Trigonometric functions: VCE Math Methods Unit 2 Year 11

What this dot point is asking

VCAA wants you to define trigonometric functions using the unit circle, know exact values at standard angles, sketch and transform $y = \sin(x)$, $y = \cos(x)$ and $y = \tan(x)$, and solve trig equations. Unit 2 is the first complete introduction; Unit 3 will use these for calculus.

The unit circle

The unit circle has radius 1, centred at the origin. A point on the circle at angle $\theta$ from the positive $x$-axis (measured anticlockwise) has coordinates $(\cos\theta, \sin\theta)$.

Definitions:

• $\cos\theta$ = $x$-coordinate.
• $\sin\theta$ = $y$-coordinate.
• $\tan\theta = \sin\theta / \cos\theta$.

Exact values

Memorise these:

Paper 1 expects exact values; substitute $\pi = 3.14$ at your peril.

Signs in each quadrant

The CAST or ASTC mnemonic:

Symmetry identities

$\sin(-\theta) = -\sin\theta$ (odd function).

$\cos(-\theta) = \cos\theta$ (even function).

$\sin(\pi - \theta) = \sin\theta$.

$\cos(\pi - \theta) = -\cos\theta$.

$\sin(\theta + 2\pi) = \sin\theta$ (periodic).

Pythagorean identity

This holds for all $\theta$. Rearranging: $\sin^2\theta = 1 - \cos^2\theta$ and similar.

Graphs

$y = \sin(x)$

Wave with amplitude 1, period $2\pi$, $y$-intercept 0. Maxima at $x = \pi/2 + 2\pi k$, minima at $x = 3\pi/2 + 2\pi k$, zeros at $x = \pi k$.

$y = \cos(x)$

Same shape as $\sin$ but shifted: $\cos(x) = \sin(x + \pi/2)$. Amplitude 1, period $2\pi$, $y$-intercept 1.

$y = \tan(x)$

Period $\pi$. Vertical asymptotes at $x = \pi/2 + \pi k$. Zero at $x = \pi k$. Increasing in each period.

Transformations

$y = a \sin(b(x - h)) + k$:

• Amplitude. $|a|$. Vertical stretch.
• Period. $2\pi / |b|$. Horizontal stretch / compression.
• Phase shift. $h$. Horizontal translation.
• Vertical shift. $k$.

Example. $y = 3 \sin(2 x - \pi)$ rewritten as $y = 3 \sin(2(x - \pi/2))$: amplitude 3, period $\pi$, phase shift $\pi/2$ right.

Solving trig equations

To solve $\sin(x) = k$ in a given range:

1. Find principal solutions. Use the inverse sine (calculator or exact-value table) to find one solution.
2. Use symmetry / periodicity to find all others in the range.

For $\sin(x) = k$ with $|k| \leq 1$:

• Principal solution $x_1 = \arcsin(k)$ in $[-\pi/2, \pi/2]$.
• Second solution $x_2 = \pi - x_1$ (sin symmetry).
• All others by adding multiples of $2\pi$.

For $\cos(x) = k$:

• Principal $x_1 = \arccos(k)$ in $[0, \pi]$.
• Second $x_2 = -x_1$ (or $2\pi - x_1$).

For $\tan(x) = k$:

• Principal $x_1 = \arctan(k)$ in $(-\pi/2, \pi/2)$.
• All others by adding multiples of $\pi$.

Equations with composite argument (like $\sin(2x) = 1/2$): solve for the composite first, then divide by the coefficient and adjust the range accordingly.

Examples in context

Example 1. Daylight hours over a year. The daily hours of daylight at a location are modelled by $H = 12 + 3\sin\!\left(\frac{2\pi}{365}d\right)$, where $d$ is the day number from the equinox. The amplitude $3$ means daylight ranges from $9$ to $15$ hours, with period $365$ days. The longest day ($H = 15$) occurs when $\sin = 1$, i.e. $\frac{2\pi}{365}d = \frac{\pi}{2}$, giving $d \approx 91$ days after the equinox.

Example 2. Exact value from the unit circle. To evaluate $\cos\!\left(\frac{5\pi}{6}\right)$, note $\frac{5\pi}{6}$ is in the second quadrant with reference angle $\frac{\pi}{6}$. Cosine is negative there, so $\cos\!\left(\frac{5\pi}{6}\right) = -\cos\!\left(\frac{\pi}{6}\right) = -\frac{\sqrt 3}{2}$. No calculator needed; the CAST rule fixes the sign.

Try this

Q1. State the exact values of $\sin\!\left(\frac{\pi}{3}\right)$ and $\tan\!\left(\frac{\pi}{4}\right)$. [2 marks]

• Cue. $\sin\!\left(\frac{\pi}{3}\right) = \frac{\sqrt 3}{2}$; $\tan\!\left(\frac{\pi}{4}\right) = 1$.

Q2. Solve $\cos(x) = -\frac{1}{2}$ for $x \in [0, 2\pi]$. [3 marks]

• Cue. Reference angle $\frac{\pi}{3}$; cosine negative in Q2, Q3: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$.

Q3. A wheel's height is $h = 5 - 5\cos\!\left(\frac{\pi}{4}t\right)$ m. (a) State the period and amplitude. (b) Find the first time $t > 0$ that $h = 10$. [2+2 marks]

• Cue. (a) Period $\frac{2\pi}{\pi/4} = 8$, amplitude $5$. (b) $\cos\!\left(\frac{\pi}{4}t\right) = -1 \Rightarrow \frac{\pi}{4}t = \pi \Rightarrow t = 4$.