| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.8 This question requires proving a non-standard identity involving reciprocal and double angle functions (requiring multiple substitutions and algebraic manipulation), then solving a trigonometric equation using the proven identity. Part (a) demands strategic use of double angle formulas and reciprocal identities beyond routine application. Part (b) requires substitution, rearrangement into a quadratic in cos x, and careful consideration of domain restrictions. This is more challenging than standard P3 trigonometric identity questions but doesn't require exceptional insight. |
| 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^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\cosec^2 2\theta(1-\cos 2\theta) = \dfrac{2-2\cos 2\theta}{\sin^2 2\theta}\) | M1 | Uses \(\cosec 2\theta = \frac{1}{\sin 2\theta}\) at some stage |
| \(= \dfrac{2-2(1-2\sin^2\theta)}{4\sin^2\theta\cos^2\theta}\) | M1dM1 | M1: Attempts double angle identity; dM1: Uses both \(\cos 2\theta = 1-2\sin^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(= \sec^2\theta = 1 + \tan^2\theta \equiv \text{RHS}\) * | A1* | Fully correct proof with sufficient stages; must see \(\sec^2\theta\) before final answer; conclusion required if working from both sides |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sec^2 x - 3\sec x - 4 = 0 \Rightarrow \sec x = \ldots\) | M1 | Uses part (a) to form 3-term quadratic in \(\sec x\) or \(\cos x\); reference quadratic in \(\cos\): \(4\cos^2 x + 3\cos x - 1 = 0\) |
| \(\cos x = \dfrac{1}{4}\) (ignore \(-1\)) | A1 | Correct cosine; ignore reference to \(\cos x = -1\) |
| \(\cos x = \dfrac{1}{4} \Rightarrow x = \ldots\) | dM1 | Correct method to find \(x\) from \(\cos x = k\) where \( |
| \(x = 75.5°,\ 284.5°\) | A1 | awrt both values; allow with or without \(180°\) included but no other angles in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1+\tan^2 x = 4+3\sec x \Rightarrow 9\sec^2 x = \tan^4 x - 6\tan^2 x + 9 \Rightarrow \tan^4 x - 15\tan^2 x = 0\) | M1 | Makes \(\sec x\) subject, squares, solves quadratic in \(\tan^2 x\) |
| \(\tan x = (\pm)\sqrt{15}\) | A1 | Correct value; need not give both |
| \(\tan x = k \neq 0 \Rightarrow x = \ldots\) | dM1 | Solves from \(\tan x = k,\ k\neq 0\) |
| \(x = 75.5°,\ 284.5°\) | A1 | Both correct; must reject extra solutions |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\cosec^2 2\theta(1-\cos 2\theta) = \dfrac{2-2\cos 2\theta}{\sin^2 2\theta}$ | M1 | Uses $\cosec 2\theta = \frac{1}{\sin 2\theta}$ at some stage |
| $= \dfrac{2-2(1-2\sin^2\theta)}{4\sin^2\theta\cos^2\theta}$ | M1dM1 | M1: Attempts double angle identity; dM1: Uses both $\cos 2\theta = 1-2\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $= \sec^2\theta = 1 + \tan^2\theta \equiv \text{RHS}$ * | A1* | Fully correct proof with sufficient stages; must see $\sec^2\theta$ before final answer; conclusion required if working from both sides |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sec^2 x - 3\sec x - 4 = 0 \Rightarrow \sec x = \ldots$ | M1 | Uses part (a) to form 3-term quadratic in $\sec x$ or $\cos x$; reference quadratic in $\cos$: $4\cos^2 x + 3\cos x - 1 = 0$ |
| $\cos x = \dfrac{1}{4}$ (ignore $-1$) | A1 | Correct cosine; ignore reference to $\cos x = -1$ |
| $\cos x = \dfrac{1}{4} \Rightarrow x = \ldots$ | dM1 | Correct method to find $x$ from $\cos x = k$ where $|k|<1$ |
| $x = 75.5°,\ 284.5°$ | A1 | awrt both values; allow with or without $180°$ included but no other angles in range |
### Part (b) Alternative:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1+\tan^2 x = 4+3\sec x \Rightarrow 9\sec^2 x = \tan^4 x - 6\tan^2 x + 9 \Rightarrow \tan^4 x - 15\tan^2 x = 0$ | M1 | Makes $\sec x$ subject, squares, solves quadratic in $\tan^2 x$ |
| $\tan x = (\pm)\sqrt{15}$ | A1 | Correct value; need not give both |
| $\tan x = k \neq 0 \Rightarrow x = \ldots$ | dM1 | Solves from $\tan x = k,\ k\neq 0$ |
| $x = 75.5°,\ 284.5°$ | A1 | Both correct; must reject extra solutions |
---\begin{enumerate}
\item (a) Prove that
\end{enumerate}
$$2 \operatorname { cosec } ^ { 2 } 2 \theta ( 1 - \cos 2 \theta ) \equiv 1 + \tan ^ { 2 } \theta$$
(b) Hence solve for $0 < x < 360 ^ { \circ }$, where $x \neq ( 90 n ) ^ { \circ } , n \in \mathbb { N }$, the equation
$$2 \operatorname { cosec } ^ { 2 } 2 x ( 1 - \cos 2 x ) = 4 + 3 \sec x$$
giving your answers to one decimal place.\\
(Solutions relying entirely on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel P3 2023 Q8 [8]}}