Trigonometric Functions — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC Maths Extension 1 trigonometric equations. 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 HSC Maths Extension 1 — Trigonometric Functions. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach 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 — Solving with the double-angle identity
Question
Solve $\sin 2x = \cos x$ for $0 \leq x \leq 2\pi$.
Solution
Use the double angle identity $\sin 2x = 2\sin x\cos x$:
$$2\sin x\cos x = \cos x \implies \cos x(2\sin x - 1) = 0.$$
So $\cos x = 0$ or $\sin x = \tfrac12$.
$\cos x = 0$: $x = \dfrac{\pi}{2},\ \dfrac{3\pi}{2}$.
$\sin x = \tfrac12$: $x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6}$.
So $x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$. Never divide both sides by $\cos x$ — you'd lose the $\cos x = 0$ solutions.
The two sides involve different angles ($2x$ and $x$), so step one is to make them match. The double-angle identity $\sin 2x = 2\sin x\cos x$ does exactly that:
$$2\sin x\cos x = \cos x.$$
Now move everything to one side and factor — resist the urge to divide by $\cos x$, because that throws away solutions:
$$2\sin x\cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0.$$
A product is zero when a factor is zero, giving two cases. From $\cos x = 0$ we get $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$. From $\sin x = \dfrac12$ we get $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$.
All four lie in $[0, 2\pi]$: $x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$. Factoring rather than dividing is what keeps every solution.
- $\sin 2x = 2\sin x\cos x$ ⇒ $2\sin x\cos x = \cos x$
- $\cos x(2\sin x - 1) = 0$
- $\cos x = 0$: $x = \frac{\pi}{2},\ \frac{3\pi}{2}$
- $\sin x = \frac12$: $x = \frac{\pi}{6},\ \frac{5\pi}{6}$
- $x = \frac{\pi}{6},\ \frac{\pi}{2},\ \frac{5\pi}{6},\ \frac{3\pi}{2}$
Where the marks go
- 1 mark: Applies $\sin 2x = 2\sin x\cos x$ and factorises
- 1 mark: Solves $\cos x = 0$ for $x = \frac{\pi}{2}, \frac{3\pi}{2}$
- 1 mark: Solves $\sin x = \frac12$ for $x = \frac{\pi}{6}, \frac{5\pi}{6}$
Key idea
Convert $\sin 2x$ with the double-angle identity, then factor (don't divide by $\cos x$) so no solutions are lost.
Example 2 — Auxiliary-angle equation
Question
Express $\sqrt{3}\sin x + \cos x$ in the form $R\sin(x + \alpha)$ with $R > 0$ and $0 < \alpha < \frac{\pi}{2}$, then solve $\sqrt{3}\sin x + \cos x = 1$ for $0 \leq x \leq 2\pi$.
Solution
Match $R\sin(x+\alpha) = R\cos\alpha\sin x + R\sin\alpha\cos x$ to the coefficients:
$R\cos\alpha = \sqrt{3}$, $R\sin\alpha = 1$. So $R = \sqrt{3 + 1} = 2$ and $\tan\alpha = \dfrac{1}{\sqrt3}$, giving $\alpha = \dfrac{\pi}{6}$.
Thus $\sqrt3\sin x + \cos x = 2\sin\!\left(x + \dfrac{\pi}{6}\right)$.
Solve $2\sin\!\left(x+\dfrac{\pi}{6}\right) = 1 \implies \sin\!\left(x+\dfrac{\pi}{6}\right) = \dfrac12$.
With $\dfrac{\pi}{6} \leq x+\dfrac{\pi}{6} \leq 2\pi+\dfrac{\pi}{6}$: $x+\dfrac{\pi}{6} = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{13\pi}{6}$ (the lower value $\tfrac{\pi}{6}$ is the endpoint, so keep it).
So $x = 0,\ \dfrac{2\pi}{3},\ 2\pi$. Shift the domain when you solve for the bracket — that's where marks are lost.
The auxiliary-angle method rewrites a sum of sine and cosine as a single sine wave. Expand the target form: $R\sin(x+\alpha) = R\cos\alpha\,\sin x + R\sin\alpha\,\cos x$. Matching the $\sin x$ and $\cos x$ coefficients:
$$R\cos\alpha = \sqrt3, \qquad R\sin\alpha = 1.$$
Squaring and adding gives $R^2 = 3 + 1 = 4$, so $R = 2$. Dividing gives $\tan\alpha = \dfrac{1}{\sqrt3}$, so $\alpha = \dfrac{\pi}{6}$ (it sits in the required range). Hence
$$\sqrt3\sin x + \cos x = 2\sin\!\left(x+\tfrac{\pi}{6}\right).$$
Now the equation becomes $2\sin\!\left(x+\tfrac{\pi}{6}\right) = 1$, i.e. $\sin\!\left(x+\tfrac{\pi}{6}\right) = \dfrac12$. Here's the careful bit: since $0 \leq x \leq 2\pi$, the bracket runs over $\tfrac{\pi}{6} \leq x+\tfrac{\pi}{6} \leq \tfrac{13\pi}{6}$. The values with sine $\tfrac12$ in that window are $\dfrac{\pi}{6}$, $\dfrac{5\pi}{6}$ and $\dfrac{13\pi}{6}$ — the first sits right on the lower endpoint, so it counts.
Subtracting $\tfrac{\pi}{6}$: $x = 0$, $x = \dfrac{2\pi}{3}$ and $x = 2\pi$. Adjusting the domain for the shifted angle is what makes all three genuine answers appear — and don't drop the endpoint $x = 0$.
- $R\cos\alpha = \sqrt3$, $R\sin\alpha = 1$ ⇒ $R = 2$, $\tan\alpha = \frac{1}{\sqrt3}$ ⇒ $\alpha = \frac{\pi}{6}$
- $\sqrt3\sin x + \cos x = 2\sin(x+\frac{\pi}{6})$
- $\sin(x+\frac{\pi}{6}) = \frac12$, bracket in $[\frac{\pi}{6}, \frac{13\pi}{6}]$
- $x+\frac{\pi}{6} = \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6}$
- $x = 0,\ \frac{2\pi}{3},\ 2\pi$
Where the marks go
- 1 mark: Finds $R = 2$
- 1 mark: Finds $\alpha = \frac{\pi}{6}$ and states the $R\sin(x+\alpha)$ form
- 1 mark: Reduces to $\sin(x+\frac{\pi}{6}) = \frac12$ over the shifted domain
- 1 mark: Correct solutions $x = 0, \frac{2\pi}{3}, 2\pi$
Key idea
Auxiliary angle: match coefficients to get $R$ and $\alpha$, rewrite as a single sine, then solve over the shifted domain.
Frequently asked questions
Step-by-step solutions to exam-style questions on Trigonometric Functions in HSC Maths Extension 1, 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 in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Extension 1 syllabus for Trigonometric Functions, using the methods and notation expected in the exam.