| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Applied context with trigonometry |
| Difficulty | Standard +0.3 This is a standard C3 question on expressing trigonometric expressions in R cos(θ-α) form and finding maxima/minima. Part (a) is routine application of the addition formula, parts (b-c) use standard optimization techniques, and part (d) requires solving a trigonometric equation. While multi-part with an applied context, each step follows textbook procedures without requiring novel insight or complex problem-solving. |
| Spec | 1.02z Models in context: use functions in modelling1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = \sqrt{7^2 + 24^2} = 25\) | B1 | Accept 25.0 but not \(\sqrt{625}\) or values not exactly 25 |
| \(\tan\alpha = \dfrac{24}{7} \Rightarrow \alpha = \text{awrt } 73.74°\) | M1A1 | For \(\tan\alpha = \pm\dfrac{24}{7}\) or \(\pm\dfrac{7}{24}\). Must be in degrees (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum of \(24\sin x + 7\cos x = 25\), so \(V_{\min} = \dfrac{21}{25} = (0.84)\) | M1A1 | Calculates \(V = \dfrac{21}{\text{their } R}\), NOT R. Correct answer \(\dfrac{21}{25}\), accept 0.84 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(AB = \dfrac{7}{\sin\theta}\), with \(\theta = \alpha\) | M1, B1 | |
| Distance \(= 7.29\text{m}\) \(\left(= \dfrac{175}{24}\text{ m}\right)\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\cos(\theta - \alpha) = \dfrac{21}{1.68} \Rightarrow \cos(\theta - \alpha) = 0.5\) | M1, A1 | |
| \(\theta - \alpha = 60° \Rightarrow \theta = \ldots\,;\quad \theta - \alpha = -60° \Rightarrow \theta = \ldots\) | dM1, dM1 | |
| \(\theta = \text{awrt } 133.7°,\ 13.7°\) | A1, A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Uses \(\sin\theta = \dfrac{7}{AB}\) with a numerical \(\theta\) to find \(AB = \ldots\) | M1 | |
| Uses \(\theta =\) their value of \(\alpha\) in a trig calculation involving sin | B1 | \(\sin\alpha = \dfrac{AB}{7}\) is condoned |
| Obtains \(\dfrac{175}{24}\) or awrt \(7.29\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Substitutes \(V = 1.68\) and answer to part (a) in \(V = \dfrac{21}{24\sin\theta + 7\cos\theta}\) to get equation of form \(R\cos(\theta \pm \alpha) = \dfrac{21}{1.68}\) or \(1.68R\cos(\theta \pm \alpha) = 21\) or \(\cos(\theta \pm \alpha) = \dfrac{21}{1.68R}\) | M1 | Follow through on their \(R\) and \(\alpha\) |
| Obtains \(\cos(\theta \pm \alpha) = 0.5\) oe | A1 | Follow through on their \(\alpha\); may be implied by later working |
| Obtains one value of \(\theta\) in the range \(0 < \theta < 150\) from inverse cos + their \(\alpha\) | dM1 | Dependent on first M1 being scored |
| Obtains second value of \(\theta\) in the range \(0 < \theta < 150\) from inverse cos + their \(\alpha\) | dM1 | Dependent on first M1 being scored |
| One correct answer awrt \(\theta = 133.7\) or \(13.7\) to 1dp | A1 | |
| Both correct answers awrt \(\theta = 133.7\) and \(13.7\) to 1dp | A1 | Extra solutions in range loses last A1; answers in radians lose mark first time it occurs; answers must be to 3dp |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{7^2 + 24^2} = 25$ | B1 | Accept 25.0 but not $\sqrt{625}$ or values not exactly 25 |
| $\tan\alpha = \dfrac{24}{7} \Rightarrow \alpha = \text{awrt } 73.74°$ | M1A1 | For $\tan\alpha = \pm\dfrac{24}{7}$ or $\pm\dfrac{7}{24}$. Must be in degrees **(3)** |
---
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum of $24\sin x + 7\cos x = 25$, so $V_{\min} = \dfrac{21}{25} = (0.84)$ | M1A1 | Calculates $V = \dfrac{21}{\text{their } R}$, NOT R. Correct answer $\dfrac{21}{25}$, accept 0.84 **(2)** |
---
# Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $AB = \dfrac{7}{\sin\theta}$, with $\theta = \alpha$ | M1, B1 | |
| Distance $= 7.29\text{m}$ $\left(= \dfrac{175}{24}\text{ m}\right)$ | A1 | **(3)** |
---
# Question 8(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\cos(\theta - \alpha) = \dfrac{21}{1.68} \Rightarrow \cos(\theta - \alpha) = 0.5$ | M1, A1 | |
| $\theta - \alpha = 60° \Rightarrow \theta = \ldots\,;\quad \theta - \alpha = -60° \Rightarrow \theta = \ldots$ | dM1, dM1 | |
| $\theta = \text{awrt } 133.7°,\ 13.7°$ | A1, A1 | **(6)** |
## Question 8 (continued):
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Uses $\sin\theta = \dfrac{7}{AB}$ with a numerical $\theta$ to find $AB = \ldots$ | M1 | |
| Uses $\theta =$ their value of $\alpha$ in a trig calculation involving sin | B1 | $\sin\alpha = \dfrac{AB}{7}$ is condoned |
| Obtains $\dfrac{175}{24}$ or awrt $7.29$ | A1 | |
---
### Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Substitutes $V = 1.68$ and answer to part (a) in $V = \dfrac{21}{24\sin\theta + 7\cos\theta}$ to get equation of form $R\cos(\theta \pm \alpha) = \dfrac{21}{1.68}$ or $1.68R\cos(\theta \pm \alpha) = 21$ or $\cos(\theta \pm \alpha) = \dfrac{21}{1.68R}$ | M1 | Follow through on their $R$ and $\alpha$ |
| Obtains $\cos(\theta \pm \alpha) = 0.5$ oe | A1 | Follow through on their $\alpha$; may be implied by later working |
| Obtains one value of $\theta$ **in the range** $0 < \theta < 150$ from inverse cos + their $\alpha$ | dM1 | Dependent on first M1 being scored |
| Obtains second value of $\theta$ **in the range** $0 < \theta < 150$ from inverse cos + their $\alpha$ | dM1 | Dependent on first M1 being scored |
| One correct answer awrt $\theta = 133.7$ or $13.7$ to 1dp | A1 | |
| Both correct answers awrt $\theta = 133.7$ **and** $13.7$ to 1dp | A1 | Extra solutions in range loses last A1; answers in radians lose mark first time it occurs; answers must be to 3dp |
> For info: $\alpha = 1.287$, $\theta_1 = 2.334$, $\theta_2 = 0.240$8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e29d66c-c3c6-4e4b-acfb-c73c60604d93-11_453_1225_255_369}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Kate crosses a road, of constant width 7 m , in order to take a photograph of a marathon runner, John, approaching at $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
Kate is 24 m ahead of John when she starts to cross the road from the fixed point $A$. John passes her as she reaches the other side of the road at a variable point $B$, as shown in Figure 2.\\
Kate's speed is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and she moves in a straight line, which makes an angle $\theta$, $0 < \theta < 150 ^ { \circ }$, with the edge of the road, as shown in Figure 2.
You may assume that $V$ is given by the formula
$$V = \frac { 21 } { 24 \sin \theta + 7 \cos \theta } , \quad 0 < \theta < 150 ^ { \circ }$$
\begin{enumerate}[label=(\alph*)]
\item Express $24 \sin \theta + 7 \cos \theta$ in the form $R \cos ( \theta - \alpha )$, where $R$ and $\alpha$ are constants and where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$, giving the value of $\alpha$ to 2 decimal places.
Given that $\theta$ varies,
\item find the minimum value of $V$.
Given that Kate's speed has the value found in part (b),
\item find the distance $A B$.
Given instead that Kate's speed is $1.68 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item find the two possible values of the angle $\theta$, given that $0 < \theta < 150 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q8 [14]}}