Worked Solutions
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.
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.