| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.8 This is a multi-part question requiring manipulation of double angle formulas with reciprocal trig functions, algebraic proof, equation solving with substitution, and integration. Part (a) requires systematic use of double angle identities and algebraic manipulation to prove an identity. Part (b) involves substituting the proven identity and solving a trigonometric equation. Part (c) requires recognizing that the integrand simplifies to cosec x ยท cot x and integrating to get an exact answer. While each individual step uses standard techniques, the question requires sustained reasoning across multiple parts and careful algebraic manipulation throughout, placing it moderately above average difficulty. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\cos 2x}{\sin x} + \frac{\sin 2x}{\cos x} = \frac{\cos 2x \cos x + \sin 2x \sin x}{\sin x \cos x}\) | B1 | Applying at least one correct double/compound angle identity, or forming correct single fraction |
| \(= \frac{1-2\sin^2 x}{\sin x} + \frac{2\sin x \cos x}{\cos x}\) | M1 | Correct overall strategy applying double angle identities to reduce to single angle arguments |
| \(= \frac{1}{\sin x} - \frac{2\sin^2 x}{\sin x} + 2\sin x = \frac{1}{\sin x} = \cosec x\) | A1* | Fully correct proof; must see \(\frac{1}{\sin x} \to \cosec x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cosec^2\theta = 6\cot\theta - 4 \Rightarrow 1 + \cot^2\theta = 6\cot\theta - 4\) | M1 | Correctly applies result of (a) and uses relevant identities to produce equation in \(\cot\theta\) only |
| \(\cot^2\theta - 6\cot\theta + 5 = 0\) | A1 | Correct quadratic \(\cot^2\theta - 6\cot\theta + 5 = 0\) or \(5\tan^2\theta - 6\tan\theta + 1 = 0\) |
| \(\tan\theta = \frac{1}{5}, 1\) | dM1 | Attempts to solve quadratic for at least one value of trig term |
| \(\theta = 0.197, \frac{\pi}{4}\) | A1, A1 | Each correct value; accept awrt 0.197; both values and no others in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_{\pi/6}^{\pi/4} \cosec x \cot x \, dx = \left[-\cosec x\right]_{\pi/6}^{\pi/4}\) | M1 | Uses part (a) and achieves \(\pm k\cosec x\); may arise from longer methods |
| \(= 2 - \sqrt{2}\) | A1 | Must have scored M1; answer only with no integration shown is M0A0 |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\cos 2x}{\sin x} + \frac{\sin 2x}{\cos x} = \frac{\cos 2x \cos x + \sin 2x \sin x}{\sin x \cos x}$ | B1 | Applying at least one correct double/compound angle identity, or forming correct single fraction |
| $= \frac{1-2\sin^2 x}{\sin x} + \frac{2\sin x \cos x}{\cos x}$ | M1 | Correct overall strategy applying double angle identities to reduce to single angle arguments |
| $= \frac{1}{\sin x} - \frac{2\sin^2 x}{\sin x} + 2\sin x = \frac{1}{\sin x} = \cosec x$ | A1* | Fully correct proof; must see $\frac{1}{\sin x} \to \cosec x$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cosec^2\theta = 6\cot\theta - 4 \Rightarrow 1 + \cot^2\theta = 6\cot\theta - 4$ | M1 | Correctly applies result of (a) and uses relevant identities to produce equation in $\cot\theta$ only |
| $\cot^2\theta - 6\cot\theta + 5 = 0$ | A1 | Correct quadratic $\cot^2\theta - 6\cot\theta + 5 = 0$ or $5\tan^2\theta - 6\tan\theta + 1 = 0$ |
| $\tan\theta = \frac{1}{5}, 1$ | dM1 | Attempts to solve quadratic for at least one value of trig term |
| $\theta = 0.197, \frac{\pi}{4}$ | A1, A1 | Each correct value; accept awrt 0.197; both values and no others in range |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_{\pi/6}^{\pi/4} \cosec x \cot x \, dx = \left[-\cosec x\right]_{\pi/6}^{\pi/4}$ | M1 | Uses part (a) and achieves $\pm k\cosec x$; may arise from longer methods |
| $= 2 - \sqrt{2}$ | A1 | Must have scored M1; answer only with no integration shown is M0A0 |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Show that
$$\frac { \cos 2 x } { \sin x } + \frac { \sin 2 x } { \cos x } \equiv \operatorname { cosec } x \quad x \neq \frac { n \pi } { 2 } \quad n \in \mathbb { Z }$$
(b) Hence solve, for $0 < \theta < \frac { \pi } { 2 }$
$$\left( \frac { \cos 2 \theta } { \sin \theta } + \frac { \sin 2 \theta } { \cos \theta } \right) ^ { 2 } = 6 \cot \theta - 4$$
giving your answers to 3 significant figures as appropriate.\\
(c) Using the result from part (a), or otherwise, find the exact value of
$$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 4 } } \left( \frac { \cos 2 x } { \sin x } + \frac { \sin 2 x } { \cos x } \right) \cot x d x$$
\hfill \mbox{\textit{Edexcel P3 2023 Q9 [10]}}