| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard C4 multi-part question covering routine techniques: double angle formula manipulation, quadratic in cos x, harmonic form (R sin(θ+α)), and exact trigonometric values. All parts follow textbook methods with no novel insight required. Slightly above average difficulty due to length and multiple techniques, but each individual step is straightforward. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos 2x = 2\cos^2 x - 1\) | B1 | Seen in question, in consistent variable |
| \(3(2\cos^2 x - 1) + 7\cos x + 5\) | M1 | Substitute candidate's \(\cos 2x\) in terms of \(\cos x\) |
| \(6\cos^2 x + 7\cos x + 2 (= 0)\) | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \((2\cos x + 1)(3\cos x + 2)\) | M1 | Attempt factors; formula ('a' and 'c' correct; allow one slip) |
| \(\cos x = -\frac{1}{2}\), \(\cos x = -\frac{2}{3}\) | A1 (2 marks) | Accept \(-0.5, -0.67\); \(x = \cos^{-1}\left(-\frac{1}{2}\right)\); \(\cos^{-1}\left(-\frac{2}{3}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = \sqrt{58}\) | B1 | Accept 7.6 or better |
| \(\alpha = \sin^{-1}\left(\frac{3}{\text{their } R}\right)\) | M1 | OE \(\alpha = \sin^{-1}\left(\frac{3}{7}\right)\) |
| \(= 23.2°\) | A1 (3 marks) | AWRT \(23.2°\) (23.1985...) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha + \theta = \sin^{-1}\left(\frac{4}{\text{their } R}\right)\) | M1 | Candidate's \(R\), \(\alpha\) |
| \(\theta = 8.5°\) | A1F | F on \(\alpha\), AWRT, condone 8.6 |
| \(\theta = 125.1°\) | A1 (3 marks) | Two solutions only, but ignore out of range |
| Answer | Marks | Guidance |
|---|---|---|
| \(h^2 = 1 + (2\sqrt{2})^2\) | M1 | Pythagoras with \(h\) or \(\sec x\) |
| \(h = 3 \Rightarrow \cos\beta = \frac{1}{3}\) | A1 (2 marks) | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 2\beta = 2\sin\beta\cos\beta\) | M1 | |
| \(\sin 2\beta = \frac{4}{9}\sqrt{2}\) | A1 (2 marks) | CSO; accept \(p = \frac{4}{9}\) (not 0.444...) |
# Question 6(a)(i):
$\cos 2x = 2\cos^2 x - 1$ | B1 | Seen in question, in consistent variable
$3(2\cos^2 x - 1) + 7\cos x + 5$ | M1 | Substitute candidate's $\cos 2x$ in terms of $\cos x$
$6\cos^2 x + 7\cos x + 2 (= 0)$ | A1 (3 marks) |
# Question 6(a)(ii):
$(2\cos x + 1)(3\cos x + 2)$ | M1 | Attempt factors; formula ('a' and 'c' correct; allow one slip)
$\cos x = -\frac{1}{2}$, $\cos x = -\frac{2}{3}$ | A1 (2 marks) | Accept $-0.5, -0.67$; $x = \cos^{-1}\left(-\frac{1}{2}\right)$; $\cos^{-1}\left(-\frac{2}{3}\right)$
# Question 6(b)(i):
$R = \sqrt{58}$ | B1 | Accept 7.6 or better
$\alpha = \sin^{-1}\left(\frac{3}{\text{their } R}\right)$ | M1 | OE $\alpha = \sin^{-1}\left(\frac{3}{7}\right)$
$= 23.2°$ | A1 (3 marks) | AWRT $23.2°$ (23.1985...)
# Question 6(b)(ii):
$\alpha + \theta = \sin^{-1}\left(\frac{4}{\text{their } R}\right)$ | M1 | Candidate's $R$, $\alpha$
$\theta = 8.5°$ | A1F | F on $\alpha$, AWRT, condone 8.6
$\theta = 125.1°$ | A1 (3 marks) | Two solutions only, but ignore out of range
# Question 6(c)(i):
$h^2 = 1 + (2\sqrt{2})^2$ | M1 | Pythagoras with $h$ or $\sec x$
$h = 3 \Rightarrow \cos\beta = \frac{1}{3}$ | A1 (2 marks) | AG
# Question 6(c)(ii):
$\sin 2\beta = 2\sin\beta\cos\beta$ | M1 |
$\sin 2\beta = \frac{4}{9}\sqrt{2}$ | A1 (2 marks) | CSO; accept $p = \frac{4}{9}$ (not 0.444...)
---6
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation $3 \cos 2 x + 7 \cos x + 5 = 0$ can be written in the form $a \cos ^ { 2 } x + b \cos x + c = 0$, where $a , b$ and $c$ are integers.
\item Hence find the possible values of $\cos x$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Express $7 \sin \theta + 3 \cos \theta$ in the form $R \sin ( \theta + \alpha )$, where $R > 0$ and $\alpha$ is an acute angle. Give your value of $\alpha$ to the nearest $0.1 ^ { \circ }$.
\item Hence solve the equation $7 \sin \theta + 3 \cos \theta = 4$ for all solutions in the interval $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$, giving $\theta$ to the nearest $0.1 ^ { \circ }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $\beta$ is an acute angle and that $\tan \beta = 2 \sqrt { 2 }$, show that $\cos \beta = \frac { 1 } { 3 }$.
\item Hence show that $\sin 2 \beta = p \sqrt { 2 }$, where $p$ is a rational number.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q6 [15]}}