| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| 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 harmonic form question with routine follow-through parts. Part (a) uses the textbook R-formula method and basic equation solving. Part (b) applies standard double angle formulas. Part (c) combines factor theorem with trigonometric identities in a structured way with significant scaffolding. While multi-part, each component follows predictable techniques without requiring novel insight. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05l 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 |
|---|---|---|
| \(R\sin(x + \alpha) = R\sin x \cos\alpha + R\cos x \sin\alpha\) | M1 | Expanding and comparing coefficients |
| \(R\cos\alpha = 3\), \(R\sin\alpha = 4\) | ||
| \(R = \sqrt{3^2 + 4^2} = 5\) | B1 | |
| \(\alpha = \arctan\left(\frac{4}{3}\right) = 53.1°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(5\sin(2\theta + 53.1°) = 5\) | M1 | Using result from (a)(i) with \(2\theta\) |
| \(\sin(2\theta + 53.1°) = 1\) | ||
| \(2\theta + 53.1° = 90°, 450°\) | M1 | |
| \(\theta = 18.4°, 198.4°\) | A1 | Both values |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}\), so \(\frac{2\tan\theta}{1-\tan^2\theta} \cdot \tan\theta = 2\) | M1 | Using double angle formula |
| \(\frac{2\tan^2\theta}{1-\tan^2\theta} = 2\) | ||
| \(2\tan^2\theta = 2(1-\tan^2\theta)\) | ||
| \(4\tan^2\theta = 2\), hence \(2\tan^2\theta = 1\) | A1 | Completion, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan^2\theta = \frac{1}{2}\), \(\tan\theta = \pm\frac{1}{\sqrt{2}}\) | M1 | |
| \(\theta = 35.3°, 144.7°\) | A1 | Both values |
| Answer | Marks | Guidance |
|---|---|---|
| \(f\left(\frac{1}{2}\right) = 8\left(\frac{1}{8}\right) - 4\left(\frac{1}{2}\right) + 1 = 1 - 2 + 1 = 0\) | B1 | Correct substitution and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\cos 2\theta\cos\theta + 1 = 4(2\cos^2\theta - 1)\cos\theta + 1\) | M1 | Using \(\cos 2\theta = 2\cos^2\theta - 1\) |
| \(= 8\cos^3\theta - 4\cos\theta + 1\) | A1 | Setting \(x = \cos\theta\) gives \(8x^3 - 4x + 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = 72°\) is a solution so \(x = \cos 72°\) satisfies \(8x^3 - 4x + 1 = 0\) | M1 | |
| From (c)(i), \((2x-1)\) is a factor, so \(8x^3 - 4x + 1 = (2x-1)(4x^2 + 2x - 1)\) | M1 | Factorising |
| \(\cos 72° \neq \frac{1}{2}\), so \(4x^2 + 2x - 1 = 0\) | ||
| \(x = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}\) | M1 | Using quadratic formula |
| Since \(\cos 72° > 0\): \(\cos 72° = \frac{\sqrt{5}-1}{4}\), so \(p = 4\) | A1 |
# Question 5:
## Part (a)(i):
| $R\sin(x + \alpha) = R\sin x \cos\alpha + R\cos x \sin\alpha$ | M1 | Expanding and comparing coefficients |
|---|---|---|
| $R\cos\alpha = 3$, $R\sin\alpha = 4$ | | |
| $R = \sqrt{3^2 + 4^2} = 5$ | B1 | |
| $\alpha = \arctan\left(\frac{4}{3}\right) = 53.1°$ | A1 | |
## Part (a)(ii):
| $5\sin(2\theta + 53.1°) = 5$ | M1 | Using result from (a)(i) with $2\theta$ |
|---|---|---|
| $\sin(2\theta + 53.1°) = 1$ | | |
| $2\theta + 53.1° = 90°, 450°$ | M1 | |
| $\theta = 18.4°, 198.4°$ | A1 | Both values |
## Part (b)(i):
| $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$, so $\frac{2\tan\theta}{1-\tan^2\theta} \cdot \tan\theta = 2$ | M1 | Using double angle formula |
|---|---|---|
| $\frac{2\tan^2\theta}{1-\tan^2\theta} = 2$ | | |
| $2\tan^2\theta = 2(1-\tan^2\theta)$ | | |
| $4\tan^2\theta = 2$, hence $2\tan^2\theta = 1$ | A1 | Completion, no errors |
## Part (b)(ii):
| $\tan^2\theta = \frac{1}{2}$, $\tan\theta = \pm\frac{1}{\sqrt{2}}$ | M1 | |
|---|---|---|
| $\theta = 35.3°, 144.7°$ | A1 | Both values |
## Part (c)(i):
| $f\left(\frac{1}{2}\right) = 8\left(\frac{1}{8}\right) - 4\left(\frac{1}{2}\right) + 1 = 1 - 2 + 1 = 0$ | B1 | Correct substitution and conclusion |
## Part (c)(ii):
| $4\cos 2\theta\cos\theta + 1 = 4(2\cos^2\theta - 1)\cos\theta + 1$ | M1 | Using $\cos 2\theta = 2\cos^2\theta - 1$ |
|---|---|---|
| $= 8\cos^3\theta - 4\cos\theta + 1$ | A1 | Setting $x = \cos\theta$ gives $8x^3 - 4x + 1$ |
## Part (c)(iii):
| $\theta = 72°$ is a solution so $x = \cos 72°$ satisfies $8x^3 - 4x + 1 = 0$ | M1 | |
|---|---|---|
| From (c)(i), $(2x-1)$ is a factor, so $8x^3 - 4x + 1 = (2x-1)(4x^2 + 2x - 1)$ | M1 | Factorising |
| $\cos 72° \neq \frac{1}{2}$, so $4x^2 + 2x - 1 = 0$ | | |
| $x = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}$ | M1 | Using quadratic formula |
| Since $\cos 72° > 0$: $\cos 72° = \frac{\sqrt{5}-1}{4}$, so $p = 4$ | A1 | |
---5
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $3 \sin x + 4 \cos x$ in the form $R \sin ( x + \alpha )$ where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, giving your value of $\alpha$ to the nearest $0.1 ^ { \circ }$.
\item Hence solve the equation $3 \sin 2 \theta + 4 \cos 2 \theta = 5$ in the interval $0 ^ { \circ } < \theta < 360 ^ { \circ }$, giving your solutions to the nearest $0.1 ^ { \circ }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation $\tan 2 \theta \tan \theta = 2$ can be written as $2 \tan ^ { 2 } \theta = 1$.
\item Hence solve the equation $\tan 2 \theta \tan \theta = 2$ in the interval $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$, giving your solutions to the nearest $0.1 ^ { \circ }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $2 x - 1$ is a factor of $8 x ^ { 3 } - 4 x + 1$.
\item Show that $4 \cos 2 \theta \cos \theta + 1$ can be written as $8 x ^ { 3 } - 4 x + 1$ where $x = \cos \theta$.
\item Given that $\theta = 72 ^ { \circ }$ is a solution of $4 \cos 2 \theta \cos \theta + 1 = 0$, use the results from parts (c)(i) and (c)(ii) to show that the exact value of $\cos 72 ^ { \circ }$ is $\frac { ( \sqrt { 5 } - 1 ) } { p }$ where $p$ is an integer.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2014 Q5 [15]}}