| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.3 This is a standard harmonic form question with straightforward application. Part (a) is routine bookwork using R cos(θ+α) = R cos α cos θ - R sin α sin θ to find R = √(5²+3²) and α = arctan(3/5). Part (b) applies this to a real context but simply requires substituting the harmonic form and solving a basic trigonometric equation. The multi-step nature and context add slight complexity, but this remains a typical C3/C4 textbook exercise requiring no novel insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05n 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 |
| \(R = \sqrt{5^2 + 3^2} = \sqrt{34}\) | B1 | \(R = \sqrt{34}\) (\(R = \pm\sqrt{34}\) is B0) |
| \(\tan\alpha = \pm\frac{3}{5}\), \(\tan\alpha = \pm\frac{5}{3} \Rightarrow \alpha = \ldots\) Also allow \(\cos\alpha = \pm\frac{5}{\sqrt{34}}\) or \(\pm\frac{3}{\sqrt{34}}\), \(\sin\alpha = \pm\frac{3}{\sqrt{34}}\) or \(\pm\frac{5}{\sqrt{34}} \Rightarrow \alpha = \ldots\) where "\(\sqrt{34}\)" is their \(R\) | M1 | Correct method for \(\alpha\) |
| \(\alpha = \arctan\left(\frac{3}{5}\right) = \text{awrt } 0.5404\) | A1 | Anything that rounds to 0.5404 (Degrees is 30.96… scores A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6 + 2.5\cos\left(\frac{4\pi t}{25}\right) - 1.5\sin\left(\frac{4\pi t}{25}\right) = 4.6 \Rightarrow \frac{\sqrt{34}}{2}\cos\left(\frac{4\pi t}{25} + 0.5404\right) = -1.4 \Rightarrow \cos\left(\frac{4\pi t}{25} + \text{"0.5404"}\right) = \ldots\) | M1 | Uses part (a) and proceeds as far as \(\cos\left(\frac{4\pi t}{25} \pm \text{their } 0.5404\right) = k\) or \(\cos\theta \pm \text{their } 0.5404 = k\) or \(\cos t \pm \text{their } 0.5404 = k\) where \( |
| \(\cos\left(\frac{4\pi t}{25} + \text{"0.5404"}\right) = -0.48\) | A1 | Allow: \(\cos\left(\frac{4\pi t}{25} \pm \text{their } 0.5404\right) = \text{awrt} -0.48\); May see \(-\frac{7\sqrt{34}}{85}\) or \(-\frac{2.8}{\sqrt{34}}\) for \(-0.48\) |
| \(\frac{4\pi t}{25} + \text{"0.5404"} = 2.07 \Rightarrow t = \ldots\) or \(\frac{4\pi t}{25} + \text{"0.5404"} = 2\pi - 2.07 = 4.21 \Rightarrow t = \ldots\) NB 2.07 may be seen as \(\pi - 1.07\) and 4.21 may be seen as \(\pi + 1.07\) | dM1 | Takes invcos then adds or subtracts their 0.5404 and applies \(\frac{4\pi t}{25}\) to obtain a value for \(t\). Dependent on previous M mark and may be implied by obtaining a value for \(t\) of awrt 3 or awrt 7 |
| awrt 3.05 or awrt 7.3 | A1 | Allow awrt 3.05 or awrt 7.3 |
| \(\frac{4\pi t}{25} \pm \text{"0.5404"} = 2\pi - 2.07 \Rightarrow t = \ldots\) and \(\frac{4\pi t}{25} \pm \text{"0.5404"} = 2.07 \Rightarrow t = \ldots\) | ddM1 | For a correct method to find a different value of \(t\) in the range. Dependent on both previous method marks |
| \(3:03\) or \(15:03\) or 3 hrs 3 min or 183 minutes and \(7:18\) or \(19:18\) or 7 hrs 18 min or 438 minutes | A1 | Both values required |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{5^2 + 3^2} = \sqrt{34}$ | B1 | $R = \sqrt{34}$ ($R = \pm\sqrt{34}$ is B0) |
| $\tan\alpha = \pm\frac{3}{5}$, $\tan\alpha = \pm\frac{5}{3} \Rightarrow \alpha = \ldots$ Also allow $\cos\alpha = \pm\frac{5}{\sqrt{34}}$ or $\pm\frac{3}{\sqrt{34}}$, $\sin\alpha = \pm\frac{3}{\sqrt{34}}$ or $\pm\frac{5}{\sqrt{34}} \Rightarrow \alpha = \ldots$ where "$\sqrt{34}$" is their $R$ | M1 | Correct method for $\alpha$ |
| $\alpha = \arctan\left(\frac{3}{5}\right) = \text{awrt } 0.5404$ | A1 | Anything that rounds to 0.5404 (Degrees is 30.96… scores A0) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 + 2.5\cos\left(\frac{4\pi t}{25}\right) - 1.5\sin\left(\frac{4\pi t}{25}\right) = 4.6 \Rightarrow \frac{\sqrt{34}}{2}\cos\left(\frac{4\pi t}{25} + 0.5404\right) = -1.4 \Rightarrow \cos\left(\frac{4\pi t}{25} + \text{"0.5404"}\right) = \ldots$ | M1 | Uses part (a) and proceeds as far as $\cos\left(\frac{4\pi t}{25} \pm \text{their } 0.5404\right) = k$ or $\cos\theta \pm \text{their } 0.5404 = k$ or $\cos t \pm \text{their } 0.5404 = k$ where $|k| < 1$ |
| $\cos\left(\frac{4\pi t}{25} + \text{"0.5404"}\right) = -0.48$ | A1 | Allow: $\cos\left(\frac{4\pi t}{25} \pm \text{their } 0.5404\right) = \text{awrt} -0.48$; May see $-\frac{7\sqrt{34}}{85}$ or $-\frac{2.8}{\sqrt{34}}$ for $-0.48$ |
| $\frac{4\pi t}{25} + \text{"0.5404"} = 2.07 \Rightarrow t = \ldots$ **or** $\frac{4\pi t}{25} + \text{"0.5404"} = 2\pi - 2.07 = 4.21 \Rightarrow t = \ldots$ NB 2.07 may be seen as $\pi - 1.07$ and 4.21 may be seen as $\pi + 1.07$ | dM1 | Takes invcos then adds or subtracts their 0.5404 and applies $\frac{4\pi t}{25}$ to obtain a value for $t$. **Dependent on previous M mark and may be implied by obtaining a value for $t$ of awrt 3 or awrt 7** |
| awrt 3.05 or awrt 7.3 | A1 | Allow awrt 3.05 **or** awrt 7.3 |
| $\frac{4\pi t}{25} \pm \text{"0.5404"} = 2\pi - 2.07 \Rightarrow t = \ldots$ **and** $\frac{4\pi t}{25} \pm \text{"0.5404"} = 2.07 \Rightarrow t = \ldots$ | ddM1 | For a correct method to find a different value of $t$ in the range. **Dependent on both previous method marks** |
| $3:03$ or $15:03$ or 3 hrs 3 min or 183 minutes **and** $7:18$ or $19:18$ or 7 hrs 18 min or 438 minutes | A1 | Both values required |
---7. (a) Express $5 \cos \theta - 3 \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$ Give the exact value of $R$ and give the value of $\alpha$, in radians, to 4 decimal places.
The height of sea water, $H$ metres, on a harbour wall is modelled by the equation
$$H = 6 + 2.5 \cos \left( \frac { 4 \pi t } { 25 } \right) - 1.5 \sin \left( \frac { 4 \pi t } { 25 } \right) , \quad 0 \leqslant t < 12$$
where $t$ is the number of hours after midday.\\
(b) Calculate the times at which the model predicts that the height of sea water on the harbour wall will be 4.6 metres. Give your answers to the nearest minute.\\
\includegraphics[max width=\textwidth, alt={}, center]{a9870c94-0910-46ec-a54a-44a431cb324e-18_2257_54_314_1977}
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C34 2019 Q7 [9]}}