Trigonometric functions $y = \sin(x)$, $y = \cos(x)$ and $y = \tan(x)$, the unit circle, exact values at standard angles, transformations of trig graphs, and solving trigonometric equations

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) $\cos\!\left(\frac{\pi}{4}t\right) = -1 \Rightarrow \frac{\pi}{4}t = \pi \Rightarrow t = 4$.