| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Deduce related equation solution |
| Difficulty | Standard +0.8 This is a multi-part harmonic form question requiring: (a) standard R-α conversion with exact and decimal answers, (b) solving a rational equation using the harmonic form, and (c) a deduction step requiring insight about the relationship between sin θ and -sin θ solutions. Part (c) elevates this above routine C3 questions as students must recognize the transformation and apply it correctly, though the individual techniques are standard. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.05n 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 | Mark | Guidance |
| \(R = \sqrt{5}\) | B1 | Condone \(R = \pm\sqrt{5}\). Ignore decimals |
| \(\tan\alpha = \frac{1}{2} \Rightarrow \alpha = 26.57°\) | M1A1 | \(\tan\alpha = \pm\frac{1}{2}\), \(\tan\alpha = \pm\frac{2}{1} \Rightarrow \alpha = \ldots\) If \(R\) used to find \(\alpha\), only accept \(\cos\alpha = \pm\frac{2}{R}\) OR \(\sin\alpha = \pm\frac{1}{R}\). A1: \(\alpha =\) awrt \(26.57°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2}{2\cos\theta - \sin\theta - 1} = 15 \Rightarrow \frac{2}{\sqrt{5}\cos(\theta + 26.6°) - 1} = 15\) \(\Rightarrow \cos(\theta + 26.6°) = \frac{17}{15\sqrt{5}}\) (awrt 0.507) | M1A1 | Attempts to use part (a) \(\Rightarrow \cos(\theta \pm \text{their } 26.6°) = K\), \( |
| \(\theta + 26.57° = 59.54° \Rightarrow \theta =\) awrt \(33.0°\) or awrt \(273.9°\) | A1 | One solution correct: \(\theta =\) awrt \(33.0°\) or \(\theta =\) awrt \(273.9°\). Do not accept 33 for 33.0 |
| \(\theta + 26.6° = 360° - \text{their } 59.5°\) | dM1 | Obtains second solution in range. Dependent on having scored previous M |
| \(\Rightarrow \theta =\) awrt \(273.9°\) and awrt \(33.0°\) | A1 | Both solutions \(\theta =\) awrt \(33.0°\) and awrt \(273.9°\). Extra solutions inside range withhold this A1. Ignore solutions outside range \(0 \leq \theta < 360°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\theta - \text{their } 26.57° = \text{their } 59.54° \Rightarrow \theta = \ldots\) | M1 | Alternatively \(-\theta + \text{their } 26.6° = -\text{their } 59.5° \Rightarrow \theta = \ldots\) If incorrect sign for \(\alpha\) used in (b), e.g. \(\cos(\theta - 26.57°)\), score for \(\theta + \text{their } 26.57° = \text{their } 59.54°\) |
| \(\theta =\) awrt \(86.1°\) | A1 | awrt \(86.1°\) ONLY. Allow both marks following correct (a) and (b). Do not award if 86.1 was their smallest answer in (b) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{5}$ | B1 | Condone $R = \pm\sqrt{5}$. Ignore decimals |
| $\tan\alpha = \frac{1}{2} \Rightarrow \alpha = 26.57°$ | M1A1 | $\tan\alpha = \pm\frac{1}{2}$, $\tan\alpha = \pm\frac{2}{1} \Rightarrow \alpha = \ldots$ If $R$ used to find $\alpha$, only accept $\cos\alpha = \pm\frac{2}{R}$ OR $\sin\alpha = \pm\frac{1}{R}$. A1: $\alpha =$ awrt $26.57°$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{2\cos\theta - \sin\theta - 1} = 15 \Rightarrow \frac{2}{\sqrt{5}\cos(\theta + 26.6°) - 1} = 15$ $\Rightarrow \cos(\theta + 26.6°) = \frac{17}{15\sqrt{5}}$ (awrt 0.507) | M1A1 | Attempts to use part (a) $\Rightarrow \cos(\theta \pm \text{their } 26.6°) = K$, $|K| \leq 1$. A1: $\cos(\theta \pm \text{their } 26.6°) = \frac{17}{15\sqrt{5}}$ (awrt 0.507), can be implied by $(\theta \pm \text{their } 26.6°) =$ awrt $59.5°/59.6°$ |
| $\theta + 26.57° = 59.54° \Rightarrow \theta =$ awrt $33.0°$ or awrt $273.9°$ | A1 | One solution correct: $\theta =$ awrt $33.0°$ or $\theta =$ awrt $273.9°$. Do not accept 33 for 33.0 |
| $\theta + 26.6° = 360° - \text{their } 59.5°$ | dM1 | Obtains second solution in range. Dependent on having scored previous M |
| $\Rightarrow \theta =$ awrt $273.9°$ and awrt $33.0°$ | A1 | Both solutions $\theta =$ awrt $33.0°$ and awrt $273.9°$. Extra solutions inside range withhold this A1. Ignore solutions outside range $0 \leq \theta < 360°$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\theta - \text{their } 26.57° = \text{their } 59.54° \Rightarrow \theta = \ldots$ | M1 | Alternatively $-\theta + \text{their } 26.6° = -\text{their } 59.5° \Rightarrow \theta = \ldots$ If incorrect sign for $\alpha$ used in (b), e.g. $\cos(\theta - 26.57°)$, score for $\theta + \text{their } 26.57° = \text{their } 59.54°$ |
| $\theta =$ awrt $86.1°$ | A1 | awrt $86.1°$ ONLY. Allow both marks following correct (a) and (b). Do not award if 86.1 was their smallest answer in (b) |
*Note: Answers in radians — withhold only one A mark, the first time a solution in radians appears. FYI: (a) $\alpha = 0.46$ (b) $\theta_1 =$ awrt 0.58 and $\theta_2 =$ awrt 4.78 (c) $\theta_3 =$ awrt 1.50. Require 2 dp accuracy.*
---\begin{enumerate}
\item (a) Express $2 \cos \theta - \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90 ^ { \circ }$. Give the exact value of $R$ and give the value of $\alpha$ to 2 decimal places.\\
(b) Hence solve, for $0 \leqslant \theta < 360 ^ { \circ }$,
\end{enumerate}
$$\frac { 2 } { 2 \cos \theta - \sin \theta - 1 } = 15$$
Give your answers to one decimal place.\\
(c) Use your solutions to parts (a) and (b) to deduce the smallest positive value of $\theta$ for which
$$\frac { 2 } { 2 \cos \theta + \sin \theta - 1 } = 15$$
Give your answer to one decimal place.
\hfill \mbox{\textit{Edexcel C3 2016 Q3 [10]}}