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VCE Mathematical Methods (Year 12) · Functions, Relations and Graphs · 19m read

Circular (Trigonometric) Functions

Ever wondered why the same waves show up in ocean tides, music, and even your heartbeat? Circular functions (sine, cosine, and tangent) are the mathematical tools that describe these repeating patterns. You'll learn to sketch their graphs, understand how stretching, shifting, and flipping changes their shape, and solve equations to find exact moments when these functions hit specific values — skills you'll definitely see on the VCE exam and use whenever you need to predict or analyse anything cyclical in the real world.

Learning objective: Sketch sine, cosine and tangent graphs with transformations and solve trigonometric equations over a given interval.

By the end of this lesson you can:

  • I identify amplitude, period, phase shift and vertical shift from y = a·sin(n(x − h)) + k.
  • I solve trig equations by finding all solutions in the given interval.
  • I convert between radians and degrees.

Circular functions ride a Ferris wheel. Height above the centre follows sine; horizontal position follows cosine as the wheel turns.

Amplitude, period and phase

For y = a·sin(n(x − h)) + k: amplitude is |a|, period is 2π/|n|, phase shift is h (right if positive), and vertical shift is k. The graph oscillates between k − |a| and k + |a|.

When solving trig equations, you must find every solution inside the given interval. Always sketch a quick picture. For sin x = c, both x = arcsin(c) and x = π − arcsin(c) are base solutions in [0, 2π]. For cos x = c, x = arccos(c) and x = 2π − arccos(c). Then add 2π repeatedly to extend the interval.

Trig toolkit

  • Radians ↔ degrees: 180° = π radians.
  • Period of sin/cos: 2π/|n|.
  • Period of tan: π/|n|.
  • Amplitude: |a| (vertical stretch).
  • Pythagoras identity: sin²x + cos²x = 1.

Quick reference

| y = a sin(n(x − h)) + k | Value |
|---|---|
| Amplitude | Size of a |
| Period | 2π/n |
| Mean (centre) line | y = k |
| Range | [k − a, k + a] for a > 0 |
| tan period | π/n |

Worked example (Medium)

Solve 2 sin(x) = 1 for x ∈ [0, 2π].

  1. Isolate sin x. Divide by 2. sin(x) = 1/2
  2. Base solution. sin(π/6) = 1/2. x = π/6
  3. Second solution. Sin is positive in Q1 and Q2. x = π − π/6 = 5π/6
  4. Check interval. Both lie in [0, 2π]. No further additions needed.

Answer: x = π/6 or x = 5π/6.

Worked example (Hard)

Sketch $y = 2 \sin\!\left(\dfrac{\pi x}{3}\right) + 1$ over $0 \le x \le 6$. State amplitude, period, range and the y-intercept.

  1. Amplitude. Coefficient of sine. A = 2
  2. Period. $T = 2\pi/b$ where $b = \pi/3$. T = 2\pi \div (\pi/3) = 6
  3. Vertical shift / range. Centre line $y = 1$, max $= 1 + 2 = 3$, min $= 1 - 2 = -1$. \text{range} = [-1,\, 3]
  4. y-intercept. Set $x = 0$. y = 2 \sin 0 + 1 = 1

Answer: Amplitude 2, period 6, range $[-1, 3]$, y-intercept $(0, 1)$. One full cycle fits within $0 \le x \le 6$.

Common mistakes

  • Losing solutions by not considering all quadrants For $\sin(x) = \frac{1}{2}$ on $[0, 2\pi]$, students often find only $x = \frac{\pi}{6}$ and forget the second-quadrant solution $x = \frac{5\pi}{6}$. Each trig ratio is positive in two quadrants and negative in two. Fix: Use the ASTC (All, Sin, Tan, Cos) rule systematically. After finding the base angle, generate solutions in every relevant quadrant before filtering by the given domain.
  • Forgetting to adjust the domain when the argument is $nx + c$ instead of $x$ For $\sin(2x) = \frac{1}{2}$ on $[0, 2\pi]$, you need solutions for $2x \in [0, 4\pi]$, doubling the search interval. Missing this yields only half the solutions. Fix: Let $u = nx + c$, transform the interval for $u$, solve in the wider interval, then convert back with $x = \frac{u - c}{n}$.
  • Mixing up the period formulas for $\sin$/$\cos$ versus $\tan$ The period of $\sin(nx)$ and $\cos(nx)$ is $\frac{2\pi}{|n|}$, but the period of $\tan(nx)$ is $\frac{\pi}{|n|}$. Using the wrong formula shifts every solution. Fix: Memorise: $\sin$ and $\cos$ repeat every $\frac{2\pi}{|n|}$; $\tan$ repeats every $\frac{\pi}{|n|}$.

The unit circle, radians, and the connection to arc length

Radians are not an arbitrary unit — they arise from the unit circle. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. For a circle of radius $r$, an arc of length $s$ subtends an angle $\theta = \frac{s}{r}$ radians, so arc length $s = r\theta$. On the unit circle ($r = 1$), the arc length equals the angle itself, which is why radians are the "natural" angle measure for calculus.

This connection explains why $\frac{d}{dx}[\sin x] = \cos x$ only works in radians. If $x$ is in degrees, the derivative picks up a conversion factor: $\frac{d}{dx}[\sin(x°)] = \frac{\pi}{180}\cos(x°)$. The clean derivative in radians is what makes $\sin x$ and $\cos x$ fundamental in mathematical analysis.

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ is a direct consequence of the unit-circle definition: $\cos\theta$ and $\sin\theta$ are the $x$- and $y$-coordinates of a point on $x^2 + y^2 = 1$. Dividing through by $\cos^2\theta$ or $\sin^2\theta$ produces the two companion identities $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$ studied in Specialist.

Practice questions

  1. State the period of y = 3 cos(2x) − 1.

    1. π
    2. 2π
    3. 3
    4. π/2

    (2 marks)

  2. Convert 135° to radians in exact form.

    1. 3π/4
    2. π/4
    3. 2π/3
    4. 5π/6

    (2 marks)

  3. Solve cos(x) = −√3/2 for x ∈ [0, 2π].

    (3 marks)

  4. Solve $\cos x = -\dfrac{1}{2}$ for $0 \le x \le 2\pi$.

    (3 marks)

Summary

  • Trig exact values: $\sin 30° = \cos 60° = \frac{1}{2}$; $\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}$; $\sin 60° = \cos 30° = \frac{\sqrt{3}}{2}$.
  • Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$.
  • Transformations of $y = a\sin(n(x-h))+k$: amplitude $|a|$, period $2\pi/n$, phase shift $h$, vertical shift $k$.

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