$ model starts at its minimum; a $\sin$ model starts on the midline, rising. A Ferris wheel boarding at the bottom is a natural $-a\cos$.
§2 — Amplitude & principal axis from max/min SL 3.5
The fastest way into any model is the highest and lowest values it reaches. From that single pair you get two of the four constants immediately.
Amplitude:$a=\dfrac{\text{max}-\text{min}}{2}$ — half the vertical distance between peak and trough
Principal axis:$d=\dfrac{\text{max}+\text{min}}{2}$ — the average of the peak and trough
For a tide running between $0.8$ m and $4.2$ m: $a=\dfrac{4.2-0.8}{2}=1.7$ m and $d=\dfrac{4.2+0.8}{2}=2.5$ m — the sea rocks $1.7$ m either side of a $2.5$ m mean.
TrapAmplitude is half the peak-to-trough distance — forgetting the $\div 2$ is the single most common dropped mark on this topic. It is also always positive: subtract min from max, then halve.
§3 — Period & finding $b$ SL 3.5
The constant $b$ controls how tightly the wave is packed. You almost never read $b$ directly — you read the period from the context, then convert.
Period from $b$:$\text{period}=\dfrac{360^\circ}{b}$ (degrees) or $\dfrac{2\pi}{b}$ (radians)
$b$ from period:$b=\dfrac{360^\circ}{\text{period}}$ or $b=\dfrac{2\pi}{\text{period}}$
TrickSpot one full cycle in the context — one rotation, one day, one wavelength — and that duration is the period. A Ferris wheel turning once every $4$ minutes has period $4$, so $b=\dfrac{360^\circ}{4}=90^\circ$ per minute.
TrapPeriod and $b$ are inversely related: a larger $b$ squeezes the graph into a shorter period. If your period comes out bigger when $b$ grows, you have divided the wrong way round.
§4 — Maximum & minimum values SL 3.5
Once you know the midline and the amplitude, the extreme values need no calculus at all — just add and subtract.
Maximum:$y_{\max}=d+a$
Minimum:$y_{\min}=d-a$
NoteBoth $\sin$ and $\cos$ run between $-1$ and $1$, so the model runs between $d-a$ and $d+a$. Add the amplitude to the midline for the peak, subtract it for the trough — differentiating a sinusoid to find its maximum is wasted effort in AI.
§5 — Evaluating the model SL 3.5
To find the output at a given input, substitute and let the GDC do the arithmetic — working from the inside of the bracket outwards.
Evaluate:compute $b(x-c)$ first, take $\sin$ or $\cos$, then multiply by $a$ and add $d$
Peak time (cos):$d+a\cos\big(b(t-c)\big)$ peaks when the bracket is $0$, i.e. at $t=c$, where $y=d+a$
A yearly temperature model $T(t)=d+a\cos\!\left(\dfrac{\pi}{6}(t-c)\right)$ has period $\dfrac{2\pi}{\pi/6}=12$ months and reaches its hottest value $T=d+a$ at month $t=c$ — read straight off, no working.
TrapMatch the GDC mode to the model. A model built with $\dfrac{\pi}{6}$ or $\dfrac{2\pi}{b}$ is in radians; one built with $\dfrac{360^\circ}{b}$ is in degrees. A stray degree setting on a $\pi$-model is the number-one error on these questions.
§6 — Reading $a,b,c,d$ off a graph or data SL 3.5
The reverse task — given a plotted curve or a table of readings, recover the model — is a staple of AI Paper 2. Work through the four constants in order.
Constant
How to read it from the graph
$d$ (midline)
$\dfrac{\text{max}+\text{min}}{2}$ — the level halfway between the highest and lowest points
$a$ (amplitude)
$\dfrac{\text{max}-\text{min}}{2}$ — the swing from the midline to a peak
$b$
measure the period (peak-to-peak or trough-to-trough), then $b=\dfrac{360^\circ}{\text{period}}$ or $\dfrac{2\pi}{\text{period}}$
$c$ (shift)
the horizontal distance the first peak (cos) or first rising midline-crossing (sin) sits from $x=0$
TrickAlways in the order $d,a,b,c$. Nail the midline and amplitude from the max and min, then the period gives $b$, and only then read the shift $c$ — because $c$ is measured relative to the wave you have already pinned down.
§7 — Solving trig equations in an interval AHL 3.7
The commonest HL-style question asks when the model equals a value — "at what times is the tide exactly $3$ m?" Solve on the GDC over the stated interval.
Principal value:$bx=\arcsin\!\left(\dfrac{k-d}{a}\right)$, then divide by $b$ for the first solution
GDC-first:graph $y_1=$ model and $y_2=k$, then use intersect across the whole interval
TrickGo GDC-first: plot both curves, set the window to the given interval, and read every intersection. This finds all solutions without juggling reference angles by hand.
Trap$\arcsin$ returns only one angle, but a sine equation has two solutions per cycle — the partner is $180^\circ-\arcsin s$ (degrees). Never quote the single principal value when the question asks for solutions "in the interval".
§8 — Duration above a threshold AHL 3.8
"For how long each cycle is the rider above $40$ m?" or "how many hours is the tide above $3$ m?" — find the two crossing times, then subtract.
Crossings:solve $\sin(bx)=s$ with $s=\dfrac{k-d}{a}$ — the two roots are $\dfrac{\arcsin s}{b}$ and $\dfrac{180^\circ-\arcsin s}{b}$
Duration:the difference of the two crossing times
TrickThe two crossings are symmetric about the peak. On the GDC, find both intersection $x$-values in one period and subtract — the gap is the time spent above the level.
TrapThe answer is the difference of the two times, not the larger value. Reading off only the second crossing (and calling it the duration) is the classic slip here.
§9 — Synthesis: probability over one cycle AHL 3.8
Mixed topic — Trigonometric Graphs × Probability. If an event happens at a random instant during one cycle, the chance it meets a condition is a geometric probability: the favourable fraction of that cycle.
The condition $a\sin(bt)+d\ge k$ holds while $\sin(bt)\ge s$, with $s=\dfrac{k-d}{a}$. Over one full turn this is the symmetric arc from $\arcsin s$ to $180^\circ-\arcsin s$, so:
NoteThe constant $b$ cancels. The duration above the level depends on $b$ (a longer period means more real time above the line), but the probability does not — it is set by the shape of one cycle, not its length.
TrapTurn the requirement into $\sin(bt)\ge s$ first, take the arc $180^\circ-2\arcsin s$, then divide by the full $360^\circ$. Do not divide a duration in hours by the period unless the two are in the same units.
A Ferris wheel has diameter $60$ m and its lowest point is $2$ m above the ground. It completes one revolution every $4$ minutes, and a rider boards at the lowest point at time $t=0$. The height, in metres, is modelled by $h(t)=d-a\cos(bt)$ with $t$ in minutes and the calculator in degree mode. (a) Write down the amplitude $a$ and the principal axis $d$. (b) Show that $b=90$. (c) Find the height of the rider $1$ minute after boarding. (d) Find the length of time, during each revolution, that the rider is above $47$ m.
Solution
(a) The radius is half the diameter, so $a=\dfrac{60}{2}=30$ m. The lowest point is $2$ m and the highest is $2+60=62$ m, so $d=\dfrac{62+2}{2}=32$ m. (A1)(A1)
(b) One revolution is the period, $4$ minutes, so $b=\dfrac{360^\circ}{\text{period}}=\dfrac{360}{4}=90$. (M1)(A1)
(c) $h(1)=32-30\cos(90\times 1)=32-30\cos 90^\circ=32-30(0)=32$ m. (M1)(A1)(At a quarter-turn the rider is level with the midline — reassuringly exact.)
(d) Solve $32-30\cos(90t)=47\Rightarrow\cos(90t)=-0.5$, giving $90t=120^\circ$ or $240^\circ$, so $t=\dfrac{120}{90}=1.33$ and $t=\dfrac{240}{90}=2.67$ min. The rider is above $47$ m for $2.67-1.33=\dfrac{4}{3}\approx1.33$ min (i.e. $80$ s). (M1)(A1)
Examiner's note: in AI you would normally graph $h(t)$ and $y=47$ on the GDC and use intersect for part (d) — the algebra above just confirms the two times. Keep the calculator in degree mode throughout (the model uses $90^\circ$ per minute), give the duration as the difference of the two times, and round to 3 s.f. with units.
Common Student Questions
Why is the amplitude half the distance between the maximum and minimum?
Because the curve swings the same distance above and below the midline. The full peak-to-trough gap is therefore $2a$, so the amplitude is half of it: $a=\tfrac{\text{max}-\text{min}}{2}$. Quoting the whole gap (forgetting the $\div2$) is the most common error on this topic — and it throws off the maximum, minimum and every later part.
Should my GDC be in degrees or radians?
Match the mode to the model. If the constant $b$ was found from $\tfrac{360^\circ}{\text{period}}$, or the model contains $^\circ$, work in degrees. If it contains $\pi$ — such as $\tfrac{\pi}{6}$ or $\tfrac{2\pi}{b}$ — work in radians. A wrong mode gives numbers that look plausible but are completely off, and it is the single biggest source of lost marks here. Check the mode before every trig calculation.
Does it matter whether I use a sine or a cosine model?
Both describe the same wave, so either can fit — they differ only by a horizontal shift. Pick the one that matches the starting point: a $+a\cos$ model starts at its maximum, a $-a\cos$ model starts at its minimum, and a $\sin$ model starts on the midline, rising. If a rider boards a Ferris wheel at the bottom, $-a\cos$ needs no phase shift, which is why it is the natural choice.
Why does a trig equation have two answers in each cycle?
Because a horizontal line cuts one hump of the wave twice — once going up and once coming down. Your calculator's $\arcsin$ (or $\arccos$) only returns one of them; the partner in degrees is $180^\circ-\arcsin s$ for sine. The safe method is to graph the model and the line $y=k$ on the GDC and read every intersection inside the stated interval, so no solution is missed.
Does the period affect the probability of being above a certain level?
No — and that surprises students. The duration above a level does depend on the period (a slower wave spends more real time up there), but the probability at a random instant depends only on the fraction of one cycle spent above the line: $P=\tfrac{180^\circ-2\arcsin s}{360^\circ}$ with $s=\tfrac{k-d}{a}$. The constant $b$ cancels, because stretching the cycle stretches the favourable part and the whole cycle by the same factor.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every sinusoidal-model method in full: amplitude, period, phase, evaluating and solving, with the exact GDC keystrokes for degrees and radians.
Tutorial booklet — 25+ AI HL-style questions from reading a graph to full tide, Ferris-wheel and temperature models.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every trig-graph method.
Predicted-paper questions — the exact sinusoidal-modelling question types most likely in the next session.
