| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Solve with multiple compound angles |
| Difficulty | Standard +0.3 This question requires systematic application of compound angle formulae (sin(A±B) and cos(A±B)) followed by simplification and solving tan x = constant. While it involves multiple steps, the techniques are standard P3 material with no novel insight required—students apply memorized formulae, simplify using exact values (sin π/6, cos π/3), and solve a basic trigonometric equation. Slightly easier than average due to the mechanical nature of the approach. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct trig formulae and obtain an equation in \(\sin x\) and \(\cos x\) | *M1 | Allow one sign error |
| Obtain a correct equation in any form | A1 | e.g. \(2\cos x\sin\frac{\pi}{6} = -2\sin x\sin\frac{\pi}{3}\) |
| Substitute exact trig ratios and obtain an expression for \(\tan x\) | DM1 | Allow one sign error |
| Obtain answer \(\tan x = -\dfrac{1}{\sqrt{3}}\) | A1 | Or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain answer, e.g. \(x = \dfrac{5\pi}{6}\) | B1 | |
| Obtain second answer, e.g. \(x = \dfrac{11\pi}{6}\) and no others in the interval | B1FT | FT first answer \(+ \pi\) (provided \(0 \leq\) first answer \(\leq \pi\)). Or FT first answer \(- \pi\) (provided \(\pi \leq\) first answer \(\leq 2\pi\)). Ignore any answers outside interval. |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct trig formulae and obtain an equation in $\sin x$ and $\cos x$ | *M1 | Allow one sign error |
| Obtain a correct equation in any form | A1 | e.g. $2\cos x\sin\frac{\pi}{6} = -2\sin x\sin\frac{\pi}{3}$ |
| Substitute exact trig ratios and obtain an expression for $\tan x$ | DM1 | Allow one sign error |
| Obtain answer $\tan x = -\dfrac{1}{\sqrt{3}}$ | A1 | Or exact equivalent |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain answer, e.g. $x = \dfrac{5\pi}{6}$ | B1 | |
| Obtain second answer, e.g. $x = \dfrac{11\pi}{6}$ and no others in the interval | B1FT | FT first answer $+ \pi$ (provided $0 \leq$ first answer $\leq \pi$). Or FT first answer $- \pi$ (provided $\pi \leq$ first answer $\leq 2\pi$). Ignore any answers outside interval. |
---5
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\sin \left( x + \frac { 1 } { 6 } \pi \right) - \sin \left( x - \frac { 1 } { 6 } \pi \right) = \cos \left( x + \frac { 1 } { 3 } \pi \right) - \cos \left( x - \frac { 1 } { 3 } \pi \right)$$
find the exact value of $\tan x$.
\item Hence find the exact roots of the equation
$$\sin \left( x + \frac { 1 } { 6 } \pi \right) - \sin \left( x - \frac { 1 } { 6 } \pi \right) = \cos \left( x + \frac { 1 } { 3 } \pi \right) - \cos \left( x - \frac { 1 } { 3 } \pi \right)$$
for $0 \leqslant x \leqslant 2 \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q5 [6]}}