| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard C4 harmonic form question with routine techniques: converting to R sin(θ+α) using Pythagorean theorem and arctan, solving the resulting equation, and proving a standard trigonometric identity. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| 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 intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 10\) | B1 | |
| \(\tan \alpha = \frac{8}{6}\), \(\alpha = 53.1\) | B1 F | For \(\alpha\): ft incorrect \(R\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin(2x + 53.1) = 0.7\) | M1 | |
| \(2x + 53.1 = 44.4\) | A1 F | one correct answer; ft \(\alpha\) and \(R\) |
| \(135.6\) or \(135.7, 404.4, 495.6\) or \(495.7\) | A1 | 3 other correct answers – ignore extras |
| \(x = 41.2\) or \(41.3, 175.6\) or \(175.7, 221.2\) or \(221.3, 355.6\) or \(355.7\) | A1 | four solutions; CAO (with decimal place discrepancies); Answers only: 0/4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 2x = 2\sin x \cos x\) | B1 | identities for \(\sin 2x\) and \(\cos 2x\) in any correct form |
| \(\cos 2x = \cos^2 x - \sin^2 x\) | B1 | |
| \(\frac{\sin 2x}{1 - \cos 2x} = \frac{2\sin x \cos x}{1 - (1 - 2\sin^2 x)}\) | M1 | use of candidate's double angle formulae |
| \(= \frac{2\sin x \cos x}{2\sin^2 x} = \frac{\cos x}{\sin x} = \frac{1}{\tan x}\) | A1 | AG, CSO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{\tan x} = \tan x\), \(\tan x = \pm 1\) | M1 A1 | (see * below) |
| \(x = 45, 135, 225, 315\) | B1 | |
| A1 | 4: if answers given without working, B1 max; if \(\frac{1}{\tan x} = \tan x\) seen and followed by correct answers without working 4 out of 4 |
**7(a)(i)**
| $R = 10$ | B1 | |
| $\tan \alpha = \frac{8}{6}$, $\alpha = 53.1$ | B1 F | For $\alpha$: ft incorrect $R$ |
**7(a)(ii)**
| $\sin(2x + 53.1) = 0.7$ | M1 | |
| $2x + 53.1 = 44.4$ | A1 F | one correct answer; ft $\alpha$ and $R$ |
| $135.6$ or $135.7, 404.4, 495.6$ or $495.7$ | A1 | 3 other correct answers – ignore extras |
| $x = 41.2$ or $41.3, 175.6$ or $175.7, 221.2$ or $221.3, 355.6$ or $355.7$ | A1 | four solutions; CAO (with decimal place discrepancies); Answers only: 0/4 |
**7(b)(i)**
| $\sin 2x = 2\sin x \cos x$ | B1 | identities for $\sin 2x$ and $\cos 2x$ in any correct form |
| $\cos 2x = \cos^2 x - \sin^2 x$ | B1 | |
| $\frac{\sin 2x}{1 - \cos 2x} = \frac{2\sin x \cos x}{1 - (1 - 2\sin^2 x)}$ | M1 | use of candidate's double angle formulae |
| $= \frac{2\sin x \cos x}{2\sin^2 x} = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$ | A1 | AG, CSO |
**7(b)(ii)**
| $\frac{1}{\tan x} = \tan x$, $\tan x = \pm 1$ | M1 A1 | (see * below) |
| $x = 45, 135, 225, 315$ | B1 | |
| | A1 | 4: if answers given without working, B1 max; if $\frac{1}{\tan x} = \tan x$ seen and followed by correct answers without working 4 out of 4 |
**Total: 14 marks**
---7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $6 \sin \theta + 8 \cos \theta$ in the form $R \sin ( \theta + \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. Give your value for $\alpha$ to the nearest $0.1 ^ { \circ }$.
\item Hence solve the equation $6 \sin 2 x + 8 \cos 2 x = 7$, giving all solutions to the nearest $0.1 ^ { \circ }$ in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Prove the identity $\frac { \sin 2 x } { 1 - \cos 2 x } = \frac { 1 } { \tan x }$.
\item Hence solve the equation
$$\frac { \sin 2 x } { 1 - \cos 2 x } = \tan x$$
giving all solutions in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q7 [14]}}