Trigonometry — Worked Solutions (Preliminary Maths Advanced)
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Worked examples for Preliminary Maths Advanced trigonometry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Maths Advanced — Trigonometry. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Exact values
Question
Find the exact value of $\sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ$.
Solution
Use the exact-value triangle: $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\cos 60^\circ = \frac{1}{2}$, $\sin 30^\circ = \frac{1}{2}$.
First product: $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4}$. Second product: $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.
Subtract: $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$.
Answer $\frac{1}{2}$. Know the exact-value triangles cold — they save you every time these appear.
Everything here is a standard angle, so we lean on the exact-value triangles rather than a calculator. From the $30$–$60$–$90$ triangle: $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$, while $\cos 60^\circ = \frac{1}{2}$ and $\sin 30^\circ = \frac{1}{2}$.
Now work each product carefully. $\sin 60^\circ \cos 30^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4}$, because $\sqrt{3} \times \sqrt{3} = 3$. And $\cos 60^\circ \sin 30^\circ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Taking the difference, $\frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.
So the value is $\frac{1}{2}$. (If you've met the angle formulas, this expression is exactly $\sin(60^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$ — a nice consistency check.)
Exact values from the standard triangle.
- $\sin 60^\circ = \frac{\sqrt{3}}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2}$
- $\cos 60^\circ = \frac{1}{2},\ \sin 30^\circ = \frac{1}{2}$
- $\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{3}{4}$
- $\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$
- $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$
Value: $\frac{1}{2}$.
Where the marks go
- 1 mark: Correct exact values for all four ratios
- 1 mark: Correct products $\frac{3}{4}$ and $\frac{1}{4}$
- 1 mark: Correct final value $\frac{1}{2}$
Key idea
Memorise the $30$–$60$–$90$ and $45$–$45$–$90$ triangles so exact values come straight out; substitute and simplify carefully.
Example 2 — Solving a trig equation
Question
Solve $2\sin\theta - 1 = 0$ for $0^\circ \le \theta \le 360^\circ$.
Solution
Rearrange to isolate the ratio: $\sin\theta = \frac{1}{2}$.
The base angle is $30^\circ$ since $\sin 30^\circ = \frac{1}{2}$. Sine is positive in the first and second quadrants.
First quadrant: $\theta = 30^\circ$. Second quadrant: $\theta = 180^\circ - 30^\circ = 150^\circ$.
So $\theta = 30^\circ$ or $150^\circ$. Always check the quadrants — there are usually two solutions in a full revolution, not one.
First get $\sin\theta$ on its own: add $1$ and divide by $2$ to get $\sin\theta = \frac{1}{2}$.
Next find the base (reference) angle — the acute angle whose sine is $\frac{1}{2}$. That's $30^\circ$.
Now think about where sine is positive. Using the ASTC rule, sine is positive in the first and second quadrants. The first-quadrant solution is just the base angle, $30^\circ$. The second-quadrant solution is $180^\circ - 30^\circ = 150^\circ$.
Both lie in our interval $0^\circ$ to $360^\circ$, so $\theta = 30^\circ$ or $\theta = 150^\circ$. The key is to never stop at the first answer — the sign of the ratio tells you which quadrants to search.
Isolate the ratio.
- $2\sin\theta = 1 \Rightarrow \sin\theta = \frac{1}{2}$
- Base angle: $30^\circ$
- $\sin$ positive in quadrants 1 and 2
Solutions in $[0^\circ, 360^\circ]$:
- Q1: $\theta = 30^\circ$
- Q2: $\theta = 180^\circ - 30^\circ = 150^\circ$
$\theta = 30^\circ,\ 150^\circ$.
Where the marks go
- 1 mark: Rearranges to $\sin\theta = \frac{1}{2}$
- 1 mark: Identifies base angle $30^\circ$ and positive quadrants
- 1 mark: Both solutions $\theta = 30^\circ$ and $150^\circ$
Key idea
Isolate the trig ratio, find the base angle, then use ASTC to place every solution within the given interval.
Frequently asked questions
Step-by-step solutions to Trigonometry questions in Preliminary Maths Advanced, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Advanced, using the methods and notation expected in exams and assessments.