Circular Functions

An observation wheel has centre $P$ and completes one full rotation every 30 minutes, moving anticlockwise at constant speed. At the lowest point (point $A$) a pod is 15 metres above the ground, and the wheel has radius 60 metres. The height of a pod originally at $A$ after $t$ minutes is $h(t) = -60\cos(bt) + c$. Show that $b = \dfrac{\pi}{15}$ and $c = 75$.

1. 1

   One full rotation takes 30 minutes, so the period of $h$ is 30. Use period $= \dfrac{2\pi}{b}$.

   $$\frac{2\pi}{b} = 30 \implies b = \frac{2\pi}{30} = \frac{\pi}{15}$$
2. 2

   The centre height is the lowest point plus the radius. For a cosine starting at its minimum, this midline equals $c$.

   $$c = 15 + 60 = 75$$

VCAA 2023 Mathematical Methods Exam 2, Section B Q2a

Consider $f:\left[0, \dfrac{5\pi}{2}\right] \to R$, $f(x) = \sin(x) + 1$. There exist real numbers $a$ and $k$ in $\left(0, \dfrac{5\pi}{2}\right)$ such that $f(x + k) = f(x)$ for all $x \in [0, a]$. Find $k$ and the largest possible value of $a$.

1. 1

   For $f(x + k) = f(x)$ to hold, $k$ must be a full period of the function. The period of $\sin$ is $2\pi$.

   $$k = 2\pi$$
2. 2

   The shifted input $x + k$ must stay within the domain, so we need $x + 2\pi \le \dfrac{5\pi}{2}$ for every $x \in [0, a]$. The tightest case is $x = a$.

   $$a + 2\pi \le \frac{5\pi}{2} \implies a \le \frac{\pi}{2}$$
3. 3

   The largest $a$ takes the boundary value.

   $$a = \frac{\pi}{2}$$

VCAA 2025 Mathematical Methods Exam 2, Section B Q4c

For $y = 3\sin(2x)$, state the amplitude and the period, then give the range.

1. 1

   The amplitude is the number multiplying the $\sin$, taken as a size. Here that number is $3$.

   $$\text{amplitude} = |3| = 3$$
2. 2

   The period is $\dfrac{2\pi}{n}$, where $n$ is the number multiplying $x$ inside the bracket. Here $n = 2$.

   $$\text{period} = \frac{2\pi}{2} = \pi$$
3. 3

   A plain $\sin$ swings between $-1$ and $1$, so multiplying by $3$ swings between $-3$ and $3$.

   $$\text{range} = [-3,\ 3]$$

Describe the graph of $y = -2\cos\!\left(x - \dfrac{\pi}{3}\right) + 1$: amplitude, period, phase shift, range, and where its first maximum sits.

1. 1

   Amplitude is the size of the front number, ignoring the minus sign.

   $$\text{amplitude} = |-2| = 2$$
2. 2

   The number inside multiplying $x$ is $1$, so the period is unchanged.

   $$\text{period} = \frac{2\pi}{1} = 2\pi$$
3. 3

   Inside reads $x - \frac{\pi}{3}$, which shifts the whole curve to the right.

   $$\text{phase shift} = \frac{\pi}{3} \text{ to the right}$$
4. 4

   The midline is $y = 1$ and the amplitude is $2$, so add and subtract $2$ from $1$.

   $$\text{range} = [-1,\ 3]$$
5. 5

   The minus sign flips a cosine upside down, so it starts at its lowest point. The maximum (value $3$) comes half a period later, at $x = \frac{\pi}{3} + \pi$.

   $$\text{first maximum at } \left(\frac{4\pi}{3},\ 3\right)$$

Solve $2\sin(2x) = \sqrt{3}$ for $x \in [0,\ 2\pi]$.

1. 1

   Get the trig term on its own first.

   $$\sin(2x) = \frac{\sqrt{3}}{2}$$
2. 2

   Let $X = 2x$. As $x$ runs over $[0,\ 2\pi]$, the new angle $X$ runs over $[0,\ 4\pi]$. This is the step that stops you losing solutions.

   $$\sin(X) = \frac{\sqrt{3}}{2}, \quad X \in [0,\ 4\pi]$$
3. 3

   $\sin$ is positive in the first and second quadrants. The base angle is $\frac{\pi}{3}$, so over two full turns $X = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{7\pi}{3}$ or $\frac{8\pi}{3}$.

   $$X = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{7\pi}{3},\ \frac{8\pi}{3}$$
4. 4

   Divide every answer by $2$ to undo $X = 2x$.

   $$x = \frac{\pi}{6},\ \frac{\pi}{3},\ \frac{7\pi}{6},\ \frac{4\pi}{3}$$