Worked Solutions
Trigonometric Functions — Worked Solutions (HSC Maths Extension 1)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our HSC Maths Extension 1 course →
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.
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 $\dfrac{\pi}{6} \leq x+\dfrac{\pi}{6} \leq 2\pi+\dfrac{\pi}{6}$: $x+\dfrac{\pi}{6} = \dfrac{5\pi}{6},\ \dfrac{13\pi}{6}$ (reject $\tfrac{\pi}{6}$, below range).
So $x = \dfrac{2\pi}{3},\ 2\pi$. 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 $\dfrac{5\pi}{6}$ and $\dfrac{13\pi}{6}$ (the usual $\tfrac{\pi}{6}$ is just outside the lower end).
Subtracting $\tfrac{\pi}{6}$: $x = \dfrac{2\pi}{3}$ and $x = 2\pi$. Adjusting the domain for the shifted angle is why both genuine answers appear and the spurious one is rejected.
- $R\cos\alpha = \sqrt3$, $R\sin\alpha = 1$ ⇒ $R = 2$, $\tan\alpha = \frac{1}{\sqrt3}$ ⇒ $\alpha = \frac{\pi}{6}$
- $\sqrt3\sin x + \cos x = 2\sin(x+\frac{\pi}{6})$
- $\sin(x+\frac{\pi}{6}) = \frac12$, bracket in $[\frac{\pi}{6}, \frac{13\pi}{6}]$
- $x+\frac{\pi}{6} = \frac{5\pi}{6},\ \frac{13\pi}{6}$
- $x = \frac{2\pi}{3},\ 2\pi$
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 $x = \frac{2\pi}{3}, 2\pi$
Key idea
Auxiliary angle: match coefficients to get $R$ and $\alpha$, rewrite as a single sine, then solve over the shifted domain.