| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Challenging +1.3 This is a Further Maths question requiring triple angle expansions, algebraic manipulation to prove an identity, then solving a resulting equation. While it involves multiple steps and reciprocal trig functions, the techniques are standard for Further Maths students: expand cos 3θ and sin 3θ, combine fractions, apply double angle formulas. The 'hence' part is straightforward once the identity is proven. More demanding than typical A-level questions but routine for Further Maths. |
| 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 intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{\text{LHS}=\}\dfrac{\cos3\theta\cos\theta+\sin3\theta\sin\theta}{\sin\theta\cos\theta}\) | M1 | 3.1a — Correct valid method forming common denominator of \(\sin\theta\cos\theta\) |
| \(=\dfrac{\cos(3\theta-\theta)}{\sin\theta\cos\theta}\left\{=\dfrac{\cos2\theta}{\sin\theta\cos\theta}\right\}\) | A1 | 2.1 — Numerator simplifies to \(\cos(3\theta-\theta)\) or \(\cos2\theta\) |
| \(=\dfrac{\cos2\theta}{\frac{1}{2}\sin2\theta}=2\cot2\theta\) * | dM1, A1* | 1.1b, 2.1 — Applies \(\sin2\theta\equiv2\sin\theta\cos\theta\); correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{\text{LHS}=\}\dfrac{\cos2\theta\cos^2\theta-\sin2\theta\sin\theta\cos\theta+\sin2\theta\cos\theta\sin\theta+\cos2\theta\sin^2\theta}{\sin\theta\cos\theta}\) | M1 | 3.1a |
| \(=\dfrac{\cos2\theta(\cos^2\theta+\sin^2\theta)}{\sin\theta\cos\theta}\left\{=\dfrac{\cos2\theta}{\sin\theta\cos\theta}\right\}\) | A1 | 2.1 |
| \(=\dfrac{\cos2\theta}{\frac{1}{2}\sin2\theta}=2\cot2\theta\) * | dM1, A1* | 1.1b, 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{\text{RHS}=\}\dfrac{2\cos2\theta}{\sin2\theta}=\dfrac{2\cos(3\theta-\theta)}{\sin2\theta}=\dfrac{2(\cos3\theta\cos\theta+\sin3\theta\sin\theta)}{\sin2\theta}\) | M1, A1 | 3.1a, 2.1 |
| \(=\dfrac{2(\cos3\theta\cos\theta+\sin3\theta\sin\theta)}{2\sin\theta\cos\theta}\) | dM1 | 1.1b |
| \(=\dfrac{\cos3\theta}{\sin\theta}+\dfrac{\sin3\theta}{\cos\theta}\) * | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left\{\dfrac{\cos3\theta}{\sin\theta}+\dfrac{\sin3\theta}{\cos\theta}=4\Rightarrow\right\}\ 2\cot2\theta=4\Rightarrow2\left(\dfrac{1}{\tan2\theta}\right)=4\) | M1 | 1.1b — Evidence of applying \(\cot2\theta=\dfrac{1}{\tan2\theta}\) |
| Rearranges to give \(\tan2\theta=k\); \(k\neq0\) and applies \(\arctan k\) | dM1 | 1.1b |
| \(\left\{90°<\theta<180°,\ \tan2\theta=\dfrac{1}{2}\Rightarrow\right\}\) Only one solution \(\theta=103.3°\) (1 dp) or awrt \(103.3°\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2\cot2\theta=4\Rightarrow\dfrac{2}{\tan2\theta}=4\) | M1 | 1.1b |
| Applies \(\tan2\theta\equiv\dfrac{2\tan\theta}{1-\tan^2\theta}\), forms correct method for solving 3TQ to give \(\tan\theta=k\), \(k\neq0\), applies \(\arctan k\) | dM1 | 1.1b — \(\Rightarrow\tan^2\theta+4\tan\theta-1=0\Rightarrow\tan\theta=\dfrac{-4\pm\sqrt{(4)^2-4(1)(-1)}}{2(1)}\) |
| \(\{90°<\theta<180°,\ \tan\theta=-2-\sqrt{5}\Rightarrow\}\) Only one solution \(\theta=103.3°\) (1 dp) or awrt \(103.3°\) | A1 | 2.2a |
# Question 12:
## Part (a) — Way 1
| Working | Mark | Guidance |
|---------|------|----------|
| $\{\text{LHS}=\}\dfrac{\cos3\theta\cos\theta+\sin3\theta\sin\theta}{\sin\theta\cos\theta}$ | M1 | 3.1a — Correct valid method forming common denominator of $\sin\theta\cos\theta$ |
| $=\dfrac{\cos(3\theta-\theta)}{\sin\theta\cos\theta}\left\{=\dfrac{\cos2\theta}{\sin\theta\cos\theta}\right\}$ | A1 | 2.1 — Numerator simplifies to $\cos(3\theta-\theta)$ or $\cos2\theta$ |
| $=\dfrac{\cos2\theta}{\frac{1}{2}\sin2\theta}=2\cot2\theta$ * | dM1, A1* | 1.1b, 2.1 — Applies $\sin2\theta\equiv2\sin\theta\cos\theta$; correct proof |
## Part (a) — Way 2
| Working | Mark | Guidance |
|---------|------|----------|
| $\{\text{LHS}=\}\dfrac{\cos2\theta\cos^2\theta-\sin2\theta\sin\theta\cos\theta+\sin2\theta\cos\theta\sin\theta+\cos2\theta\sin^2\theta}{\sin\theta\cos\theta}$ | M1 | 3.1a |
| $=\dfrac{\cos2\theta(\cos^2\theta+\sin^2\theta)}{\sin\theta\cos\theta}\left\{=\dfrac{\cos2\theta}{\sin\theta\cos\theta}\right\}$ | A1 | 2.1 |
| $=\dfrac{\cos2\theta}{\frac{1}{2}\sin2\theta}=2\cot2\theta$ * | dM1, A1* | 1.1b, 2.1 |
## Part (a) — Way 3
| Working | Mark | Guidance |
|---------|------|----------|
| $\{\text{RHS}=\}\dfrac{2\cos2\theta}{\sin2\theta}=\dfrac{2\cos(3\theta-\theta)}{\sin2\theta}=\dfrac{2(\cos3\theta\cos\theta+\sin3\theta\sin\theta)}{\sin2\theta}$ | M1, A1 | 3.1a, 2.1 |
| $=\dfrac{2(\cos3\theta\cos\theta+\sin3\theta\sin\theta)}{2\sin\theta\cos\theta}$ | dM1 | 1.1b |
| $=\dfrac{\cos3\theta}{\sin\theta}+\dfrac{\sin3\theta}{\cos\theta}$ * | A1* | 2.1 |
**Notes for (a):**
- M1: Correct valid method forming common denominator $\sin\theta\cos\theta$, i.e. correct process of $\dfrac{(...)\cos\theta+(...){\sin\theta}}{\cos\theta\sin\theta}$
- A1: Numerator simplifies to $\cos(3\theta-\theta)$ or $\cos2\theta$
- dM1: Dependent on previous M; applies $\sin2\theta\equiv2\sin\theta\cos\theta$ to common denominator $\sin\theta\cos\theta$
- Allow $2^\text{nd}$ M1 for stating correct $\sin2\theta\equiv2\sin\theta\cos\theta$ and attempting to apply it
## Part (b) — Way 1
| Working | Mark | Guidance |
|---------|------|----------|
| $\left\{\dfrac{\cos3\theta}{\sin\theta}+\dfrac{\sin3\theta}{\cos\theta}=4\Rightarrow\right\}\ 2\cot2\theta=4\Rightarrow2\left(\dfrac{1}{\tan2\theta}\right)=4$ | M1 | 1.1b — Evidence of applying $\cot2\theta=\dfrac{1}{\tan2\theta}$ |
| Rearranges to give $\tan2\theta=k$; $k\neq0$ and applies $\arctan k$ | dM1 | 1.1b |
| $\left\{90°<\theta<180°,\ \tan2\theta=\dfrac{1}{2}\Rightarrow\right\}$ **Only one solution** $\theta=103.3°$ (1 dp) or awrt $103.3°$ | A1 | 2.2a |
**Notes for (b) Way 1:**
- Note: Give M0M0A0 for writing e.g. $\tan2\theta=2$ with no evidence of applying $\cot2\theta=\dfrac{1}{\tan2\theta}$
- 1st M1 can be implied by seeing $\tan2\theta=\dfrac{1}{2}$
- Condone 2nd M1 for applying $\dfrac{1}{2}\arctan\!\left(\dfrac{1}{2}\right)\{=13.28...\}$
## Part (b) — Way 2
| Working | Mark | Guidance |
|---------|------|----------|
| $2\cot2\theta=4\Rightarrow\dfrac{2}{\tan2\theta}=4$ | M1 | 1.1b |
| Applies $\tan2\theta\equiv\dfrac{2\tan\theta}{1-\tan^2\theta}$, forms correct method for solving 3TQ to give $\tan\theta=k$, $k\neq0$, applies $\arctan k$ | dM1 | 1.1b — $\Rightarrow\tan^2\theta+4\tan\theta-1=0\Rightarrow\tan\theta=\dfrac{-4\pm\sqrt{(4)^2-4(1)(-1)}}{2(1)}$ |
| $\{90°<\theta<180°,\ \tan\theta=-2-\sqrt{5}\Rightarrow\}$ **Only one solution** $\theta=103.3°$ (1 dp) or awrt $103.3°$ | A1 | 2.2a |
**Notes for (b):**
- Give M1 dM1 A1 for no working leading to $\theta=$ awrt $103.3°$ and no other solutions
- Give M1 dM1 A0 for no working leading to $\theta=$ awrt $103.3°$ and other solutions outside or inside $90°<\theta<180°$
---\begin{enumerate}
\item (a) Prove
\end{enumerate}
$$\frac { \cos 3 \theta } { \sin \theta } + \frac { \sin 3 \theta } { \cos \theta } \equiv 2 \cot 2 \theta \quad \theta \neq ( 90 n ) ^ { \circ } , n \in \mathbb { Z }$$
(b) Hence solve, for $90 ^ { \circ } < \theta < 180 ^ { \circ }$, the equation
$$\frac { \cos 3 \theta } { \sin \theta } + \frac { \sin 3 \theta } { \cos \theta } = 4$$
giving any solutions to one decimal place.
\hfill \mbox{\textit{Edexcel Paper 2 2019 Q12 [7]}}