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Circular (trig) functions and graphs: VCE Math Methods Unit 3

Syllabus dot point

“Graphs of circular functions f(x)=sin⁡(x)f(x) = \sin(x), f(x)=cos⁡(x)f(x) = \cos(x) and f(x)=tan⁡(x)f(x) = \tan(x), their key features (period, amplitude, asymptotes), exact values at standard angles, and graphs of the form f(x)=asin⁡(b(x−h))+kf(x) = a\sin(b(x - h)) + k”

9 min read

Quick answer

The circular functions sine, cosine and tangent are the trigonometric backbone of VCE Math Methods, with sine and cosine bounded between −1-1 and 11 with period 2π2\pi, tangent unbounded with period π\pi and vertical asymptotes, and all three transforming under the standard form asin⁡(b(x−h))+ka\sin(b(x-h)) + k to amplitude ∣a∣|a|, period 2π∣b∣\frac{2\pi}{|b|}, and midline y=ky = k.

Jump to a section
  1. What this dot point is asking
  2. Sine and cosine
  3. Tangent
  4. Exact values at standard angles
  5. Transformed circular functions
  6. The Pythagorean identity
  7. Examples in context
  8. Try this
  9. Exam-style questions

What this dot point is asking

VCAA wants you to graph and analyse the three primary circular functions sin⁡\sin, cos⁡\cos and tan⁡\tan in radians, recognise their key features under transformation, and quote exact values at standard unit-circle angles without a calculator. This is core Paper 1 content.

Sine and cosine

f(x)=sin⁡(x)f(x) = \sin(x) and f(x)=cos⁡(x)f(x) = \cos(x) are the two foundational trig functions.

Sine.

  • Domain: R\mathbb{R}. Range: [−1,1][-1, 1].
  • Period: 2π2\pi. Amplitude: 11.
  • x-intercepts at x=nπx = n\pi for n∈Zn \in \mathbb{Z}.
  • Starts at sin⁡(0)=0\sin(0) = 0, rises to maximum 11 at x=π/2x = \pi/2, returns to 00 at x=πx = \pi, minimum −1-1 at x=3π/2x = 3\pi/2, back to 00 at x=2πx = 2\pi.

Cosine.

  • Domain: R\mathbb{R}. Range: [−1,1][-1, 1].
  • Period: 2π2\pi. Amplitude: 11.
  • x-intercepts at x=π/2+nπx = \pi/2 + n\pi.
  • Starts at cos⁡(0)=1\cos(0) = 1, drops to 00 at x=π/2x = \pi/2, minimum −1-1 at x=πx = \pi, back to 00 at x=3π/2x = 3\pi/2, max 11 at x=2πx = 2\pi.

Cosine is sine shifted left by π/2\pi/2: cos⁡(x)=sin⁡(x+π/2)\cos(x) = \sin(x + \pi/2).

Tangent

f(x)=tan⁡(x)=sin⁡(x)cos⁡(x)f(x) = \tan(x) = \frac{\sin(x)}{\cos(x)}.

  • Domain: R∖{π/2+nπ:n∈Z}\mathbb{R} \setminus \{\pi/2 + n\pi : n \in \mathbb{Z}\} (excluded where cos⁡(x)=0\cos(x) = 0).
  • Range: R\mathbb{R}.
  • Period: π\pi (not 2π2\pi).
  • Vertical asymptotes at x=π/2+nπx = \pi/2 + n\pi.
  • x-intercepts at x=nπx = n\pi.

Within one period, the tangent graph rises steeply from negative infinity through the origin to positive infinity.

Exact values at standard angles

These are Paper 1 essential. Memorise the unit-circle table:

angle sin⁡\sin cos⁡\cos tan⁡\tan
00 00 11 00
π/6\pi/6 1/21/2 3/2\sqrt{3}/2 1/31/\sqrt{3}
π/4\pi/4 2/2\sqrt{2}/2 2/2\sqrt{2}/2 11
π/3\pi/3 3/2\sqrt{3}/2 1/21/2 3\sqrt{3}
π/2\pi/2 11 00 undefined

For angles beyond [0,π/2][0, \pi/2], use the ASTC quadrant rule:

  • Quadrant 1 (00 to π/2\pi/2): all positive.
  • Quadrant 2 (π/2\pi/2 to π\pi): sine positive only.
  • Quadrant 3 (π\pi to 3π/23\pi/2): tangent positive only.
  • Quadrant 4 (3π/23\pi/2 to 2π2\pi): cosine positive only.

The reference angle (acute angle to the x-axis) gives the magnitude; the quadrant gives the sign.

Transformed circular functions

The standard form is f(x)=asin⁡(b(x−h))+kf(x) = a \sin(b(x - h)) + k (and similarly for cos⁡\cos).

  • Amplitude =∣a∣= |a|. Vertical dilation by factor ∣a∣|a|; reflection in the x-axis if a<0a < 0.
  • Period =2π∣b∣= \frac{2\pi}{|b|}. Horizontal dilation by factor 1∣b∣\frac{1}{|b|}.
  • Horizontal translation =h= h (right if h>0h > 0).
  • Midline (vertical translation) y=ky = k.
  • Range: [k−∣a∣,k+∣a∣][k - |a|, k + |a|].

For tan, the period under tan⁡(b(x−h))\tan(b(x - h)) is π∣b∣\frac{\pi}{|b|} instead of 2π/∣b∣2\pi/|b|.

The Pythagorean identity

sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1

Useful corollaries:

  • tan⁡2(x)+1=sec⁡2(x)\tan^2(x) + 1 = \sec^2(x) where sec⁡(x)=1cos⁡(x)\sec(x) = \frac{1}{\cos(x)} (used in ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x).
  • cos⁡2(x)=1−sin⁡2(x)\cos^2(x) = 1 - \sin^2(x) and sin⁡2(x)=1−cos⁡2(x)\sin^2(x) = 1 - \cos^2(x) for simplifying.
Worked example: analysing a transformed sine function

Step 1: Rewrite in standard form

f(x)=−3sin⁡(2x−π/3)+1=−3sin⁡ ⁣(2 ⁣(x−π6))+1f(x) = -3 \sin(2x - \pi/3) + 1 = -3 \sin\!\left(2\!\left(x - \frac{\pi}{6}\right)\right) + 1. Factoring the coefficient of xx out of the argument reveals the horizontal shift h=π/6h = \pi/6 and makes it straightforward to read off the period and phase; working from the un-factored form leads to errors in period and range calculations.

Step 2: State the amplitude, period and range

Amplitude: ∣−3∣=3|-3| = 3. Period: 2π2=π\frac{2\pi}{2} = \pi. Midline: y=1y = 1. Range: [1−3,1+3]=[−2,4][1 - 3, 1 + 3] = [-2, 4]. The amplitude is always the absolute value of the coefficient, not the signed value; the range is then found by adding and subtracting the amplitude from the midline.

Step 3: Find the y-intercept

f(0)=−3sin⁡(−π/3)+1=−3⋅(−3/2)+1=332+1f(0) = -3 \sin(-\pi/3) + 1 = -3 \cdot (-\sqrt{3}/2) + 1 = \frac{3\sqrt{3}}{2} + 1. Substituting x=0x = 0 and using sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta) with the exact value sin⁡(π/3)=3/2\sin(\pi/3) = \sqrt{3}/2 gives the exact y-intercept without a calculator.

Common Paper 1 traps
Period mismatch for tan⁡\tan
tan⁡(bx)\tan(b x) has period πb\frac{\pi}{b}, not 2πb\frac{2\pi}{b}. Easy 1-mark loss.
Wrong sign from the quadrant
sin⁡(7π/6)\sin(7\pi/6) is negative (quadrant 3), not positive. Always sketch the unit circle or use ASTC.
Degrees instead of radians
All VCE Math Methods trig is in radians. Writing sin⁡(30)\sin(30) instead of sin⁡(π/6)\sin(\pi/6) will be marked wrong.
Forgetting the horizontal dilation reciprocal
b=2b = 2 compresses horizontally by factor 2, halving the period.
Including too many or too few solutions
When solving sin⁡(2x)=1/2\sin(2x) = 1/2 for x∈[0,2π]x \in [0, 2\pi], you have 2x∈[0,4π]2x \in [0, 4\pi], so four solutions for 2x2x, hence four solutions for xx.

Examples in context

Example 1. Modelling temperature over a day. A city's temperature is modelled by T=18−6cos⁡ ⁣(π12t)T = 18 - 6\cos\!\left(\frac{\pi}{12}t\right) degrees Celsius, where tt is hours after midnight. The amplitude is 66 about a mean of 1818, with period 2ππ/12=24\frac{2\pi}{\pi/12} = 24 hours. The minimum (T=12T = 12) occurs at t=0t = 0 (midnight, where cos⁡=1\cos = 1) and the maximum (T=24T = 24) at t=12t = 12 (noon).

Example 2. Exact value with the ASTC rule. To evaluate tan⁡ ⁣(4π3)\tan\!\left(\frac{4\pi}{3}\right): the angle is in the third quadrant (π+π3\pi + \frac{\pi}{3}) with reference angle π3\frac{\pi}{3}. Tangent is positive in quadrant 3, and tan⁡ ⁣(π3)=3\tan\!\left(\frac{\pi}{3}\right) = \sqrt 3, so tan⁡ ⁣(4π3)=3\tan\!\left(\frac{4\pi}{3}\right) = \sqrt 3.

Try this

Q1. State the amplitude, period and range of y=4sin⁡(3x)+2y = 4\sin(3x) + 2. [3 marks]

  • Cue. Amplitude 44; period 2π3\frac{2\pi}{3}; range [−2,6][-2, 6].

Q2. Find the exact value of cos⁡ ⁣(5π4)\cos\!\left(\frac{5\pi}{4}\right). [2 marks]

  • Cue. Quadrant 3, reference π4\frac{\pi}{4}, cosine negative: −22-\frac{\sqrt 2}{2}.

Q3. For y=2sin⁡ ⁣(2 ⁣(x−π4))y = 2\sin\!\left(2\!\left(x - \frac{\pi}{4}\right)\right), state (a) the period and (b) the phase shift. [2+1 marks]

  • Cue. (a) 2π2=π\frac{2\pi}{2} = \pi. (b) π4\frac{\pi}{4} to the right.

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.

2023 VCAA-style (Exam 1)3 marks
Sketch the graph of y=2cos⁡ ⁣(x2)−1y = 2 \cos\!\left(\frac{x}{2}\right) - 1 for x∈[0,4π]x \in [0, 4\pi], showing the amplitude, period, and any axis intercepts.
Show worked answer →

Compare to the standard form y=acos⁡(b(x−h))+ky = a \cos(b(x - h)) + k with a=2a = 2, b=1/2b = 1/2, h=0h = 0, k=−1k = -1.

Amplitude: ∣a∣=2|a| = 2. Period: 2π∣b∣=2π1/2=4π\frac{2\pi}{|b|} = \frac{2\pi}{1/2} = 4\pi.

Midline (centre line): y=−1y = -1. Maximum: y=1y = 1 (at x=0x = 0); minimum: y=−3y = -3 (at x=2πx = 2\pi).

x-intercepts: 2cos⁡(x/2)−1=02\cos(x/2) - 1 = 0 gives cos⁡(x/2)=1/2\cos(x/2) = 1/2, so x/2=π/3x/2 = \pi/3 or x/2=5π/3x/2 = 5\pi/3 (within one period). So x=2π/3x = 2\pi/3 and x=10π/3x = 10\pi/3.

Sketch one full cycle on [0,4π][0, 4\pi] starting at (0,1)(0, 1), descending through (2π/3,0)(2\pi/3, 0), reaching minimum (2π,−3)(2\pi, -3), rising through (10π/3,0)(10\pi/3, 0), back to (4π,1)(4\pi, 1).

Markers reward correct amplitude, period, intercepts and the labelled max/min.

2024 VCAA-style (Exam 1)2 marks
Find the exact value of sin⁡ ⁣(7π6)\sin\!\left(\frac{7\pi}{6}\right).
Show worked answer →

7π6=π+π6\frac{7\pi}{6} = \pi + \frac{\pi}{6}, which lies in the third quadrant. In quadrant 3, sine is negative.

The reference angle is π6\frac{\pi}{6}, and sin⁡(π/6)=1/2\sin(\pi/6) = 1/2.

So sin⁡(7π/6)=−12\sin(7\pi/6) = -\frac{1}{2}.

Markers expect exact-form answer with correct sign from the unit circle quadrant.

Practise this

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