Skip to main content
VICMath MethodsSyllabus dot point

How are circular functions extended to model periodic phenomena?

Sketch and analyse trigonometric functions y=asin(b(xh))+ky = a\sin(b(x - h)) + k and y=acos(b(xh))+ky = a\cos(b(x - h)) + k, identifying amplitude, period, phase and vertical translation, and solve trig equations over a specified interval

A focused answer to the VCE Maths Methods Unit 2 dot point on circular functions. Sketches transformed sine and cosine graphs, identifies amplitude a|a|, period 2π/b2\pi/|b|, phase shift hh and vertical translation kk, and works the VCAA SAC-style trig equation sin(2x)=1/2\sin(2x) = 1/2 on a stated interval.

Generated by Claude Opus 4.87 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Parent functions
  3. Transformed form y=asin(b(xh))+ky = a\sin(b(x - h)) + k
  4. Solving trig equations over a stated interval
  5. Reference angles and quadrant signs
  6. Sketching transformed graphs
  7. In one sentence
  8. Examples in context
  9. Try this

What this dot point is asking

VCAA wants you to sketch and analyse transformed sine and cosine functions, identifying amplitude, period, phase shift and vertical translation, and to solve trigonometric equations over a specified interval.

Parent functions

sinx\sin x and cosx\cos x:

  • Domain: all real xx. Range: [1,1][-1, 1].
  • Period: 2π2\pi. Amplitude: 11.

Transformed form y=asin(b(xh))+ky = a\sin(b(x - h)) + k

Parameter Effect
a\|a\| Amplitude. Range [ka,k+a][k - \|a\|, k + \|a\|]. If a<0a < 0, vertical reflection.
bb Period = 2π/b2\pi/\|b\|. Larger b\|b\| compresses horizontally.
hh Phase shift; graph moves hh right (if positive).
kk Vertical translation; centre line y=ky = k.

Same parameters apply to y=acos(b(xh))+ky = a\cos(b(x - h)) + k.

Solving trig equations over a stated interval

  1. Isolate the trig function.
  2. Find the principal solution.
  3. Use symmetry and periodicity to find all solutions in the stated interval.
  4. If the equation involves bxbx, list all solutions for bxbx in the expanded interval and divide by bb.

Reference angles and quadrant signs

The engine of trig equation solving is the reference angle, the acute angle whose sine, cosine or tangent has the same magnitude as the required value. Find it from the positive value, then place solutions in the correct quadrants using the sign of the original equation: sine is positive in quadrants 1 and 2, cosine in quadrants 1 and 4, and tangent in quadrants 1 and 3. The standard placements are πθ\pi - \theta (quadrant 2), π+θ\pi + \theta (quadrant 3) and 2πθ2\pi - \theta (quadrant 4). Exact values from the special triangles (π6,π4,π3\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}) appear constantly, so commit them to memory for tech-free Exam 1.

Sketching transformed graphs

A reliable sketch shows the centre line y=ky = k, the maximum y=k+ay = k + |a| and minimum y=kay = k - |a|, and one full period of width 2πb\dfrac{2\pi}{|b|} starting from the phase shift hh. Mark the five key points across a period (the quarter, half and three-quarter points), since these locate the maxima, minima and crossings of the centre line. Reading aa, bb, hh and kk directly off the equation, then plotting these features, is exactly the method VCAA rewards.

In one sentence

Transformed circular functions y=asin(b(xh))+ky = a\sin(b(x - h)) + k have amplitude a|a|, period 2π/b2\pi/|b|, phase shift hh and centre line y=ky = k; solving trig equations over an interval uses the reference angle plus quadrant symmetry, and equations involving bxbx require finding solutions for bxbx in the expanded interval before dividing by bb.

Examples in context

Example 1. Tidal height model. A harbour's water depth (metres) is modelled by D=2sin ⁣(π6t)+5D = 2\sin\!\left(\frac{\pi}{6}t\right) + 5, where tt is hours after midnight. The amplitude is 22 m about a mean depth of 55 m, and the period is 2ππ/6=12\frac{2\pi}{\pi/6} = 12 hours. High tide (D=7D = 7) first occurs when sin ⁣(π6t)=1\sin\!\left(\frac{\pi}{6}t\right) = 1, i.e. π6t=π2\frac{\pi}{6}t = \frac{\pi}{2}, giving t=3t = 3 hours (3 am).

Example 2. Ferris wheel height. A rider's height is h=10cos ⁣(π30t)(1)+12h = 10\cos\!\left(\frac{\pi}{30}t\right)\cdot(-1) + 12, i.e. h=10cos ⁣(π30t)+12h = -10\cos\!\left(\frac{\pi}{30}t\right) + 12 metres, with tt in seconds. The period is 2ππ/30=60\frac{2\pi}{\pi/30} = 60 s per rotation, amplitude 1010 m about a centre 1212 m. At t=0t = 0, h=10+12=2h = -10 + 12 = 2 m (boarding height); the maximum 2222 m occurs at t=30t = 30 s.

Try this

Q1. State the amplitude, period and range of y=3sin(2x)1y = 3\sin(2x) - 1. [3 marks]

  • Cue. Amplitude 33; period 2π2=π\frac{2\pi}{2} = \pi; range [4,2][-4, 2].

Q2. Solve 2sin(x)=32\sin(x) = \sqrt{3} for x[0,2π]x \in [0, 2\pi]. [3 marks]

  • Cue. sinx=32\sin x = \frac{\sqrt 3}{2}; x=π3,2π3x = \frac{\pi}{3}, \frac{2\pi}{3}.

Q3. A signal is modelled by y=4cos ⁣(π2t)y = 4\cos\!\left(\frac{\pi}{2}t\right). (a) State its period. (b) Find the first time t>0t > 0 when y=0y = 0. [2+2 marks]

  • Cue. (a) Period 2ππ/2=4\frac{2\pi}{\pi/2} = 4. (b) cos ⁣(π2t)=0π2t=π2t=1\cos\!\left(\frac{\pi}{2}t\right) = 0 \Rightarrow \frac{\pi}{2}t = \frac{\pi}{2} \Rightarrow t = 1.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

VCAA 2022 Exam 14 marksSolve sin(2x)=1/2\sin(2x) = 1/2 for x[0,2π]x \in [0, 2\pi].
Show worked answer →

Let u=2xu = 2x. Then u[0,4π]u \in [0, 4\pi] and sinu=1/2\sin u = 1/2.

Solutions for sinu=1/2\sin u = 1/2 in [0,4π][0, 4\pi]: u=π/6,5π/6,13π/6,17π/6u = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6.

Back-substitute u=2xu = 2x: x=u/2=π/12,5π/12,13π/12,17π/12x = u/2 = \pi/12, 5\pi/12, 13\pi/12, 17\pi/12.

Markers reward the substitution, the expanded interval for uu, the four solutions in that interval, and back-substitution.

VCAA 2023 Exam 25 marksThe temperature in a greenhouse is modelled by T(t)=186cos ⁣(π12t)T(t) = 18 - 6\cos\!\left(\dfrac{\pi}{12}t\right) degrees Celsius, where tt is hours after midnight, 0t240 \le t \le 24. (a) State the maximum and minimum temperatures and the times they occur. (b) Find the times at which the temperature first reaches 2121 degrees.
Show worked answer →

(a) The amplitude is 66 about the centre line 1818, so the maximum is 18+6=2418 + 6 = 24 degrees and the minimum is 186=1218 - 6 = 12 degrees. The minimum occurs where cos ⁣(π12t)=1\cos\!\left(\frac{\pi}{12}t\right) = 1, i.e. t=0t = 0 (midnight); the maximum where cos ⁣(π12t)=1\cos\!\left(\frac{\pi}{12}t\right) = -1, i.e. π12t=π\frac{\pi}{12}t = \pi, so t=12t = 12 (noon).

(b) Set T=21T = 21: 186cos ⁣(π12t)=2118 - 6\cos\!\left(\frac{\pi}{12}t\right) = 21, so cos ⁣(π12t)=12\cos\!\left(\frac{\pi}{12}t\right) = -\dfrac{1}{2}. Let u=π12tu = \frac{\pi}{12}t, with u[0,2π]u \in [0, 2\pi]. Then u=2π3u = \frac{2\pi}{3} or u=4π3u = \frac{4\pi}{3}, giving t=12π2π3=8t = \frac{12}{\pi}\cdot\frac{2\pi}{3} = 8 and t=12π4π3=16t = \frac{12}{\pi}\cdot\frac{4\pi}{3} = 16. The temperature first reaches 2121 degrees at t=8t = 8 (8 am).

Markers reward reading amplitude and centre line, the times of extremes, and solving the trig equation across the interval.

Related dot points