| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Equation with half angles |
| Difficulty | Standard +0.8 This is a Further Maths question requiring the t-substitution (Weierstrass substitution) with tan(θ/2), which is a specialized technique beyond standard A-level. Part (a) is algebraically routine once the substitution formulas are known (3 marks for showing a given result). Part (b) requires solving the quadratic, then reversing the substitution using θ = 2arctan(t) and finding the general solution with periodicity—moderately challenging but follows a standard procedure for this topic. The technique itself elevates this above typical C3/C4 trigonometry questions. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Substituting \(\sin\theta = \frac{2t}{1+t^2}\) and \(\cos\theta = \frac{1-t^2}{1+t^2}\) | M1 | |
| \(4 \times \frac{2t}{1+t^2} + 5 \times \frac{1-t^2}{1+t^2} = 3\) | A1 | Removal of fractions |
| \(4 \times 2t + 5(1-t^2) = 3(1+t^2)\) oe | A1 | convincing |
| \(8t + 5 - 5t^2 = 3 + 3t^2\) | A1 | |
| \(4t^2 - 4t - 1 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Solving \(4t^2 - 4t - 1 = 0\) | M1 | M0A0 no working |
| \(t = \frac{1 \pm \sqrt{2}}{2}\) \((-0.207106...\) or \(1.207106...)\) | A1 | |
| Attempting to solve for \(\theta\) | M1 | FT their \(t\) |
| \(\tan\frac{\theta}{2} = \frac{1-\sqrt{2}}{2}\) or \(\tan\frac{\theta}{2} = \frac{1+\sqrt{2}}{2}\) | A1 | |
| \(\frac{\theta}{2} = -11.7...\) \((+180n)\) | A1 | \(\frac{\theta}{2} = -0.2...\) \((+\pi n)\) |
| or | (A1) | \(\frac{\theta}{2} = 0.87...\) \((+\pi n)\) |
| \(\frac{\theta}{2} = 50.36...\) \((+180n)\) | A1 | |
| Then, the general solution, | A1 | \(\theta = (-0.408... + 2\pi n)°\) |
| \(\theta = (-23.4(018...) + 360n)°\) oe | A1 | \(\theta = (1.758... + 2\pi n)°\) |
| or | M0 M0 for -23.4... and 100.7... without working | |
| \(\theta = (100.7(214...) + 360n)°\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | Correct notation required | |
| \(\pi\int_1^3 \sin^2 y \, dy\) | M1 | Integrable form with no more than 1 slip |
| Volume = \(\pi\int_1^3 \frac{1-\cos 2y}{2} \, dy\) | A1 | oe cao |
| \(\pi\left[\frac{1}{2}y - \frac{1}{4}\sin 2y\right]_1^3\) | m1 | Attempt to substitute in correct limits |
| \(\pi\left[\left(\frac{3}{2} - \frac{1}{4}\sin 6\right) - \left(\frac{1}{2} - \frac{1}{4}\sin 2\right)\right]\) | A1 | cao |
| Volume = 4.08 | A1 | cao |
**Part a)**
Substituting $\sin\theta = \frac{2t}{1+t^2}$ and $\cos\theta = \frac{1-t^2}{1+t^2}$ | M1 |
$4 \times \frac{2t}{1+t^2} + 5 \times \frac{1-t^2}{1+t^2} = 3$ | A1 | Removal of fractions
$4 \times 2t + 5(1-t^2) = 3(1+t^2)$ oe | A1 | convincing
$8t + 5 - 5t^2 = 3 + 3t^2$ | A1 |
$4t^2 - 4t - 1 = 0$ | A1 |
**Part b)**
Solving $4t^2 - 4t - 1 = 0$ | M1 | M0A0 no working
$t = \frac{1 \pm \sqrt{2}}{2}$ $(-0.207106...$ or $1.207106...)$ | A1 |
Attempting to solve for $\theta$ | M1 | FT their $t$
$\tan\frac{\theta}{2} = \frac{1-\sqrt{2}}{2}$ or $\tan\frac{\theta}{2} = \frac{1+\sqrt{2}}{2}$ | A1 |
$\frac{\theta}{2} = -11.7...$ $(+180n)$ | A1 | $\frac{\theta}{2} = -0.2...$ $(+\pi n)$
or | (A1) | $\frac{\theta}{2} = 0.87...$ $(+\pi n)$
$\frac{\theta}{2} = 50.36...$ $(+180n)$ | A1 |
Then, the general solution, | A1 | $\theta = (-0.408... + 2\pi n)°$
$\theta = (-23.4(018...) + 360n)°$ oe | A1 | $\theta = (1.758... + 2\pi n)°$
or | | M0 M0 for -23.4... and 100.7... without working
$\theta = (100.7(214...) + 360n)°$ oe | A1 |
**Part c)**
| B1 | Correct notation required
$\pi\int_1^3 \sin^2 y \, dy$ | M1 | Integrable form with no more than 1 slip
Volume = $\pi\int_1^3 \frac{1-\cos 2y}{2} \, dy$ | A1 | oe cao
$\pi\left[\frac{1}{2}y - \frac{1}{4}\sin 2y\right]_1^3$ | m1 | Attempt to substitute in correct limits
$\pi\left[\left(\frac{3}{2} - \frac{1}{4}\sin 6\right) - \left(\frac{1}{2} - \frac{1}{4}\sin 2\right)\right]$ | A1 | cao
Volume = 4.08 | A1 | cao
---\begin{enumerate}[label=(\alph*)]
\item By putting $t = \tan\left(\frac{\theta}{2}\right)$, show that the equation
$$4\sin\theta + 5\cos\theta = 3$$
can be written in the form
$$4t^2 - 4t - 1 = 0.$$ [3]
\item Hence find the general solution of the equation
$$4\sin\theta + 5\cos\theta = 3.$$ [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q3 [9]}}