Worked Solutions
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.
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.