| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard C3 harmonic form question following a predictable template: convert to R cos(x+α), find minimum, then solve equation. The arithmetic involves a 7-24-25 Pythagorean triple making calculations straightforward. While it requires multiple techniques (harmonic form, inverse trig, equation solving), these are routine applications with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = \sqrt{7^2 + 24^2} = 25\) | B1 | \(R = 25\) |
| \(\tan\alpha = \frac{24}{7} \Rightarrow \alpha = 1.287002218...\) | M1 | \(\tan\alpha = \frac{24}{7}\) or \(\tan\alpha = \frac{7}{24}\) |
| \(\alpha = 1.287...\) | A1 | awrt 1.287 |
| Hence \(7\cos x - 24\sin x = 25\cos(x + 1.287)\) | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Minimum value \(= -25\) | B1ft | \(-25\) or \(-R\) (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos(x + 1.287) = \frac{10}{25}\) | M1 | \(\cos(x \pm \text{their } \alpha) = \frac{10}{\text{their } R}\) |
| \(\text{PV} = 1.159279481...\) or \(66.42182152...°\) | M1 | Applying \(\cos^{-1}\!\left(\frac{10}{\text{their } R}\right)\) |
| \(x + 1.287 = \{1.159279...,\ 5.123906...,\ 7.442465...\}\) | M1 | Either \(2\pi +\) or \(-\) their PV or \(360° +\) or \(-\) their PV |
| \(x = \{3.836906...,\ 6.155465...\}\) | A1 | awrt 3.84 OR 6.16 |
| A1 | awrt 3.84 AND 6.16 (5 marks total) [9] |
## Question 1:
**Part (a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = \sqrt{7^2 + 24^2} = 25$ | B1 | $R = 25$ |
| $\tan\alpha = \frac{24}{7} \Rightarrow \alpha = 1.287002218...$ | M1 | $\tan\alpha = \frac{24}{7}$ or $\tan\alpha = \frac{7}{24}$ |
| $\alpha = 1.287...$ | A1 | awrt 1.287 |
| Hence $7\cos x - 24\sin x = 25\cos(x + 1.287)$ | | (3 marks total) |
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Minimum value $= -25$ | B1ft | $-25$ or $-R$ (1 mark) |
**Part (c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos(x + 1.287) = \frac{10}{25}$ | M1 | $\cos(x \pm \text{their } \alpha) = \frac{10}{\text{their } R}$ |
| $\text{PV} = 1.159279481...$ or $66.42182152...°$ | M1 | Applying $\cos^{-1}\!\left(\frac{10}{\text{their } R}\right)$ |
| $x + 1.287 = \{1.159279...,\ 5.123906...,\ 7.442465...\}$ | M1 | Either $2\pi +$ or $-$ their PV or $360° +$ or $-$ their PV |
| $x = \{3.836906...,\ 6.155465...\}$ | A1 | awrt 3.84 OR 6.16 |
| | A1 | awrt 3.84 AND 6.16 (5 marks total) **[9]** |
---\begin{enumerate}
\item (a) Express $7 \cos x - 24 \sin x$ in the form $R \cos ( x + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$. Give the value of $\alpha$ to 3 decimal places.\\
(b) Hence write down the minimum value of $7 \cos x - 24 \sin x$.\\
(c) Solve, for $0 \leqslant x < 2 \pi$, the equation
\end{enumerate}
$$7 \cos x - 24 \sin x = 10$$
giving your answers to 2 decimal places.\\
\hfill \mbox{\textit{Edexcel C3 2011 Q1 [9]}}