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Circular functions extended (VCE Maths Methods Unit 2)

Syllabus dot point

“Sketch and analyse trigonometric functions 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 and vertical translation, and solve trig equations over a specified interval”

7 min read

Quick answer

For y=asin⁡(b(x−h))+ky = a\sin(b(x-h)) + k (and the cosine version): amplitude is ∣a∣|a|, period is 2π/∣b∣2\pi/|b|, hh is the phase shift and kk is the centre line. To solve equations like sin⁡(2x)=c\sin(2x) = c on an interval, expand the interval for bxbx, find all solutions, then divide by bb.

Jump to a section
  1. What this dot point is asking
  2. Parent functions
  3. Transformed form y=asin⁡(b(x−h))+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
  10. Exam-style questions

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

sin⁡x\sin x and cos⁡x\cos x:

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

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

Parameter Effect
∥a∥\|a\| Amplitude. Range [k−∥a∥,k+∥a∥][k - \|a\|, k + \|a\|]. If a<0a < 0, vertical reflection.
bb Period = 2π/∥b∥2\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(x−h))+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 π−θ\pi - \theta (quadrant 2), π+θ\pi + \theta (quadrant 3) and 2π−θ2\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+∣a∣y = k + |a| and minimum y=k−∣a∣y = k - |a|, and one full period of width 2π∣b∣\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.

Solving a trig equation on an interval

Step 1: Isolate the trigonometric function

We are given 2cos⁡(x)+1=02\cos(x) + 1 = 0 for x∈[0,2π]x \in [0, 2\pi]. Before we can use any trig identity or reference angle, we must rearrange so that cos⁡x\cos x stands alone on one side. Subtracting 11 and dividing by 22:

cos⁡x=−12\cos x = -\frac{1}{2}

Step 2: Find the reference angle

The reference angle is the acute angle (between 00 and π2\dfrac{\pi}{2}) whose cosine has the same magnitude as the value on the right-hand side. We ignore the negative sign for now and ask: for which angle is cos⁡θ=12\cos\theta = \dfrac{1}{2}? From the special triangle for 60°60°:

θref=π3\theta_{\text{ref}} = \frac{\pi}{3}

Step 3: Place solutions in the correct quadrants

Cosine is negative in quadrants 2 and 3, because in those quadrants the xx-coordinate on the unit circle is negative. Using the standard quadrant-placement formulas, the two solutions in [0,2π][0, 2\pi] are:

x=π−π3=2π3(quadrant 2)x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \quad \text{(quadrant 2)}

x=π+π3=4π3(quadrant 3)x = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \quad \text{(quadrant 3)}

Step 4: State all solutions in the interval

Both values lie within [0,2π][0, 2\pi], and cosine has period 2π2\pi so there are no further solutions in this interval. We check: cos⁡ ⁣(2π3)=−12\cos\!\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{2} and cos⁡ ⁣(4π3)=−12\cos\!\left(\dfrac{4\pi}{3}\right) = -\dfrac{1}{2}, confirming both are correct.

Final answer: x=2π3x = \dfrac{2\pi}{3} or x=4π3x = \dfrac{4\pi}{3}.

Common traps
Confusing bb with the period
The period is 2π∣b∣\dfrac{2\pi}{|b|}, not bb.
Forgetting to expand the interval
Solving sin⁡(2x)=c\sin(2x) = c for x∈[0,2π]x \in [0, 2\pi] requires all solutions for 2x2x in [0,4π][0, 4\pi] before dividing by 22, otherwise half the solutions are lost.
Missing solutions in the stated interval
Sine and cosine are periodic; list every solution that falls in the interval, not just the principal one.
Wrong quadrant from the sign
Use the sign of the original equation, not the reference value, to choose quadrants.

In one sentence

Transformed circular functions y=asin⁡(b(x−h))+ky = a\sin(b(x - h)) + k have amplitude ∣a∣|a|, period 2π/∣b∣2\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. sin⁡x=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)=0⇒π2t=π2⇒t=1\cos\!\left(\frac{\pi}{2}t\right) = 0 \Rightarrow \frac{\pi}{2}t = \frac{\pi}{2} \Rightarrow t = 1.

Exam-style questions

Questions in the style of VCAA 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.

Practice4 marks
Solve 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 sin⁡u=1/2\sin u = 1/2.

Solutions for sin⁡u=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.

Practice5 marks
The temperature in a greenhouse is modelled by T(t)=18−6cos⁡ ⁣(π12t)T(t) = 18 - 6\cos\!\left(\dfrac{\pi}{12}t\right) degrees Celsius, where tt is hours after midnight, 0≤t≤240 \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 18−6=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: 18−6cos⁡ ⁣(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=12π⋅2π3=8t = \frac{12}{\pi}\cdot\frac{2\pi}{3} = 8 and t=12π⋅4π3=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.

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