Trigonometric Identities — Worked Solutions (Preliminary Maths Extension 1)
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Worked examples for Preliminary Maths Extension 1 trigonometric identities. 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 Extension 1 — Trigonometric Identities. 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 — Proving an identity
Question
Prove that $\dfrac{1 + \cos\theta}{\sin\theta} + \dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{2}{\sin\theta}$.
Solution
Work the left-hand side. Combine over a common denominator $\sin\theta(1 + \cos\theta)$.
$\dfrac{(1 + \cos\theta)^2 + \sin^2\theta}{\sin\theta(1 + \cos\theta)}$.
Expand the numerator: $1 + 2\cos\theta + \cos^2\theta + \sin^2\theta = 1 + 2\cos\theta + 1 = 2 + 2\cos\theta = 2(1 + \cos\theta)$.
So LHS $= \dfrac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = \dfrac{2}{\sin\theta} =$ RHS.
The whole proof hinges on spotting $\cos^2\theta + \sin^2\theta = 1$ — that's the move examiners reward.
With identities we start on one side — here the left — and transform it into the other. Two fractions added means a common denominator, which is $\sin\theta(1 + \cos\theta)$.
$\dfrac{(1 + \cos\theta)(1 + \cos\theta) + \sin\theta \cdot \sin\theta}{\sin\theta(1 + \cos\theta)} = \dfrac{(1 + \cos\theta)^2 + \sin^2\theta}{\sin\theta(1 + \cos\theta)}$.
Now expand the top carefully: $(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$, and adding $\sin^2\theta$ lets us use the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$. The top becomes $1 + 2\cos\theta + 1 = 2(1 + \cos\theta)$.
The factor $(1 + \cos\theta)$ now cancels top and bottom, leaving $\dfrac{2}{\sin\theta}$, which is the right-hand side. The reason it works so neatly is that the Pythagorean identity collapses the messy squares into a clean factorable form.
Transform LHS.
- Common denominator: $\dfrac{(1 + \cos\theta)^2 + \sin^2\theta}{\sin\theta(1 + \cos\theta)}$
- Numerator: $1 + 2\cos\theta + \cos^2\theta + \sin^2\theta = 2 + 2\cos\theta = 2(1 + \cos\theta)$
- Cancel: $\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{2}{\sin\theta}$ = RHS
QED.
Where the marks go
- 1 mark: Combines the LHS over the common denominator
- 1 mark: Uses $\cos^2\theta + \sin^2\theta = 1$ to simplify the numerator to $2(1 + \cos\theta)$
- 1 mark: Cancels and concludes LHS $= \dfrac{2}{\sin\theta}$
Key idea
Prove identities by transforming one side; the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ is almost always the simplifying step.
Example 2 — Solving a trigonometric equation
Question
Solve $2\sin^2\theta + \cos\theta - 1 = 0$ for $0 \leq \theta \leq 2\pi$.
Solution
Get a single trig ratio. Replace $\sin^2\theta = 1 - \cos^2\theta$.
$2(1 - \cos^2\theta) + \cos\theta - 1 = 0 \Rightarrow -2\cos^2\theta + \cos\theta + 1 = 0 \Rightarrow 2\cos^2\theta - \cos\theta - 1 = 0$.
Factor: $(2\cos\theta + 1)(\cos\theta - 1) = 0$, so $\cos\theta = -\tfrac{1}{2}$ or $\cos\theta = 1$.
$\cos\theta = -\tfrac{1}{2}$: $\theta = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$. $\cos\theta = 1$: $\theta = 0, 2\pi$.
So $\theta = 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, 2\pi$. Don't drop the endpoints — the domain is closed.
The equation mixes $\sin^2\theta$ and $\cos\theta$, so the first job is to write everything in one ratio. Using $\sin^2\theta = 1 - \cos^2\theta$ turns it into a quadratic in $\cos\theta$.
$2(1 - \cos^2\theta) + \cos\theta - 1 = 0$ expands to $-2\cos^2\theta + \cos\theta + 1 = 0$. Multiplying by $-1$ for a tidy leading coefficient: $2\cos^2\theta - \cos\theta - 1 = 0$.
This factors as $(2\cos\theta + 1)(\cos\theta - 1) = 0$, so either $\cos\theta = -\tfrac{1}{2}$ or $\cos\theta = 1$.
Reading these off the unit circle over $0 \leq \theta \leq 2\pi$: $\cos\theta = -\tfrac{1}{2}$ in the second and third quadrants gives $\theta = \tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$, while $\cos\theta = 1$ gives the boundary values $\theta = 0$ and $2\pi$. The full set is $\theta = 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, 2\pi$.
Converting to a single ratio is what makes the equation solvable — you can't factor a mix of $\sin^2$ and $\cos$.
Single ratio, then quadratic.
- $\sin^2\theta = 1 - \cos^2\theta \Rightarrow 2\cos^2\theta - \cos\theta - 1 = 0$
- $(2\cos\theta + 1)(\cos\theta - 1) = 0$
- $\cos\theta = -\tfrac{1}{2} \Rightarrow \theta = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$
- $\cos\theta = 1 \Rightarrow \theta = 0, 2\pi$
$\theta = 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, 2\pi$.
Where the marks go
- 1 mark: Substitutes $\sin^2\theta = 1 - \cos^2\theta$ to form a quadratic in $\cos\theta$
- 1 mark: Factors to $(2\cos\theta + 1)(\cos\theta - 1) = 0$
- 1 mark: Solves $\cos\theta = -\tfrac{1}{2}$ giving $\theta = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$
- 1 mark: Includes $\cos\theta = 1$ giving $\theta = 0, 2\pi$ within the domain
Key idea
Convert to a single trig ratio using $\sin^2\theta + \cos^2\theta = 1$, factor the resulting quadratic, then read all solutions in the given domain.
Frequently asked questions
Step-by-step solutions to Trigonometric Identities questions in Preliminary 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, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Extension 1, using the methods and notation expected in exams and assessments.