| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Challenging +1.2 This is a two-part question requiring standard trigonometric identities (double and triple angle formulas) and algebraic manipulation to prove an identity, followed by solving an equation using the proven result. While it involves multiple steps and careful algebra, the techniques are all standard P3 content with no novel insight required. The identity proof is methodical application of known formulas, and part (b) is a straightforward substitution leading to a quadratic in sin θ. This is moderately above average difficulty due to the length and algebraic complexity, but remains a typical exam question testing standard techniques. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{\cos^2\theta}{\cos 2\theta - \sin 3\theta} = \frac{\cos^2\theta}{\cos 2\theta - (\sin 2\theta\cos\theta + \cos 2\theta\sin\theta)}\) | M1 | Applies compound angle formula to \(\sin 3\theta\). Accept \(\pm\sin\theta\cos 2\theta \pm \sin 2\theta\cos\theta\). Allow \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\) directly. |
| \(= \frac{\cos^2\theta}{1 - 2\sin^2\theta - 2\sin\theta\cos^2\theta - \sin\theta(1-2\sin^2\theta)}\) | M1 | Uses correct double angle formula for \(\cos 2\theta\) and \(\sin 2\theta\) to achieve all terms in single angle. |
| \(= \frac{1-\sin^2\theta}{1 - 3\sin\theta - 2\sin^2\theta + 4\sin^3\theta} = \frac{(1-\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1-2\sin\theta-4\sin^2\theta)}\) | M1 | Uses Pythagorean identity to identify factor of \(1-\sin\theta\) in numerator. Must be an intermediate step before given answer with \(1-\sin\theta\) cancelled. |
| \(= \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta}\) * | A1* | Cancels \(1-\sin\theta\) from numerator and denominator with no incorrect steps. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta} = 2\cosec\theta \Rightarrow (1+\sin\theta)\sin\theta = 2(1-2\sin\theta-4\sin^2\theta)\) | M1 | Substitutes result from (a) and cross multiplies to get equation in \(\sin\theta\) only. |
| \(\Rightarrow 9\sin^2\theta + 5\sin\theta - 2 = 0\) | A1 | Correct simplified quadratic in \(\sin\theta\). |
| \(\Rightarrow \sin\theta = \frac{-5 \pm \sqrt{25 - 4\times9\times-2}}{18} = \ldots \Rightarrow \theta = \sin^{-1}\ldots\) | M1 | Solves quadratic by formula, completing square or calculator and applies inverse sine to at least one root. |
| Two of \(\theta = 15.6°, 164.4°, 235.6°, 304.4°\) | A1 | Any two correct solutions in range, accept awrt. Answers in radians scores A0. |
| All of \(\theta = 15.6°, 164.4°, 235.6°, 304.4°\) | A1 | All four solutions correct and no others in the range. |
# Question 9:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{\cos^2\theta}{\cos 2\theta - \sin 3\theta} = \frac{\cos^2\theta}{\cos 2\theta - (\sin 2\theta\cos\theta + \cos 2\theta\sin\theta)}$ | M1 | Applies compound angle formula to $\sin 3\theta$. Accept $\pm\sin\theta\cos 2\theta \pm \sin 2\theta\cos\theta$. Allow $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ directly. |
| $= \frac{\cos^2\theta}{1 - 2\sin^2\theta - 2\sin\theta\cos^2\theta - \sin\theta(1-2\sin^2\theta)}$ | M1 | Uses correct double angle formula for $\cos 2\theta$ **and** $\sin 2\theta$ to achieve all terms in single angle. |
| $= \frac{1-\sin^2\theta}{1 - 3\sin\theta - 2\sin^2\theta + 4\sin^3\theta} = \frac{(1-\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1-2\sin\theta-4\sin^2\theta)}$ | M1 | Uses Pythagorean identity to identify factor of $1-\sin\theta$ in numerator. Must be an intermediate step before given answer with $1-\sin\theta$ cancelled. |
| $= \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta}$ * | A1* | Cancels $1-\sin\theta$ from numerator and denominator with no incorrect steps. |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta} = 2\cosec\theta \Rightarrow (1+\sin\theta)\sin\theta = 2(1-2\sin\theta-4\sin^2\theta)$ | M1 | Substitutes result from (a) and cross multiplies to get equation in $\sin\theta$ only. |
| $\Rightarrow 9\sin^2\theta + 5\sin\theta - 2 = 0$ | A1 | Correct simplified quadratic in $\sin\theta$. |
| $\Rightarrow \sin\theta = \frac{-5 \pm \sqrt{25 - 4\times9\times-2}}{18} = \ldots \Rightarrow \theta = \sin^{-1}\ldots$ | M1 | Solves quadratic by formula, completing square or calculator and applies inverse sine to at least one root. |
| Two of $\theta = 15.6°, 164.4°, 235.6°, 304.4°$ | A1 | Any two correct solutions in range, accept awrt. Answers in radians scores A0. |
| All of $\theta = 15.6°, 164.4°, 235.6°, 304.4°$ | A1 | All four solutions correct and no others in the range. |9. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Given that $\cos 2 \theta - \sin 3 \theta \neq 0$
\begin{enumerate}[label=(\alph*)]
\item prove that
$$\frac { \cos ^ { 2 } \theta } { \cos 2 \theta - \sin 3 \theta } \equiv \frac { 1 + \sin \theta } { 1 - 2 \sin \theta - 4 \sin ^ { 2 } \theta }$$
\item Hence solve, for $0 < \theta \leqslant 360 ^ { \circ }$
$$\frac { \cos ^ { 2 } \theta } { \cos 2 \theta - \sin 3 \theta } = 2 \operatorname { cosec } \theta$$
Give your answers to one decimal place.\\
\includegraphics[max width=\textwidth, alt={}, center]{83e12fa4-1abb-4bea-bff4-8d36757bd9c3-28_2257_52_309_1983}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q9 [9]}}