| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Solve equation with tan(θ ± α) |
| Difficulty | Standard +0.3 This is a straightforward application of the tan addition formula followed by algebraic manipulation to reach a given form, then solving a quadratic in tan θ. Part (a) is guided ('show that'), and part (b) requires only solving a quadratic and selecting solutions in the given range. Slightly easier than average due to the scaffolding and routine techniques involved. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct \(\tan(A+B)\) formula and obtain an equation in \(\tan\theta\) | M1 | e.g. \(\dfrac{\tan\theta + \tan 60°}{1 - \tan\theta\tan 60°} = \dfrac{2}{\tan\theta}\) |
| Use \(\tan 60° = \sqrt{3}\) and obtain a correct horizontal equation in any form | A1 | e.g. \(\tan\theta(\tan\theta + \sqrt{3}) = 2(1 - \sqrt{3}\tan\theta)\) |
| Reduce to \(\tan^2\theta + 3\sqrt{3}\tan\theta - 2 = 0\) correctly | A1 | AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve the given quadratic to obtain a value for \(\theta\) | M1 | \(\tan\theta = \dfrac{-3\sqrt{3} \pm \sqrt{35}}{2} = 0.3599,\ -5.556\) |
| Obtain one correct answer e.g. \(\theta = 19.8°\) | A1 | Accept 1 d.p. or better. If over-specified must be correct: 19.797…, 100.2029… |
| Obtain second correct answer \(\theta = 100.2°\) and no others in the given interval | A1 | Ignore answers outside the given interval. |
| 3 |
## Question 4:
### Part 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct $\tan(A+B)$ formula and obtain an equation in $\tan\theta$ | M1 | e.g. $\dfrac{\tan\theta + \tan 60°}{1 - \tan\theta\tan 60°} = \dfrac{2}{\tan\theta}$ |
| Use $\tan 60° = \sqrt{3}$ and obtain a correct horizontal equation in any form | A1 | e.g. $\tan\theta(\tan\theta + \sqrt{3}) = 2(1 - \sqrt{3}\tan\theta)$ |
| Reduce to $\tan^2\theta + 3\sqrt{3}\tan\theta - 2 = 0$ correctly | A1 | AG |
| | **3** | |
### Part 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve the given quadratic to obtain a value for $\theta$ | M1 | $\tan\theta = \dfrac{-3\sqrt{3} \pm \sqrt{35}}{2} = 0.3599,\ -5.556$ |
| Obtain one correct answer e.g. $\theta = 19.8°$ | A1 | Accept 1 d.p. or better. If over-specified must be correct: 19.797…, 100.2029… |
| Obtain second correct answer $\theta = 100.2°$ and no others in the given interval | A1 | Ignore answers outside the given interval. |
| | **3** | |
---4
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\tan \left( \theta + 60 ^ { \circ } \right) = 2 \cot \theta$ can be written in the form
$$\tan ^ { 2 } \theta + 3 \sqrt { 3 } \tan \theta - 2 = 0$$
\item Hence solve the equation $\tan \left( \theta + 60 ^ { \circ } \right) = 2 \cot \theta$, for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q4 [6]}}