| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve equation with reciprocal functions |
| Difficulty | Challenging +1.2 This is a multi-step reciprocal trig equation requiring algebraic manipulation to reach a given form, then solving a quadratic in cosec(2θ). While it involves several reciprocal functions and double angles, the path is clearly signposted in part (a), and part (b) becomes routine quadratic solving once the substitution is made. More challenging than basic trig equations but follows standard A-level techniques without requiring novel insight. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k 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 | Mark | Guidance |
| \(3\cot^2 2\theta - 7 = \frac{13}{2\sin\theta\cos\theta}\) or \(2\sin\theta\left(3(\text{cosec}^2 2\theta-1)-7\right)=13\sec\theta\) oe | M1 | Attempts EITHER: divide both sides by \(2\sin\theta\) and use \(\sec\theta=\frac{1}{\cos\theta}\) to get form \(\frac{A}{B\sin\theta\cos\theta}\); OR use \(\pm 1 \pm \cot^2 2\theta = \pm\text{cosec}^2 2\theta\) to replace \(\cot^2 2\theta\) term. |
| \(3(\text{cosec}^2 2\theta-1)-7 = \frac{13}{2\sin\theta\cos\theta}\) | dM1 | Attempts BOTH: divides by \(2\sin\theta\) AND uses \(\pm1\pm\cot^2 2\theta=\pm\text{cosec}^2 2\theta\). Dependent on M1. Condone arithmetical slips and invisible brackets. |
| \(\Rightarrow 3\text{cosec}^2 2\theta - 10 = 13\text{cosec}\, 2\theta \Rightarrow 3\text{cosec}^2 2\theta - 13\text{cosec}\,2\theta - 10 = 0\)* | ddM1 A1* | Full attempt to achieve 3TQ in \(\text{cosec}\,2\theta\) only. Must use \(\sin 2\theta = 2\sin\theta\cos\theta\) and collect terms. A1*: achieves printed answer with no arithmetical errors; must see \(\sin\theta\cos\theta\) or \(\sin 2\theta\) in working. Withhold for poor notation e.g. \(\text{cosec}^2 2\theta\) written as \(\text{cosec}\,2\theta^2\) or 2 missing from \(2\theta\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((3\text{cosec}\,2\theta+2)(\text{cosec}\,2\theta-5)=0 \Rightarrow \text{cosec}\,2\theta=5 \Rightarrow \sin 2\theta = \frac{1}{5}\) | M1 A1 | Attempts to solve quadratic and finds reciprocal of a root. Look for e.g. \(5\to\frac{1}{5}\). A1: \(\sin 2\theta = \frac{1}{5}\). Ignore reference to \(-\frac{3}{2}\); \(\frac{1}{5}\) on its own sufficient. Condone \(\sin^2 2\theta=\frac{1}{5}\) or \(\sin X=\frac{1}{5}\) or \(\sin\theta=\frac{1}{5}\). No intermediate working not acceptable. |
| \(\theta = \frac{\sin^{-1}\!\left(\frac{1}{5}\right)}{2} = \ldots\) | dM1 | Correct method: arcsin\(\left(\frac{1}{5}\right)\) divided by 2. Dependent on seeing quadratic solved or reciprocal of root stated first (either \(\geqslant1\) or \(\leqslant-1\)). Cannot score for solving e.g. \(\sin^2 2\theta=\ldots\) |
| \((\theta =)\) awrt \(0.101\), \(1.47\) | A1 | Both values, no other angles in given interval. Some intermediate working must be shown. |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\cot^2 2\theta - 7 = \frac{13}{2\sin\theta\cos\theta}$ or $2\sin\theta\left(3(\text{cosec}^2 2\theta-1)-7\right)=13\sec\theta$ oe | M1 | Attempts EITHER: divide both sides by $2\sin\theta$ and use $\sec\theta=\frac{1}{\cos\theta}$ to get form $\frac{A}{B\sin\theta\cos\theta}$; OR use $\pm 1 \pm \cot^2 2\theta = \pm\text{cosec}^2 2\theta$ to replace $\cot^2 2\theta$ term. |
| $3(\text{cosec}^2 2\theta-1)-7 = \frac{13}{2\sin\theta\cos\theta}$ | dM1 | Attempts BOTH: divides by $2\sin\theta$ AND uses $\pm1\pm\cot^2 2\theta=\pm\text{cosec}^2 2\theta$. Dependent on M1. Condone arithmetical slips and invisible brackets. |
| $\Rightarrow 3\text{cosec}^2 2\theta - 10 = 13\text{cosec}\, 2\theta \Rightarrow 3\text{cosec}^2 2\theta - 13\text{cosec}\,2\theta - 10 = 0$* | ddM1 A1* | Full attempt to achieve 3TQ in $\text{cosec}\,2\theta$ only. Must use $\sin 2\theta = 2\sin\theta\cos\theta$ and collect terms. A1*: achieves printed answer with no arithmetical errors; must see $\sin\theta\cos\theta$ or $\sin 2\theta$ in working. Withhold for poor notation e.g. $\text{cosec}^2 2\theta$ written as $\text{cosec}\,2\theta^2$ or 2 missing from $2\theta$. |
**(4 marks)**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3\text{cosec}\,2\theta+2)(\text{cosec}\,2\theta-5)=0 \Rightarrow \text{cosec}\,2\theta=5 \Rightarrow \sin 2\theta = \frac{1}{5}$ | M1 A1 | Attempts to solve quadratic and finds reciprocal of a root. Look for e.g. $5\to\frac{1}{5}$. A1: $\sin 2\theta = \frac{1}{5}$. Ignore reference to $-\frac{3}{2}$; $\frac{1}{5}$ on its own sufficient. Condone $\sin^2 2\theta=\frac{1}{5}$ or $\sin X=\frac{1}{5}$ or $\sin\theta=\frac{1}{5}$. **No intermediate working not acceptable.** |
| $\theta = \frac{\sin^{-1}\!\left(\frac{1}{5}\right)}{2} = \ldots$ | dM1 | Correct method: arcsin$\left(\frac{1}{5}\right)$ divided by 2. Dependent on seeing quadratic solved or reciprocal of root stated first (either $\geqslant1$ or $\leqslant-1$). Cannot score for solving e.g. $\sin^2 2\theta=\ldots$ |
| $(\theta =)$ awrt $0.101$, $1.47$ | A1 | Both values, no other angles in given interval. Some intermediate working must be shown. |
**(4 marks)**\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
(a) Show that the equation
$$2 \sin \theta \left( 3 \cot ^ { 2 } 2 \theta - 7 \right) = 13 \sec \theta$$
can be written as
$$3 \operatorname { cosec } ^ { 2 } 2 \theta - 13 \operatorname { cosec } 2 \theta - 10 = 0$$
(b) Hence solve, for $0 < \theta < \frac { \pi } { 2 }$, the equation
$$2 \sin \theta \left( 3 \cot ^ { 2 } 2 \theta - 7 \right) = 13 \sec \theta$$
giving your answers to 3 significant figures.
\hfill \mbox{\textit{Edexcel P3 2022 Q7 [8]}}