| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Equation with half angles |
| Difficulty | Standard +0.8 This FP1 question requires applying the half-angle substitution t = tan(x/6) to transform a trigonometric equation, which is a non-standard technique requiring careful algebraic manipulation of double angle formulae. Part (c) involves solving a quadratic and inverse trigonometric work. While systematic, it demands more sophistication than typical A-level questions and tests understanding beyond routine application. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x=0 \Rightarrow D=2\sin(0)+3\cos(0)+6=6+3=9\text{ m}\) | B1 | Correct depth of 9m (must include units) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(D=2\left(\frac{2t}{1+t^2}\right)+3\left(\frac{1-t^2}{1+t^2}\right)+6\) | M1 | Uses correct formulae to obtain \(D\) in terms of \(t\) |
| \(=\frac{4t+3-3t^2+6+6t^2}{1+t^2}\) | M1 | Correct method to obtain common denominator |
| \(=\frac{3t^2+4t+9}{1+t^2}\) * | A1* | Collects terms and simplifies to printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{3t^2+4t+9}{1+t^2}=5 \Rightarrow 3t^2+4t+9=5+5t^2\) | M1 | Uses \(D=5\) with model and multiplies up to obtain quadratic in \(t\) |
| \(t^2-2t-2=0\) | A1 | Correct 3TQ |
| \(t=\frac{2\pm\sqrt{4+8}}{2} \Rightarrow \frac{x}{6}=\tan^{-1}(1+\sqrt{3})\) or \(\frac{x}{6}=\tan^{-1}(1-\sqrt{3})\) | M1 | Solves 3TQ in \(t\), obtains values of \(\frac{x}{6}\) as suggested by model |
| \(\frac{x}{6}=\tan^{-1}(1+\sqrt{3})=1.21...\Rightarrow x=...\) | M1 | Fully correct strategy to identify required value of \(x\) from positive root |
| 0719 or 07:19 am | A1 | Allow e.g. 439 minutes after midnight |
## Question 3:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0 \Rightarrow D=2\sin(0)+3\cos(0)+6=6+3=9\text{ m}$ | B1 | Correct depth of 9m (must include units) |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $D=2\left(\frac{2t}{1+t^2}\right)+3\left(\frac{1-t^2}{1+t^2}\right)+6$ | M1 | Uses correct formulae to obtain $D$ in terms of $t$ |
| $=\frac{4t+3-3t^2+6+6t^2}{1+t^2}$ | M1 | Correct method to obtain common denominator |
| $=\frac{3t^2+4t+9}{1+t^2}$ * | A1* | Collects terms and simplifies to printed answer with no errors |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{3t^2+4t+9}{1+t^2}=5 \Rightarrow 3t^2+4t+9=5+5t^2$ | M1 | Uses $D=5$ with model and multiplies up to obtain quadratic in $t$ |
| $t^2-2t-2=0$ | A1 | Correct 3TQ |
| $t=\frac{2\pm\sqrt{4+8}}{2} \Rightarrow \frac{x}{6}=\tan^{-1}(1+\sqrt{3})$ or $\frac{x}{6}=\tan^{-1}(1-\sqrt{3})$ | M1 | Solves 3TQ in $t$, obtains values of $\frac{x}{6}$ as suggested by model |
| $\frac{x}{6}=\tan^{-1}(1+\sqrt{3})=1.21...\Rightarrow x=...$ | M1 | Fully correct strategy to identify required value of $x$ from positive root |
| 0719 or 07:19 am | A1 | Allow e.g. 439 minutes after midnight |
---\begin{enumerate}
\item On a particular day, the depth of water in a river estuary at a specific location is modelled by the equation
\end{enumerate}
$$D = 2 \sin \left( \frac { x } { 3 } \right) + 3 \cos \left( \frac { x } { 3 } \right) + 6 \quad 0 \leqslant x \leqslant 7 \pi$$
where the depth of water is $D$ metres at time $x$ hours after midnight on that day.\\
(a) Write down the depth of water at midnight, according to the model.
Using the substitution $t = \tan \left( \frac { x } { 6 } \right)$\\
(b) show that equation (I) can be re-written as
$$D = \frac { 3 t ^ { 2 } + 4 t + 9 } { 1 + t ^ { 2 } }$$
(c) Hence determine, according to the model, the time after midnight when the depth of water is 5 metres for the first time. Give your answer to the nearest minute.
\hfill \mbox{\textit{Edexcel FP1 AS 2021 Q3 [9]}}