| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Compound angle with reciprocal functions |
| Difficulty | Standard +0.8 This is a two-part question requiring proof of a non-standard identity involving reciprocal trig functions and double angles, followed by solving an equation using that result. Part (a) requires manipulating sec and tan with double angle formulas and algebraic skill to reach the given form. Part (b) requires setting up and solving a resulting equation. While systematic, it demands fluency with multiple identities and careful algebraic manipulation beyond routine exercises. |
| Spec | 1.05j 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 intervals |
| VIIIV SIUI NI JIIIM I ON OC | VIIV SIHI NI JIHM I I ON OC | VIAV SIHI NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sec 2A + \tan 2A = \frac{1}{\cos 2A} + \frac{\sin 2A}{\cos 2A}\) | B1 | A correct identity for \(\sec 2A = \frac{1}{\cos 2A}\) or \(\tan 2A = \frac{\sin 2A}{\cos 2A}\) |
| \(= \frac{1 + \sin 2A}{\cos 2A}\) | M1 | Setting expression as single fraction with correct denominator and at least two terms in numerator |
| \(= \frac{1 + 2\sin A\cos A}{\cos^2 A - \sin^2 A}\) | M1 | Getting expression in just \(\sin A\) and \(\cos A\) using double angle identities |
| \(= \frac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A - \sin^2 A} = \frac{(\cos A + \sin A)(\cos A + \sin A)}{(\cos A + \sin A)(\cos A - \sin A)}\) | M1 | Factorising numerator and denominator correctly |
| \(= \frac{\cos A + \sin A}{\cos A - \sin A}\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sec 2\theta + \tan 2\theta = \frac{1}{2} \Rightarrow \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \frac{1}{2}\) | ||
| \(\Rightarrow 2\cos\theta + 2\sin\theta = \cos\theta - \sin\theta\) | ||
| \(\Rightarrow \tan\theta = -\frac{1}{3}\) | M1 A1 | |
| \(\Rightarrow \theta =\) awrt \(2.820, 5.961\) | dM1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using part (a), cross multiplying, dividing by \(\cos\theta\) to reach \(\tan\theta = k\) | M1 | Condone \(\tan 2\theta = k\) for this mark only |
| \(\tan\theta = -\dfrac{1}{3}\) | A1 | |
| \(\tan\theta = k\) leading to at least one value (1 dp accuracy) for \(\theta\) between \(0\) and \(2\pi\) | dM1 | Use calculator to check; allow answers in degrees |
| \(\theta =\) awrt \(2.820, 5.961\) with no extra solutions within range | A1 | Condone 2.82 for 2.820; condone different/mixed variables in part (b) |
## Question 9:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sec 2A + \tan 2A = \frac{1}{\cos 2A} + \frac{\sin 2A}{\cos 2A}$ | B1 | A correct identity for $\sec 2A = \frac{1}{\cos 2A}$ **or** $\tan 2A = \frac{\sin 2A}{\cos 2A}$ |
| $= \frac{1 + \sin 2A}{\cos 2A}$ | M1 | Setting expression as single fraction with correct denominator and at least two terms in numerator |
| $= \frac{1 + 2\sin A\cos A}{\cos^2 A - \sin^2 A}$ | M1 | Getting expression in just $\sin A$ and $\cos A$ using double angle identities |
| $= \frac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A - \sin^2 A} = \frac{(\cos A + \sin A)(\cos A + \sin A)}{(\cos A + \sin A)(\cos A - \sin A)}$ | M1 | Factorising numerator and denominator correctly |
| $= \frac{\cos A + \sin A}{\cos A - \sin A}$ | A1* | |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sec 2\theta + \tan 2\theta = \frac{1}{2} \Rightarrow \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \frac{1}{2}$ | | |
| $\Rightarrow 2\cos\theta + 2\sin\theta = \cos\theta - \sin\theta$ | | |
| $\Rightarrow \tan\theta = -\frac{1}{3}$ | M1 A1 | |
| $\Rightarrow \theta =$ awrt $2.820, 5.961$ | dM1 A1 | |
## Question 9 (continued):
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using part (a), cross multiplying, dividing by $\cos\theta$ to reach $\tan\theta = k$ | M1 | Condone $\tan 2\theta = k$ for this mark only |
| $\tan\theta = -\dfrac{1}{3}$ | A1 | |
| $\tan\theta = k$ leading to at least one value (1 dp accuracy) for $\theta$ between $0$ and $2\pi$ | dM1 | Use calculator to check; allow answers in degrees |
| $\theta =$ awrt $2.820, 5.961$ with no extra solutions within range | A1 | Condone 2.82 for 2.820; condone different/mixed variables in part (b) |
---9. (a) Prove that
$$\sec 2 A + \tan 2 A \equiv \frac { \cos A + \sin A } { \cos A - \sin A } \quad A \neq \frac { ( 2 n + 1 ) \pi } { 4 } \quad n \in \mathbb { Z }$$
(b) Hence solve, for $0 \leqslant \theta < 2 \pi$
$$\sec 2 \theta + \tan 2 \theta = \frac { 1 } { 2 }$$
Give your answers to 3 decimal places.
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VIIIV SIUI NI JIIIM I ON OC & VIIV SIHI NI JIHM I I ON OC & VIAV SIHI NI JIIYM IONOO \\
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\hfill \mbox{\textit{Edexcel P3 2018 Q9 [9]}}