| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve equation using Pythagorean identities |
| Difficulty | Standard +0.3 This is a standard A-level trigonometry question testing routine techniques: part (a) uses the identity cosec²x = 1 + cot²x to reduce to a quadratic, and part (b) is the textbook R-formula method followed by straightforward equation solving. Both require multiple steps but no novel insight, making it slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cosec^2 x + \cot^2 x = 5\) | M1 | (Attempt to write in terms of one function) |
| \(1 + 2\cot^2 x = 5\) | ||
| \(\cot^2 x = 2\) | A1 | |
| \(\tan x = \pm \frac{1}{\sqrt{2}}\) | A1 | |
| \(x = 35.3°, 215.3°, 144.7°, 324.7°\) | B1, B1 | (each pair) |
| Answer | Marks |
|---|---|
| \(4\sin\theta + 3\cos\theta \equiv R(\sin\theta \cos\alpha + \cos\theta \sin\alpha)\) | B1 |
| \(R\cos\alpha = 4\) | B1 |
| \(R\sin\alpha = 3\) | |
| \(R = \sqrt{3^2 + 4^2} = 5\) | B1 |
| \(\tan\alpha = \frac{3}{4}\), \(\alpha = 36.87°\) | B1 |
| \(4\sin\theta + 3\cos\theta \equiv 5\sin(\theta + 36.87°)\) |
| Answer | Marks |
|---|---|
| \(5\sin(\theta + 36.87°) = 2\) | |
| \(\sin(\theta + 36.87°) = 0.4\) | B1 |
| \(\theta + 36.87° = 23.58°, 156.42°, 383.58°\) | |
| \(\theta = 119.5(5)°, 346.7(1)'\) | B1 |
| \(= 120°, 347°\) to the nearest degree | B1 |
### Part (a):
$\cosec^2 x + \cot^2 x = 5$ | M1 | (Attempt to write in terms of one function)
$1 + 2\cot^2 x = 5$ | |
$\cot^2 x = 2$ | A1 |
$\tan x = \pm \frac{1}{\sqrt{2}}$ | A1 |
$x = 35.3°, 215.3°, 144.7°, 324.7°$ | B1, B1 | (each pair)
### Part (b)(i):
$4\sin\theta + 3\cos\theta \equiv R(\sin\theta \cos\alpha + \cos\theta \sin\alpha)$ | B1 |
$R\cos\alpha = 4$ | B1 |
$R\sin\alpha = 3$ | |
$R = \sqrt{3^2 + 4^2} = 5$ | B1 |
$\tan\alpha = \frac{3}{4}$, $\alpha = 36.87°$ | B1 |
$4\sin\theta + 3\cos\theta \equiv 5\sin(\theta + 36.87°)$ | |
### Part (b)(ii):
$5\sin(\theta + 36.87°) = 2$ | |
$\sin(\theta + 36.87°) = 0.4$ | B1 |
$\theta + 36.87° = 23.58°, 156.42°, 383.58°$ | |
$\theta = 119.5(5)°, 346.7(1)'$ | B1 |
$= 120°, 347°$ to the nearest degree | B1 |
**Total: [12]**\begin{enumerate}[label=(\alph*)]
\item Solve the equation
$$\operatorname{cosec}^2 x + \cot^2 x = 5$$
for $0^{\circ} \leq x \leq 360^{\circ}$. [5]
\item \begin{enumerate}[label=(\roman*)]
\item Express $4\sin \theta + 3\cos \theta$ in the form $R\sin(\theta + \alpha)$, where $R > 0$ and $0^{\circ} \leq \alpha \leq 90^{\circ}$. [4]
\item Solve the equation
$$4\sin \theta + 3\cos \theta = 2$$
for $0^{\circ} \leq \theta \leq 360^{\circ}$, giving your answer correct to the nearest degree. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q13 [12]}}