| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Derive triple angle then evaluate integral |
| Difficulty | Moderate -0.3 This is a structured, multi-part question that guides students through standard derivations (double angle formulae, triple angle formula) before applying them to a definite integral. While it requires multiple steps, each part is heavily scaffolded with clear instructions, making it slightly easier than average for C4 level. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\sin 2x = 2\sin x \cos x\) | B1 | |
| (b)(i) \(A = B = x\) | M1, A1 | |
| \(\cos 2x = \cos^2 x - \sin^2 x\) | M1 | |
| (ii) \(\cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x\) | M1 | |
| \(= (\cos^2 x - \sin^2 x)\cos x - 2\sin^2 x \cos x\) | A1 | ft (a) and (b)(i) |
| \(= \cos^3 x - 3(1 - \cos^2 x)\cos x\) | M1 | Eliminate all sines |
| \(= 4\cos^3 x - 3\cos x\) | A1 | AG; convincingly obtained |
| (c) \(\cos^3 x = \frac{1}{4}(\cos 3x + 3\cos x)\) | M1 | Ignore middle sign |
| \(\int_0^{\pi/3} \cos^3 x dx = \frac{1}{4}\left[\frac{1}{3}\sin 3x + 3\sin x\right]_0^{\pi/3}\) | A1A1 | 1 for integrity each term; Use of limits |
| \(\frac{1}{4}\left(-\frac{1}{3} + 3\right) = \frac{2}{3}\) | M1, A1 | AG |
**(a)** $\sin 2x = 2\sin x \cos x$ | B1 |
**(b)(i)** $A = B = x$ | M1, A1 |
$\cos 2x = \cos^2 x - \sin^2 x$ | M1 |
**(ii)** $\cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x$ | M1 |
$= (\cos^2 x - \sin^2 x)\cos x - 2\sin^2 x \cos x$ | A1| ft (a) and (b)(i)
$= \cos^3 x - 3(1 - \cos^2 x)\cos x$ | M1 | Eliminate all sines
$= 4\cos^3 x - 3\cos x$ | A1 | AG; convincingly obtained
**(c)** $\cos^3 x = \frac{1}{4}(\cos 3x + 3\cos x)$ | M1 | Ignore middle sign
$\int_0^{\pi/3} \cos^3 x dx = \frac{1}{4}\left[\frac{1}{3}\sin 3x + 3\sin x\right]_0^{\pi/3}$ | A1A1 | 1 for integrity each term; Use of limits
$\frac{1}{4}\left(-\frac{1}{3} + 3\right) = \frac{2}{3}$ | M1, A1 | AG
**Total: 12 marks**
---6
\begin{enumerate}[label=(\alph*)]
\item Express $\sin 2 x$ in terms of $\sin x$ and $\cos x$.
\item Using the identity $\cos ( A + B ) = \cos A \cos B - \sin A \sin B$ :
\begin{enumerate}[label=(\roman*)]
\item express $\cos 2 x$ in terms of $\sin x$ and $\cos x$;
\item show, by writing $3 x$ as $( 2 x + x )$, that
$$\cos 3 x = 4 \cos ^ { 3 } x - 3 \cos x$$
\end{enumerate}\item Show that $\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 3 } x \mathrm {~d} x = \frac { 2 } { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2005 Q6 [12]}}