| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Trig equation from real-world model |
| Difficulty | Moderate -0.8 This is a straightforward application of trigonometric functions requiring direct substitution for part (a) and solving a basic sine equation for part (b). The equation is already set up, requiring only algebraic manipulation and calculator work with no conceptual challenges beyond standard C2 content. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H = 10 + 5\sin\!\left(\frac{\pi(1)}{6}\right) = 12.5\) * | B1 | 12.5 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9 = 10 + 5\sin\!\left(\frac{\pi t}{6}\right) \Rightarrow 5\sin\!\left(\frac{\pi t}{6}\right) = -1\) | M1 | Proceed to \(5\sin\!\left(\frac{\pi t}{6}\right) = k\). May be implied by e.g. \(\sin\!\left(\frac{\pi t}{6}\right) = -\frac{1}{5}\) |
| \(\sin\!\left(\frac{\pi t}{6}\right) = -\frac{1}{5} \Rightarrow \frac{\pi t}{6} = \arcsin\!\left(\pm\frac{1}{5}\right)\) | M1 | \(\arcsin\!\left(\pm\frac{k}{5}\right)\) |
| \(\alpha = \pm 0.2013\ldots\) (or \(11.536\ldots\) degrees) | B1 | May be implied. Given similarity between \(-\frac{1}{5}\) and \(\arcsin\!\left(-\frac{1}{5}\right)\) allow \(\alpha = \text{awrt } \pm 0.2\) |
| \(\frac{\pi t}{6} = \pi + 0.201\ldots\) or \(\frac{\pi t}{6} = 2\pi - 0.201\ldots\); giving \(t = 6.384565\ldots\) or \(11.615434\ldots\) | M1 | May be implied. Do not allow mixing of degrees and radians but allow working in just degrees |
| \(t = 0623, 1137\) | A1, A1 | Accept 6hrs 23mins, 11hrs 37mins or 5hrs 37mins, 23 mins before midday |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H = 10 + 5\sin\!\left(\frac{\pi(1)}{6}\right) = 12.5$ * | B1 | 12.5 oe |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9 = 10 + 5\sin\!\left(\frac{\pi t}{6}\right) \Rightarrow 5\sin\!\left(\frac{\pi t}{6}\right) = -1$ | M1 | Proceed to $5\sin\!\left(\frac{\pi t}{6}\right) = k$. May be implied by e.g. $\sin\!\left(\frac{\pi t}{6}\right) = -\frac{1}{5}$ |
| $\sin\!\left(\frac{\pi t}{6}\right) = -\frac{1}{5} \Rightarrow \frac{\pi t}{6} = \arcsin\!\left(\pm\frac{1}{5}\right)$ | M1 | $\arcsin\!\left(\pm\frac{k}{5}\right)$ |
| $\alpha = \pm 0.2013\ldots$ (or $11.536\ldots$ degrees) | B1 | May be implied. Given similarity between $-\frac{1}{5}$ and $\arcsin\!\left(-\frac{1}{5}\right)$ allow $\alpha = \text{awrt } \pm 0.2$ |
| $\frac{\pi t}{6} = \pi + 0.201\ldots$ or $\frac{\pi t}{6} = 2\pi - 0.201\ldots$; giving $t = 6.384565\ldots$ or $11.615434\ldots$ | M1 | May be implied. Do not allow mixing of degrees and radians but allow working in just degrees |
| $t = 0623, 1137$ | A1, A1 | Accept 6hrs 23mins, 11hrs 37mins or 5hrs 37mins, 23 mins before midday |
---5. The height of water, $H$ metres, in a harbour on a particular day is given by the equation
$$H = 10 + 5 \sin \left( \frac { \pi t } { 6 } \right) , \quad 0 \leqslant t < 24$$
where $t$ is the number of hours after midnight.
\begin{enumerate}[label=(\alph*)]
\item Show that the height of the water 1 hour after midnight is 12.5 metres.
\item Find, to the nearest minute, the times before midday when the height of the water is 9 metres.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q5 [7]}}