| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation |
| Difficulty | Challenging +1.2 Part (a) requires systematic application of double angle formulae (cos 2θ = 1-2sin²θ = 2cos²θ-1, sin 2θ = 2sinθcosθ) and algebraic manipulation to prove the identity—a multi-step process but following standard techniques. Part (b) applies the result with substitution θ=2x, leading to tan 2x = 3sin 2x, which requires further manipulation and solving a quadratic in sin 2x. While requiring several connected steps and careful algebra, this is a structured problem testing standard A-level Further Maths techniques without requiring novel insight. |
| Spec | 1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1-\cos 2\theta + \sin 2\theta}{1+\cos 2\theta + \sin 2\theta} = \frac{1-(1-2\sin^2\theta)+2\sin\theta\cos\theta}{1+\cos 2\theta+\sin 2\theta}\) or denominator version with \(\cos 2\theta = 2\cos^2\theta - 1\) | M1 | Attempts to use correct double angle formulae for both \(\sin 2\theta\) and \(\cos 2\theta\); the \(\cos 2\theta\) formula must cancel the "1" |
| \(\frac{1-(1-2\sin^2\theta)+2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)+2\sin\theta\cos\theta}\) | A1 | Correct application of both formulae simultaneously |
| \(= \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta} = \frac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos\theta(\cos\theta+\sin\theta)}\) | dM1 | Factorises numerator and denominator to demonstrate cancelling of \((\sin\theta + \cos\theta)\) |
| \(= \frac{\sin\theta}{\cos\theta} = \tan\theta\) | A1* | Fully correct proof; must see intermediate line showing cancellation; withhold if notational errors present |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1-\cos 4x + \sin 4x}{1+\cos 4x + \sin 4x} = 3\sin 2x \Rightarrow \tan 2x = 3\sin 2x\) | M1 | Makes connection with part (a), writes lhs as \(\tan 2x\); condone \(x \leftrightarrow \theta\) |
| \(\Rightarrow \sin 2x - 3\sin 2x\cos 2x = 0 \Rightarrow \sin 2x(1-3\cos 2x) = 0 \Rightarrow (\sin 2x = 0,)\; \cos 2x = \frac{1}{3}\) | A1 | Obtains \(\cos 2x = \frac{1}{3}\); may see \(\sin^2 x = \frac{1}{3}\) or \(\cos^2 x = \frac{2}{3}\) after use of double angle formulae |
| \(x = 90°\), awrt \(35.3°\) | A1 | Two correct values; condone accuracy of awrt \(90°\), \(35°\), \(145°\); also condone radian values |
| awrt \(144.7°\) | A1 | All correct values and no others in range |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1-\cos 2\theta + \sin 2\theta}{1+\cos 2\theta + \sin 2\theta} = \frac{1-(1-2\sin^2\theta)+2\sin\theta\cos\theta}{1+\cos 2\theta+\sin 2\theta}$ or denominator version with $\cos 2\theta = 2\cos^2\theta - 1$ | M1 | Attempts to use correct double angle formulae for both $\sin 2\theta$ and $\cos 2\theta$; the $\cos 2\theta$ formula must cancel the "1" |
| $\frac{1-(1-2\sin^2\theta)+2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)+2\sin\theta\cos\theta}$ | A1 | Correct application of both formulae simultaneously |
| $= \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta} = \frac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos\theta(\cos\theta+\sin\theta)}$ | dM1 | Factorises numerator and denominator to demonstrate cancelling of $(\sin\theta + \cos\theta)$ |
| $= \frac{\sin\theta}{\cos\theta} = \tan\theta$ | A1* | Fully correct proof; must see intermediate line showing cancellation; withhold if notational errors present |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1-\cos 4x + \sin 4x}{1+\cos 4x + \sin 4x} = 3\sin 2x \Rightarrow \tan 2x = 3\sin 2x$ | M1 | Makes connection with part (a), writes lhs as $\tan 2x$; condone $x \leftrightarrow \theta$ |
| $\Rightarrow \sin 2x - 3\sin 2x\cos 2x = 0 \Rightarrow \sin 2x(1-3\cos 2x) = 0 \Rightarrow (\sin 2x = 0,)\; \cos 2x = \frac{1}{3}$ | A1 | Obtains $\cos 2x = \frac{1}{3}$; may see $\sin^2 x = \frac{1}{3}$ or $\cos^2 x = \frac{2}{3}$ after use of double angle formulae |
| $x = 90°$, awrt $35.3°$ | A1 | Two correct values; condone accuracy of awrt $90°$, $35°$, $145°$; also condone radian values |
| awrt $144.7°$ | A1 | All correct values and no others in range |
**(4 marks)**\begin{enumerate}
\item In this question you should show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Given that $1 + \cos 2 \theta + \sin 2 \theta \neq 0$ prove that
$$\frac { 1 - \cos 2 \theta + \sin 2 \theta } { 1 + \cos 2 \theta + \sin 2 \theta } \equiv \tan \theta$$
(b) Hence solve, for $0 < x < 180 ^ { \circ }$
$$\frac { 1 - \cos 4 x + \sin 4 x } { 1 + \cos 4 x + \sin 4 x } = 3 \sin 2 x$$
giving your answers to one decimal place where appropriate.
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q10 [8]}}