| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard harmonic form question with routine application of R cos(θ - α) transformation followed by a straightforward double-angle substitution and solving. The techniques are well-practiced in C3/C4 with no novel insight required, making it slightly easier than average but still requiring multiple steps and careful execution. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = \sqrt{1^2 + 4^2} = \sqrt{17}\) | B1 | Condone \(R = \pm\sqrt{17}\) |
| \(\alpha = \arctan 4\) | M1 | For \(\alpha = \arctan(\pm 4)\) or \(\alpha = \arctan\left(\pm\frac{1}{4}\right)\) leading to a solution of \(\alpha\). Implied by \(\alpha =\) awrt \(76°\) or awrt 1.3 rads. If \(R\) used to find \(\alpha\), award only \(\alpha = \arccos\left(\pm\frac{1}{R}\right)\) or \(\alpha = \arcsin\left(\pm\frac{4}{R}\right)\) |
| \(\alpha =\) awrt \(1.326\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{17}\cos(2\theta - 1.326) = 1.2 \Rightarrow \cos(2\theta - 1.326) = \frac{1.2}{\sqrt{17}}\) | M1 | Using part (a), proceeding as far as \(\cos(2\theta \pm \text{their } 1.326) = \frac{1.2}{\text{their } R}\). Condone slips on 1.2 and miscopying their 1.326. Note: \(2\cos(\theta \pm \text{their } 1.326) = \frac{1.2}{\text{their } R}\) is M0 |
| \(\Rightarrow (2\theta - 1.326) = \pm 1.275...\Rightarrow \theta = ...\) | dM1 | Dependent on first M1. Full method to find one value of \(\theta\) in range \(0\) to \(\pi\) from principal value. Deal with "1.326" before the "2". Condone adding 1.326 instead of subtracting |
| \(\theta =\) awrt \(1.30\) or awrt \(0.03\) | A1 | Only allow 1.3 if preceded by answer rounding to 1.30 |
| \(2\theta - 1.326 = {'}1.275...{'}\ \textbf{and}\ {'-}1.275...'\) | ddM1 | Correct method for second value of \(\theta\) (for their \(\alpha\)) in range \(0\) to \(\pi\). E.g. \(2\theta + 1.326 = -\beta' \Rightarrow \theta =\) OR \(2\theta + 1.326 = 2\pi + \beta' \Rightarrow \theta =\) THEN MINUS \(\pi\) |
| \(\theta =\) awrt \(1.30\) and \(0.03\) | A1 | Withhold if extra solutions in the range. Degree solution: only lose first A mark awrt the first time |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = \sqrt{1^2 + 4^2} = \sqrt{17}$ | B1 | Condone $R = \pm\sqrt{17}$ |
| $\alpha = \arctan 4$ | M1 | For $\alpha = \arctan(\pm 4)$ or $\alpha = \arctan\left(\pm\frac{1}{4}\right)$ leading to a solution of $\alpha$. Implied by $\alpha =$ awrt $76°$ or awrt 1.3 rads. If $R$ used to find $\alpha$, award only $\alpha = \arccos\left(\pm\frac{1}{R}\right)$ or $\alpha = \arcsin\left(\pm\frac{4}{R}\right)$ |
| $\alpha =$ awrt $1.326$ | A1 | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{17}\cos(2\theta - 1.326) = 1.2 \Rightarrow \cos(2\theta - 1.326) = \frac{1.2}{\sqrt{17}}$ | M1 | Using part (a), proceeding as far as $\cos(2\theta \pm \text{their } 1.326) = \frac{1.2}{\text{their } R}$. Condone slips on 1.2 and miscopying their 1.326. Note: $2\cos(\theta \pm \text{their } 1.326) = \frac{1.2}{\text{their } R}$ is M0 |
| $\Rightarrow (2\theta - 1.326) = \pm 1.275...\Rightarrow \theta = ...$ | dM1 | Dependent on first M1. Full method to find one value of $\theta$ in range $0$ to $\pi$ from principal value. Deal with "1.326" before the "2". Condone adding 1.326 instead of subtracting |
| $\theta =$ awrt $1.30$ or awrt $0.03$ | A1 | Only allow 1.3 if preceded by answer rounding to 1.30 |
| $2\theta - 1.326 = {'}1.275...{'}\ \textbf{and}\ {'-}1.275...'$ | ddM1 | Correct method for second value of $\theta$ (for their $\alpha$) in range $0$ to $\pi$. E.g. $2\theta + 1.326 = -\beta' \Rightarrow \theta =$ OR $2\theta + 1.326 = 2\pi + \beta' \Rightarrow \theta =$ THEN MINUS $\pi$ |
| $\theta =$ awrt $1.30$ **and** $0.03$ | A1 | Withhold if extra solutions in the range. Degree solution: only lose first A mark awrt the first time |
---\begin{enumerate}
\item (a) Write $\cos \theta + 4 \sin \theta$ in the form $R \cos ( \theta - \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$. Give the exact value of $R$ and give the value of $\alpha$ in radians to 3 decimal places.\\
(b) Hence solve, for $0 \leqslant \theta < \pi$, the equation
\end{enumerate}
$$\cos 2 \theta + 4 \sin 2 \theta = 1.2$$
giving your answers to 2 decimal places.
\hfill \mbox{\textit{Edexcel C34 2018 Q1 [8]}}