| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation and evaluate integral |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard double angle formula manipulation. Part (a) is direct recall, part (b) involves routine algebraic verification and solving a quadratic in sin x, and part (c) applies the identity to a standard integration. All steps are textbook exercises with clear signposting, making it slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks |
|---|---|
| \(\cos 2x = 1 - 2\sin^2 x\) | B1 (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\sin x - \cos 2x = 3\sin x - (1 - 2\sin^2 x) = 3\sin x - 1 + 2\sin^2 x\) | M1, A1 (2 marks) | Candidate's \(\cos 2x\) or \(\sin^2 x\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\sin^2 x + 3\sin x - 2 = 0\) | M1 | Soluble quadratic form |
| \((2\sin x - 1)(\sin x + 2) = 0\) | M1 | Attempt to solve (allow one error in formula, allow sign errors) |
| \(\sin x = \frac{1}{2}\), \(x = 30°\), \(x = 150°\) | M1, A1 (4 marks) | \(\sin^{-1}\) and two solutions (\(0° < x < 360°\)); A0 if radians |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{2}(1 - \cos 2x)\,dx = \frac{x}{2} - \frac{\sin 2x}{4}\ (+c)\) | M1A1 (2 marks) | M1 – solve integral, must have 2 terms for \(\sin^2 x\) from (a) |
## Question 3:
**Part (a)**
$\cos 2x = 1 - 2\sin^2 x$ | B1 (1 mark) |
**Part (b)(i)**
$3\sin x - \cos 2x = 3\sin x - (1 - 2\sin^2 x) = 3\sin x - 1 + 2\sin^2 x$ | M1, A1 (2 marks) | Candidate's $\cos 2x$ or $\sin^2 x$; AG
**Part (b)(ii)**
$2\sin^2 x + 3\sin x - 2 = 0$ | M1 | Soluble quadratic form
$(2\sin x - 1)(\sin x + 2) = 0$ | M1 | Attempt to solve (allow one error in formula, allow sign errors)
$\sin x = \frac{1}{2}$, $x = 30°$, $x = 150°$ | M1, A1 (4 marks) | $\sin^{-1}$ and two solutions ($0° < x < 360°$); A0 if radians
Allow misread for $2\sin^2 x + 3\sin x - 1 = 0$: $\sin x = \frac{-3 \pm \sqrt{17}}{4}$ (M1); $x = 16.3°$, $163.7°$ (A1); Max 3/4
**Part (c)**
$\int \frac{1}{2}(1 - \cos 2x)\,dx = \frac{x}{2} - \frac{\sin 2x}{4}\ (+c)$ | M1A1 (2 marks) | M1 – solve integral, must have 2 terms for $\sin^2 x$ from (a)
**Total: 9 marks**
---3
\begin{enumerate}[label=(\alph*)]
\item Express $\cos 2 x$ in terms of $\sin x$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence show that $3 \sin x - \cos 2 x = 2 \sin ^ { 2 } x + 3 \sin x - 1$ for all values of $x$.
\item Solve the equation $3 \sin x - \cos 2 x = 1$ for $0 ^ { \circ } < x < 360 ^ { \circ }$.
\end{enumerate}\item Use your answer from part (a) to find $\int \sin ^ { 2 } x \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q3 [9]}}