| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Solve equation with sin2x/cos2x by substitution |
| Difficulty | Standard +0.3 This is a standard C4 trigonometric equation question requiring recall of double angle formulae and routine algebraic manipulation. Part (a) is pure recall, part (b) involves substituting the double angle formula and factorising to solve (standard technique), and part (c) requires substituting both double angle formulae and simplifying to reach the given result. While multi-part, each step follows textbook methods without requiring novel insight, making it slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 2x = 2\sin x\cos x\); \(\cos x = 0\), \(x = 90, 270\) | B1, B1 (1 mark) | OE e.g. \(\sin x\cos x + \sin x\cos x\); both required |
| Answer | Marks | Guidance |
|---|---|---|
| \(10\sin x + 3 = 0\), \(x = 197.5\), \(342.5\) | M1, A1A1 (4 marks) | CAO; if extra values in given range, max 1/2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos 2x = \cos^2 x - \sin^2 x\) | B1 | \(\cos 2x\) in any correct form |
| \(2\sin x\cos x + 1 - 2\sin^2 x = 1 + \sin x\) | M1 | \(\sin 2x\) expanded and \(\cos 2x\) in terms of \(\sin x\) used |
| A1 | ||
| \(2\sin x(\cos x - \sin x) = \sin x\); \(2(\cos x - \sin x) = 1\) | A1 (4 marks) | CSO; need to see \(\sin x\) taken out as factor or cancelled |
## Question 5:
### Part (a)
| $\sin 2x = 2\sin x\cos x$; $\cos x = 0$, $x = 90, 270$ | B1, B1 (1 mark) | OE e.g. $\sin x\cos x + \sin x\cos x$; both required |
### Part (b)
| $10\sin x + 3 = 0$, $x = 197.5$, $342.5$ | M1, A1A1 (4 marks) | CAO; if extra values in given range, max 1/2 |
### Part (c)
| $\cos 2x = \cos^2 x - \sin^2 x$ | B1 | $\cos 2x$ in any correct form |
| $2\sin x\cos x + 1 - 2\sin^2 x = 1 + \sin x$ | M1 | $\sin 2x$ expanded and $\cos 2x$ in terms of $\sin x$ used |
| | A1 | |
| $2\sin x(\cos x - \sin x) = \sin x$; $2(\cos x - \sin x) = 1$ | A1 (4 marks) | CSO; need to see $\sin x$ taken out as factor or cancelled |5
\begin{enumerate}[label=(\alph*)]
\item Express $\sin 2 x$ in terms of $\sin x$ and $\cos x$.
\item Solve the equation
$$5 \sin 2 x + 3 \cos x = 0$$
giving all solutions in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$ to the nearest $0.1 ^ { \circ }$, where appropriate.
\item Given that $\sin 2 x + \cos 2 x = 1 + \sin x$ and $\sin x \neq 0$, show that $2 ( \cos x - \sin x ) = 1$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q5 [9]}}