← Unit 3

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 $f(x) = \sin(x)$, $f(x) = \cos(x)$ and $f(x) = \tan(x)$, their key features (period, amplitude, asymptotes), exact values at standard angles, and graphs of the form $f(x) = a\sin(b(x - h)) + k$

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 OpusReviewed by Better Tuition Academy9 min answer

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

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: [βˆ’1,1][-1, 1].
  • Period: 2Ο€2\pi. Amplitude: 11.
  • x-intercepts at x=nΟ€x = n\pi for n∈Zn \in \mathbb{Z}.
  • Starts at sin⁑(0)=0\sin(0) = 0, rises to maximum 11 at x=Ο€/2x = \pi/2, returns to 00 at x=Ο€x = \pi, minimum βˆ’1-1 at x=3Ο€/2x = 3\pi/2, back to 00 at x=2Ο€x = 2\pi.

Cosine.

  • Domain: R\mathbb{R}. Range: [βˆ’1,1][-1, 1].
  • Period: 2Ο€2\pi. Amplitude: 11.
  • x-intercepts at x=Ο€/2+nΟ€x = \pi/2 + n\pi.
  • Starts at cos⁑(0)=1\cos(0) = 1, drops to 00 at x=Ο€/2x = \pi/2, minimum βˆ’1-1 at x=Ο€x = \pi, back to 00 at x=3Ο€/2x = 3\pi/2, max 11 at x=2Ο€x = 2\pi.

Cosine is sine shifted left by Ο€/2\pi/2: cos⁑(x)=sin⁑(x+Ο€/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: Rβˆ–{Ο€/2+nΟ€: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: Ο€\pi (not 2Ο€2\pi).
  • Vertical asymptotes at x=Ο€/2+nΟ€x = \pi/2 + n\pi.
  • x-intercepts at x=nΟ€x = 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 IMATH_45 IMATH_46 IMATH_47
IMATH_48 IMATH_49 IMATH_50 IMATH_51
IMATH_52 IMATH_53 IMATH_54 IMATH_55
IMATH_56 IMATH_57 IMATH_58 IMATH_59
IMATH_60 IMATH_61 IMATH_62 IMATH_63
IMATH_64 IMATH_65 IMATH_66 undefined

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

  • Quadrant 1 (00 to Ο€/2\pi/2): all positive.
  • Quadrant 2 (Ο€/2\pi/2 to Ο€\pi): sine positive only.
  • Quadrant 3 (Ο€\pi to 3Ο€/23\pi/2): tangent positive only.
  • Quadrant 4 (3Ο€/23\pi/2 to 2Ο€2\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(xβˆ’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 =2Ο€βˆ£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: [kβˆ’βˆ£a∣,k+∣a∣][k - |a|, k + |a|].

For tan, the period under tan⁑(b(xβˆ’h))\tan(b(x - h)) is Ο€βˆ£b∣\frac{\pi}{|b|} instead of 2Ο€/∣b∣2\pi/|b|.

The Pythagorean identity

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

Useful corollaries:

  • IMATH_90 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).
  • IMATH_93 and sin⁑2(x)=1βˆ’cos⁑2(x)\sin^2(x) = 1 - \cos^2(x) for simplifying.

Worked example

State the amplitude, period and range of f(x)=βˆ’3sin⁑(2xβˆ’Ο€/3)+1f(x) = -3 \sin(2x - \pi/3) + 1, and find the y-intercept.

Rewrite in standard form: f(x)=βˆ’3sin⁑ ⁣(2 ⁣(xβˆ’Ο€6))+1f(x) = -3 \sin\!\left(2\!\left(x - \frac{\pi}{6}\right)\right) + 1.

Amplitude: βˆ£βˆ’3∣=3|-3| = 3. Period: 2Ο€2=Ο€\frac{2\pi}{2} = \pi. Midline: y=1y = 1. Range: [1βˆ’3,1+3]=[βˆ’2,4][1 - 3, 1 + 3] = [-2, 4].

y-intercept: f(0)=βˆ’3sin⁑(βˆ’Ο€/3)+1=βˆ’3β‹…(βˆ’3/2)+1=332+1f(0) = -3 \sin(-\pi/3) + 1 = -3 \cdot (-\sqrt{3}/2) + 1 = \frac{3\sqrt{3}}{2} + 1.

Common Paper 1 traps

Period mismatch for tan⁑\tan. tan⁑(bx)\tan(b x) has period Ο€b\frac{\pi}{b}, not 2Ο€b\frac{2\pi}{b}. Easy 1-mark loss.

Wrong sign from the quadrant. sin⁑(7Ο€/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⁑(Ο€/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∈[0,2Ο€]x \in [0, 2\pi], you have 2x∈[0,4Ο€]2x \in [0, 4\pi], so four solutions for 2x2x, hence four solutions for xx.

In one sentence

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

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

2023 VCAA Paper 13 marksSketch the graph of $y = 2 \cos\!\left(\frac{x}{2}\right) - 1$ for $x \in [0, 4\pi]$, showing the amplitude, period, and any axis intercepts.
Show worked answer β†’

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

Amplitude: ∣a∣=2|a| = 2. Period: 2Ο€βˆ£b∣=2Ο€1/2=4Ο€\frac{2\pi}{|b|} = \frac{2\pi}{1/2} = 4\pi.

Midline (centre line): y=βˆ’1y = -1. Maximum: y=1y = 1 (at x=0x = 0); minimum: y=βˆ’3y = -3 (at x=2Ο€x = 2\pi).

x-intercepts: 2cos⁑(x/2)βˆ’1=02\cos(x/2) - 1 = 0 gives cos⁑(x/2)=1/2\cos(x/2) = 1/2, so x/2=Ο€/3x/2 = \pi/3 or x/2=5Ο€/3x/2 = 5\pi/3 (within one period). So x=2Ο€/3x = 2\pi/3 and x=10Ο€/3x = 10\pi/3.

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

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

2024 VCAA Paper 12 marksFind the exact value of $\sin\!\left(\frac{7\pi}{6}\right)$.
Show worked answer β†’

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

The reference angle is Ο€6\frac{\pi}{6}, and sin⁑(Ο€/6)=1/2\sin(\pi/6) = 1/2.

So sin⁑(7Ο€/6)=βˆ’12\sin(7\pi/6) = -\frac{1}{2}.

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

Related dot points