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Trigonometric functions and graphs (QCE Math Methods Unit 2)

Syllabus dot point

“Sketch and analyse graphs of y=asin⁡(b(x−h))+ky = a \sin(b(x - h)) + k and y=acos⁡(b(x−h))+ky = a \cos(b(x - h)) + k, identifying amplitude, period, phase shift and vertical translation”

7 min read

Quick answer

For y=asin⁡(b(x−h))+ky = a \sin(b(x-h)) + k and y=acos⁡(b(x−h))+ky = a \cos(b(x-h)) + k, the amplitude is ∣a∣|a|, the period is 2π/∣b∣2\pi/|b|, the phase shift is hh (right if positive), and the centre line is y=ky = k, with the range [k−∣a∣,k+∣a∣][k - |a|, k + |a|].

Jump to a section
  1. What this dot point is asking
  2. The parent functions y=sin⁡xy = \sin x and y=cos⁡xy = \cos x
  3. The transformed form y=asin⁡(b(x−h))+ky = a \sin(b(x - h)) + k
  4. Why each parameter does what it does
  5. Key features for sketching
  6. Periodic modelling
  7. Exam-style questions
Key fact

In the 2025 syllabus, trigonometric functions and their graphs are introduced in Unit 1 (Topic 4: Trigonometric functions). This page stays in Unit 2 so its link does not change.

What this dot point is asking

QCAA wants you to sketch and analyse transformed sine and cosine graphs, identifying the four key parameters (amplitude, period, phase shift, vertical translation) and using them to model periodic phenomena like tides, temperatures or pendulum displacement.

The parent functions y=sin⁡xy = \sin x and y=cos⁡xy = \cos x

For sin⁡x\sin x:

  • Domain: all real xx.
  • Range: [−1,1][-1, 1].
  • Period: 2π2\pi.
  • Amplitude: 11.
  • Zeros at x=nπx = n\pi for integer nn.
  • Maximum 11 at x=π/2+2nπx = \pi/2 + 2n\pi. Minimum −1-1 at x=3π/2+2nπx = 3\pi/2 + 2n\pi.

For cos⁡x\cos x:

  • Same domain, range, period, amplitude.
  • Zeros at x=π/2+nπx = \pi/2 + n\pi.
  • Maximum 11 at x=2nπx = 2n\pi. Minimum −1-1 at x=(2n+1)πx = (2n+1)\pi.

cos⁡x=sin⁡(x+π/2)\cos x = \sin(x + \pi/2), so the cosine graph is the sine graph shifted left by π/2\pi/2; the two are the same wave with a quarter-period offset. Both are bounded between −1-1 and 11 and repeat every 2π2\pi, which is why amplitude and period are the first two features to read from any transformed version.

The transformed form y=asin⁡(b(x−h))+ky = a \sin(b(x - h)) + k

Each parameter has a distinct effect.

  • ∣a∣|a|: amplitude. Vertical dilation. Range becomes [k−∣a∣,k+∣a∣][k - |a|, k + |a|]. If a<0a < 0, the graph is reflected vertically.
  • bb: angular frequency. Period =2π/∣b∣= 2\pi / |b|. Larger ∣b∣|b| compresses the graph horizontally.
  • hh: horizontal phase shift. Graph moves hh units right (if h>0h > 0).
  • kk: vertical translation. Centre line moves to y=ky = k.

The same parameters apply to y=acos⁡(b(x−h))+ky = a \cos(b(x - h)) + k.

Why each parameter does what it does

The four parameters map onto the four standard transformations of a graph. The amplitude aa is a vertical dilation, stretching the wave away from its centre line and flipping it when negative. The factor bb is a horizontal dilation, and because it acts inside the function it compresses the period to 2πb\dfrac{2\pi}{b} (a larger bb means more cycles in the same span). The phase shift hh is a horizontal translation, and kk is a vertical translation that sets the centre line. Reading these straight off the equation is the fastest route to both a sketch and a description.

Key features for sketching

For y=asin⁡(b(x−h))+ky = a \sin(b(x-h)) + k:

  • Centre line: y=ky = k.
  • Maximum: y=k+∣a∣y = k + |a|.
  • Minimum: y=k−∣a∣y = k - |a|.
  • Period: 2π/∣b∣2\pi / |b|.
  • A complete cycle goes: centre, max, centre, min, centre.
  • First zero of the parent sine occurs at x=hx = h.

For cosine, replace "centre" first with "maximum" first, because cos⁡\cos starts at its peak. Plotting these five quarter-period points and joining them with a smooth curve produces an accurate sketch over one period, which then repeats.

Sketching a transformed cosine

Sketch y=−3cos⁡ ⁣(2 ⁣(x−π4))+5y = -3\cos\!\left(2\!\left(x - \tfrac{\pi}{4}\right)\right) + 5 over one complete period, marking the amplitude, period, centre line and five key points.

Step 1: Read the four parameters

Match the equation to y=acos⁡(b(x−h))+ky = a\cos(b(x - h)) + k: a=−3a = -3, b=2b = 2, h=π4h = \tfrac{\pi}{4}, k=5k = 5.

Step 2: Compute amplitude, period and centre line

The amplitude is ∣a∣=3|a| = 3, so the curve oscillates 33 units above and below its centre. The centre line is y=k=5y = k = 5, giving a range of [5−3,5+3]=[2,8][5 - 3, 5 + 3] = [2, 8]. The period is:

2πb=2π2=π.\frac{2\pi}{b} = \frac{2\pi}{2} = \pi.

The phase shift h=π4h = \tfrac{\pi}{4} moves the graph π4\tfrac{\pi}{4} to the right.

Step 3: Account for the reflection

Because a=−3<0a = -3 < 0, the graph is reflected vertically. An ordinary cosine starts at its maximum; this one starts at its minimum. Every maximum and minimum of the standard cosine is swapped.

Step 4: Identify the five key points over one period

Starting at x=h=π4x = h = \tfrac{\pi}{4} (where a standard cosine would peak, but here it troughs because of the reflection), the five quarter-period landmarks are:

xx Feature
π4\tfrac{\pi}{4} minimum at y=2y = 2
π2\tfrac{\pi}{2} rising through centre at y=5y = 5
3π4\tfrac{3\pi}{4} maximum at y=8y = 8
π\pi falling through centre at y=5y = 5
5π4\tfrac{5\pi}{4} next minimum at y=2y = 2

Join these five points with a smooth wave to complete the sketch.

Final answer: Centre line y=5y = 5; amplitude 33; period π\pi; range [2,8][2, 8]; graph starts at minimum (π4,2)(\tfrac{\pi}{4}, 2).

Periodic modelling

Periodic functions are the natural model for any quantity that repeats: tides rising and falling, daily temperature, hours of daylight through the year, and alternating current. The skill is to read the four parameters from the described behaviour rather than from an equation. Worded scenarios (tides, temperature, alternating current) give the maximum, minimum and the time of one extremum. From those:

  • kk = (max + min) / 2 (centre line).
  • ∣a∣|a| = (max - min) / 2 (amplitude).
  • b=2π/b = 2\pi / period.
  • hh = time at which the sine starts at zero rising, or the cosine starts at maximum.
Common traps
Confusing bb with the period
Period is 2π/b2\pi / b, not bb itself.
Forgetting that hh shifts in the same direction as its sign
sin⁡(2(x−π/4))\sin(2(x - \pi/4)) shifts right by π/4\pi/4, not left.
Missing the reflection when a<0a < 0
Negative aa flips the graph; max becomes min and vice versa.
Reading the model in degrees by accident
QCAA Math Methods always uses radians.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks
Paper 2 (complex familiar). The temperature is modelled by T(t)=6sin⁡ ⁣(πt12)+18T(t) = 6\sin\!\left(\dfrac{\pi t}{12}\right) + 18, with TT in ∘^\circC and tt hours after 6 am. Determine (a) the amplitude, (b) the period, (c) the maximum and minimum, (d) the temperature at 3 pm.
Show worked answer →

(a) Amplitude =∣a∣=6∘= |a| = 6^\circC.

(b) Period =2πb=2ππ/12=24= \dfrac{2\pi}{b} = \dfrac{2\pi}{\pi/12} = 24 hours.

(c) Range =18±6= 18 \pm 6, so maximum 24∘24^\circC and minimum 12∘12^\circC.

(d) At 3 pm, t=9t = 9: T(9)=6sin⁡3π4+18=6×22+18=32+18≈22.2∘T(9) = 6\sin\dfrac{3\pi}{4} + 18 = 6 \times \dfrac{\sqrt 2}{2} + 18 = 3\sqrt 2 + 18 \approx 22.2^\circC.

Markers reward identifying the parameters, the period from 2πb\tfrac{2\pi}{b}, and the exact-value substitution.

Practice4 marks
Paper 1 (technique). For y=2sin⁡ ⁣(3x)−1y = 2\sin\!\left(3x\right) - 1, state (a) the amplitude, (b) the period, (c) the maximum value, and (d) the number of complete cycles in 0≤x≤2π0 \leq x \leq 2\pi.
Show worked answer →

(a) Amplitude =2= 2. (b) Period =2π3= \dfrac{2\pi}{3}. (c) Maximum =k+∣a∣=−1+2=1= k + |a| = -1 + 2 = 1.

(d) Number of cycles =2πperiod=2π2π/3=3= \dfrac{2\pi}{\text{period}} = \dfrac{2\pi}{2\pi/3} = 3 complete cycles.

Markers reward the amplitude, the period from 2πb\tfrac{2\pi}{b}, the maximum from the midline plus amplitude, and the cycle count.

Practise this

Sources & how we know this

ExamExplained