| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Moderate -0.3 This is a standard harmonic form question testing routine C4 techniques: converting to R cos(x-α) using Pythagorean identity and tan α = b/a, solving the resulting equation, and identifying minimum values. All parts follow predictable patterns with no novel problem-solving required, making it slightly easier than average but still requiring multiple techniques. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 5\) | B1 | |
| \(\tan\alpha = \frac{3}{4}\) (OE) | M1A1 | 3 marks |
| SC1: \(\tan\alpha = \frac{4}{3}, \alpha = 53.1°\); \(R, \alpha\) PI in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos(x-\alpha) = \frac{2}{R}\) | M1 | |
| \(x - \alpha = 66.4°\) | A1 | |
| \(x = 103.3°\) | A1F | |
| \(x = 330.4°\) | A1F | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| minimum value = –5 | B1F | |
| \(\cos(x-36.9) = -1\) | M1 | |
| \(x = 216.9°\) | A1 | 3 marks |
| Max 8/10 for work in radians |
**3(a)**
| $R = 5$ | B1 | | |
| $\tan\alpha = \frac{3}{4}$ (OE) | M1A1 | 3 marks | $\alpha = 36.9°$ (ISW 216.9) |
| | | | SC1: $\tan\alpha = \frac{4}{3}, \alpha = 53.1°$; $R, \alpha$ PI in (b) |
**3(b)**
| $\cos(x-\alpha) = \frac{2}{R}$ | M1 | | |
| $x - \alpha = 66.4°$ | A1 | | |
| $x = 103.3°$ | A1F | | |
| $x = 330.4°$ | A1F | 4 marks | accept 330.5°, –1 each extra ft on acute $\alpha$ |
**3(c)**
| minimum value = –5 | B1F | | ft on $R$ |
| $\cos(x-36.9) = -1$ | M1 | | SC $\cos(x+36.9)$ treat as miscopy |
| $x = 216.9°$ | A1 | 3 marks | 216.9 or better accept graphics calculator solution to this accuracy; SC Find max: max = 5 at $(x+36.9)$ stated 1/3 |
| | | | Max 8/10 for work in radians |3
\begin{enumerate}[label=(\alph*)]
\item Express $4 \cos x + 3 \sin x$ in the form $R \cos ( x - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 360 ^ { \circ }$, giving your value for $\alpha$ to the nearest $0.1 ^ { \circ }$.
\item Hence solve the equation $4 \cos x + 3 \sin x = 2$ in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$, giving all solutions to the nearest $0.1 ^ { \circ }$.
\item Write down the minimum value of $4 \cos x + 3 \sin x$ and find the value of $x$ in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$ at which this minimum value occurs.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q3 [10]}}