Skip to main content
VICMath MethodsSyllabus dot point

What are the graphs of the sine, cosine and tangent functions and what features do they have under transformation?

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(xh))+kf(x) = a\sin(b(x - h)) + k

A focused answer to the VCE Math Methods Unit 3 key-knowledge point on circular functions. Sine, cosine and tangent graphs, period and amplitude, exact unit-circle values, transformed trig graphs, and standard Paper 1 patterns.

Generated by Claude Opus 4.89 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. 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

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 nZn \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:nZ}\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(xh))+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 =2πb= \frac{2\pi}{|b|}. Horizontal dilation by factor 1b\frac{1}{|b|}.
  • Horizontal translation =h= h (right if h>0h > 0).
  • Midline (vertical translation) y=ky = k.
  • Range: [ka,k+a][k - |a|, k + |a|].

For tan, the period under tan(b(xh))\tan(b(x - h)) is πb\frac{\pi}{|b|} instead of 2π/b2\pi/|b|.

The Pythagorean identity

sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1

Useful corollaries:

  • tan2(x)+1=sec2(x)\tan^2(x) + 1 = \sec^2(x) where sec(x)=1cos(x)\sec(x) = \frac{1}{\cos(x)} (used in ddx(tanx)=sec2x\frac{d}{dx}(\tan x) = \sec^2 x).
  • cos2(x)=1sin2(x)\cos^2(x) = 1 - \sin^2(x) and sin2(x)=1cos2(x)\sin^2(x) = 1 - \cos^2(x) for simplifying.

Examples in context

Example 1. Modelling temperature over a day. A city's temperature is modelled by T=186cos ⁣(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 ⁣(xπ4))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 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.

2023 VCAA Paper 13 marksSketch 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(xh))+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: 2πb=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 Paper 12 marksFind 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.

Related dot points