| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.8 This is a multi-step algebraic manipulation requiring combining complex fractions with trigonometric expressions, followed by solving a resulting equation. The algebraic manipulation is non-trivial (finding common denominator, expanding, simplifying using Pythagorean identity), and the 'hence' part requires recognizing how to use the proven identity. More demanding than standard P1 questions but not requiring exceptional insight. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(\sin\theta + 2\cos\theta)(\cos\theta + 2\sin\theta) - (\sin\theta - 2\cos\theta)(\cos\theta - 2\sin\theta)}{(\cos\theta - 2\sin\theta)(\cos\theta + 2\sin\theta)}\) | *M1 | Obtain an expression with a common denominator |
| \(\frac{5\sin\theta\cos\theta + 2\cos^2\theta + 2\sin^2\theta - (5\sin\theta\cos\theta - 2\sin^2\theta - 2\cos^2\theta)}{\cos^2\theta - 4\sin^2\theta} = \frac{4(\cos^2\theta + \sin^2\theta)}{\cos^2\theta - 4\sin^2\theta}\) | A1 | |
| \(\frac{4}{\cos^2\theta - 4(1-\cos^2\theta)}\) | DM1 | Use \(\cos^2\theta + \sin^2\theta = 1\) twice |
| \(\frac{4}{5\cos^2\theta - 4}\) | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4}{5\cos^2\theta - 4} = 5\) leading to \(25\cos^2\theta = 24\), leading to \(\cos\theta = \sqrt{\frac{24}{25}}\ [= (\pm)0.9798]\) | M1 | Make \(\cos\theta\) the subject |
| \(\theta = 11.5°\) or \(168.5°\) | A1, A1 FT | FT on \(180° - \) 1st solution |
| 3 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(\sin\theta + 2\cos\theta)(\cos\theta + 2\sin\theta) - (\sin\theta - 2\cos\theta)(\cos\theta - 2\sin\theta)}{(\cos\theta - 2\sin\theta)(\cos\theta + 2\sin\theta)}$ | *M1 | Obtain an expression with a common denominator |
| $\frac{5\sin\theta\cos\theta + 2\cos^2\theta + 2\sin^2\theta - (5\sin\theta\cos\theta - 2\sin^2\theta - 2\cos^2\theta)}{\cos^2\theta - 4\sin^2\theta} = \frac{4(\cos^2\theta + \sin^2\theta)}{\cos^2\theta - 4\sin^2\theta}$ | A1 | |
| $\frac{4}{\cos^2\theta - 4(1-\cos^2\theta)}$ | DM1 | Use $\cos^2\theta + \sin^2\theta = 1$ twice |
| $\frac{4}{5\cos^2\theta - 4}$ | A1 | AG |
| | **4** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4}{5\cos^2\theta - 4} = 5$ leading to $25\cos^2\theta = 24$, leading to $\cos\theta = \sqrt{\frac{24}{25}}\ [= (\pm)0.9798]$ | M1 | Make $\cos\theta$ the subject |
| $\theta = 11.5°$ or $168.5°$ | A1, A1 FT | FT on $180° - $ 1st solution |
| | **3** | |7
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } \equiv \frac { 4 } { 5 \cos ^ { 2 } \theta - 4 }$.
\item Hence solve the equation $\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } = 5$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q7 [7]}}