| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Equation with half angles |
| Difficulty | Challenging +1.2 Part (a) is a straightforward verification using the t-substitution formulas (sin x = 2t/(1+t²), cos x = (1-t²)/(1+t²)), requiring only algebraic manipulation. Part (b) involves solving a quadratic equation in t and converting back to x, which is a standard Further Maths technique. While this requires multiple steps and careful algebra across 9 marks total, it follows a well-established procedure without requiring novel insight—typical of Further Pure questions but more routine than problems requiring geometric reasoning or proof. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\left(\frac{2t}{1+t^2}\right) + 4\left(\frac{1-t^2}{1+t^2}\right) - 2\) | M1 | |
| \(3\left(\frac{2t}{1+t^2}\right) + 4\left(\frac{1-t^2}{1+t^2}\right) - 2\left(\frac{1+t^2}{1+t^2}\right)\) | ||
| \(= \frac{6t + 2 - 6t^2}{1+t^2}\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6t + 2 - 6t^2}{1+t^2} = 3\) | M1 | |
| \(6t + 2 - 6t^2 = 3 + 3t^2\) | A1 | |
| \(9t^2 - 6t + 1 = 0\) | m1 A1 | m0 no working |
| Solving, eg \((3t-1)^2 = 0\) | ||
| \(t = \frac{1}{3}\) | ||
| \(\therefore \tan\frac{x}{2} = \frac{1}{3}\) | M1 | Attempt to solve for \(x\). FT their \(t\) |
| \(\frac{1}{2}x = 18.43° + 180°n\) | A1 | Accept radians: \(0.322 + n\pi\) |
| \(x = 36.87° + 360°n\) | A1 | \(0.644 + 2n\pi\) |
## Part a)
$3\left(\frac{2t}{1+t^2}\right) + 4\left(\frac{1-t^2}{1+t^2}\right) - 2$ | M1 |
$3\left(\frac{2t}{1+t^2}\right) + 4\left(\frac{1-t^2}{1+t^2}\right) - 2\left(\frac{1+t^2}{1+t^2}\right)$ | |
$= \frac{6t + 2 - 6t^2}{1+t^2}$ | A1 | Convincing
## Part b)
$\frac{6t + 2 - 6t^2}{1+t^2} = 3$ | M1 |
$6t + 2 - 6t^2 = 3 + 3t^2$ | A1 |
$9t^2 - 6t + 1 = 0$ | m1 A1 | m0 no working
Solving, eg $(3t-1)^2 = 0$ | |
$t = \frac{1}{3}$ | |
$\therefore \tan\frac{x}{2} = \frac{1}{3}$ | M1 | Attempt to solve for $x$. FT their $t$
$\frac{1}{2}x = 18.43° + 180°n$ | A1 | Accept radians: $0.322 + n\pi$
$x = 36.87° + 360°n$ | A1 | $0.644 + 2n\pi$\begin{enumerate}[label=(\alph*)]
\item Show that $3\sin x + 4\cos x - 2$ can be written as $\frac{6t + 2 - 6t^2}{1 + t^2}$, where $t = \tan\left(\frac{x}{2}\right)$. [2]
\item Hence, find the general solution of the equation $3\sin x + 4\cos x - 2 = 3$. [7]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q2 [9]}}