| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve equation with reciprocal functions |
| Difficulty | Standard +0.3 Part (i) requires substituting the identity sec²x = 1 + tan²x to form a quadratic in tan x, then solving—a standard P3 technique. Part (ii) is a routine trigonometric identity proof using triple angle formulas. Both parts are straightforward applications of known methods with no novel insight required, making this slightly easier than average. |
| 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^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| VIAV SIHI NI III IM I ON OC | VIIIV SIHI NI JIIIM I ON OO | VAYV SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(1 + \tan^2 x) - 3\tan x = 2\) | M1 | Attempts to use \(\sec^2 x = \pm 1 \pm \tan^2 x\); condone slips |
| \(2\tan^2 x = 3\tan x \Rightarrow \tan x = \frac{3}{2},\ (\tan x = 0)\) | dM1 | Valid method solving quadratic in \(\tan x\); condone division by \(\tan x\) |
| \(x = \text{awrt } 0.983\) | A1 | Only solution to \(\tan x = \frac{3}{2}\) in \(\left(0, \pi\right]\) |
| \(x = \pi\) (or awrt 3.14) | B1 | Only solution for \(\tan x = 0\) in \(\left(0, \pi\right]\); condone inclusion of 0; condone 180° if both angles in degrees |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\sin 3\theta}{\sin\theta} - \frac{\cos 3\theta}{\cos\theta} \equiv \frac{\sin 3\theta\cos\theta - \cos 3\theta\sin\theta}{\sin\theta\cos\theta} \equiv \frac{\sin(3\theta - \theta)}{\sin\theta\cos\theta}\) | M1 | Forms single fraction and uses compound angle formula on numerator; allow \(\sin(3\theta \pm \theta)\); or uses double angle on denominator allowing \(\sin\theta\cos\theta = k\sin 2\theta\) |
| \(\equiv \frac{\sin(3\theta-\theta)}{\frac{1}{2}\sin 2\theta}\) | dM1, A1 | Uses both compound angle on numerator (must be \(\sin(3\theta-\theta)\)) and double angle on denominator; allow \(\sin\theta\cos\theta = k\sin 2\theta\) |
| \(\equiv \frac{\sin 2\theta}{\frac{1}{2}\sin 2\theta} \equiv 2\) * | A1* | Fully complete proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Write as \(\frac{\sin(2\theta+\theta)}{\sin\theta} - \frac{\cos(2\theta+\theta)}{\cos\theta}\) | M1 | Positive step towards given answer |
| Expand using compound angle formulae: \(= \frac{\sin 2\theta\cos\theta + \cos 2\theta\sin\theta}{\sin\theta} - \frac{\cos 2\theta\cos\theta - \sin 2\theta\sin\theta}{\cos\theta}\) | dM1 | All correct steps; use double angle formula for \(\sin 2\theta\) (condone \(\sin 2\theta = k\sin\theta\cos\theta\)); divide and cancel \(\cos 2\theta\) terms to get expression in \(a\cos^2\theta \pm b\sin^2\theta\) |
| Correct intermediate line in single angles e.g. \(\frac{2\sin\theta\cos\theta\cos^2\theta}{\sin\theta} + \frac{2\sin\theta\cos\theta\sin^2\theta}{\cos\theta}\) or \(2\cos^2\theta+1-2\sin^2\theta+1-2\cos^2\theta+2\sin^2\theta\) or \(\frac{2\sin\theta\cos^3\theta+2\sin^3\theta\cos\theta}{\sin\theta\cos\theta}\) | A1 | Correct intermediate line leading to given answer; \(\cos 2\theta\) terms must visibly cancel |
| \(= 2\) | A1* | Fully complete proof, no errors; withhold if obvious/repeated notational errors |
# Question 9:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(1 + \tan^2 x) - 3\tan x = 2$ | M1 | Attempts to use $\sec^2 x = \pm 1 \pm \tan^2 x$; condone slips |
| $2\tan^2 x = 3\tan x \Rightarrow \tan x = \frac{3}{2},\ (\tan x = 0)$ | dM1 | Valid method solving quadratic in $\tan x$; condone division by $\tan x$ |
| $x = \text{awrt } 0.983$ | A1 | Only solution to $\tan x = \frac{3}{2}$ in $\left(0, \pi\right]$ |
| $x = \pi$ (or awrt 3.14) | B1 | Only solution for $\tan x = 0$ in $\left(0, \pi\right]$; condone inclusion of 0; condone 180° if both angles in degrees |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin 3\theta}{\sin\theta} - \frac{\cos 3\theta}{\cos\theta} \equiv \frac{\sin 3\theta\cos\theta - \cos 3\theta\sin\theta}{\sin\theta\cos\theta} \equiv \frac{\sin(3\theta - \theta)}{\sin\theta\cos\theta}$ | M1 | Forms single fraction and uses compound angle formula on numerator; allow $\sin(3\theta \pm \theta)$; or uses double angle on denominator allowing $\sin\theta\cos\theta = k\sin 2\theta$ |
| $\equiv \frac{\sin(3\theta-\theta)}{\frac{1}{2}\sin 2\theta}$ | dM1, A1 | Uses both compound angle on numerator (must be $\sin(3\theta-\theta)$) and double angle on denominator; allow $\sin\theta\cos\theta = k\sin 2\theta$ |
| $\equiv \frac{\sin 2\theta}{\frac{1}{2}\sin 2\theta} \equiv 2$ * | A1* | Fully complete proof |
# Trigonometric Proof Question:
## Part (a) — Proof that $\frac{\sin 3\theta}{\sin \theta} - \frac{\cos 3\theta}{\cos \theta} = 2$
| Working | Mark | Guidance |
|---------|------|----------|
| Write as $\frac{\sin(2\theta+\theta)}{\sin\theta} - \frac{\cos(2\theta+\theta)}{\cos\theta}$ | M1 | Positive step towards given answer |
| Expand using compound angle formulae: $= \frac{\sin 2\theta\cos\theta + \cos 2\theta\sin\theta}{\sin\theta} - \frac{\cos 2\theta\cos\theta - \sin 2\theta\sin\theta}{\cos\theta}$ | dM1 | All correct steps; use double angle formula for $\sin 2\theta$ (condone $\sin 2\theta = k\sin\theta\cos\theta$); divide and cancel $\cos 2\theta$ terms to get expression in $a\cos^2\theta \pm b\sin^2\theta$ |
| Correct intermediate line in single angles e.g. $\frac{2\sin\theta\cos\theta\cos^2\theta}{\sin\theta} + \frac{2\sin\theta\cos\theta\sin^2\theta}{\cos\theta}$ or $2\cos^2\theta+1-2\sin^2\theta+1-2\cos^2\theta+2\sin^2\theta$ or $\frac{2\sin\theta\cos^3\theta+2\sin^3\theta\cos\theta}{\sin\theta\cos\theta}$ | A1 | Correct intermediate line leading to given answer; $\cos 2\theta$ terms must visibly cancel |
| $= 2$ | A1* | Fully complete proof, no errors; withhold if obvious/repeated notational errors |
---9. In this question you must show detailed reasoning. Solutions relying entirely on calculator technology are not acceptable.\\
(i) Solve, for $0 < x \leqslant \pi$, the equation
$$2 \sec ^ { 2 } x - 3 \tan x = 2$$
giving the answers, as appropriate, to 3 significant figures.\\
(ii) Prove that
$$\frac { \sin 3 \theta } { \sin \theta } - \frac { \cos 3 \theta } { \cos \theta } \equiv 2$$
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VIAV SIHI NI III IM I ON OC & VIIIV SIHI NI JIIIM I ON OO & VAYV SIHI NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel P3 2022 Q9 [8]}}