| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring de Moivre's theorem to derive a trigonometric identity, then solving a non-standard equation. Part (a) involves binomial expansion and algebraic manipulation to extract cos(5x) in the required form. Part (b) requires substituting the derived identity, factoring, and solving a quartic in sin²θ. The multi-step nature, algebraic complexity, and need to connect parts elevates this above typical A-level questions, though it follows a recognizable FM2 pattern. |
| 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 intervals4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\cos 5x \equiv)\cos^5 x + \binom{5}{2}\cos^3 x(i\sin x)^2 + \binom{5}{4}\cos x(i\sin x)^4\) | M1 | Identifies correct real terms of binomial expansion of \((\cos x + i\sin x)^5\) with correct binomial coefficients and correct powers |
| \((\cos 5x \equiv)\cos^5 x - 10\cos^3 x\sin^2 x + 5\cos x\sin^4 x\) | A1 | Correct simplified expression |
| Applies \(\cos^2 x = 1-\sin^2 x\) inside bracket | M1 | To obtain expression in terms of \(\sin x\) inside bracket |
| \(\equiv \cos x(16\sin^4 x - 12\sin^2 x + 1)\) | A1 | Correct expression. Must be in terms of \(x\). "\(\cos 5x =\)" not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 5\theta = \sin 2\theta \sin\theta - \cos\theta \Rightarrow \cos\theta(16\sin^4\theta - 12\sin^2\theta+1) = 2\sin^2\theta\cos\theta - \cos\theta \Rightarrow \cos\theta(16\sin^4\theta-14\sin^2\theta+2)=0\) | M1 | Uses result from (a) with \(\sin 2\theta = 2\sin\theta\cos\theta\) and collects terms |
| \(16\sin^4\theta - 14\sin^2\theta+2=0 \Rightarrow \sin^2\theta = \frac{7\pm\sqrt{17}}{16} \Rightarrow \sin\theta = \ldots\) | dM1 | Solves for \(\sin^2\theta\) and takes square root; depends on first mark |
| \(\sin\theta = \sqrt{\frac{7\pm\sqrt{17}}{16}} \Rightarrow \theta = \ldots\) | ddM1 | Full method to reach at least one value of \(\theta\); depends on previous mark |
| \(\theta = 0.986,\ 0.438\) | A1 | Correct values and no others in range |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Uses \(\sin^2\theta = 1-\cos^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) and collects terms | M1 | Uses part (a) |
| Solves for \(\cos^2\theta\) by any method including calculator, takes square root to obtain at least one value for \(\cos\theta\). \(\frac{9\pm\sqrt{17}}{16} = 0.82019...,\ 0.30480...\) | dM1 | Depends on first mark. May be implied by values of \(\cos\theta\) or \(\theta\) |
| Full method to reach at least one value for \(\theta\). \(\sqrt{\frac{9\pm\sqrt{17}}{16}} = 0.905645...,\ 0.552092...\) | dM1 | Depends on previous mark. May be implied by values of \(\theta\) |
| \((\theta =)\ 0.986,\ 0.438\) | A1 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cos 5x \equiv)\cos^5 x + \binom{5}{2}\cos^3 x(i\sin x)^2 + \binom{5}{4}\cos x(i\sin x)^4$ | M1 | Identifies correct real terms of binomial expansion of $(\cos x + i\sin x)^5$ with correct binomial coefficients and correct powers |
| $(\cos 5x \equiv)\cos^5 x - 10\cos^3 x\sin^2 x + 5\cos x\sin^4 x$ | A1 | Correct simplified expression |
| Applies $\cos^2 x = 1-\sin^2 x$ inside bracket | M1 | To obtain expression in terms of $\sin x$ inside bracket |
| $\equiv \cos x(16\sin^4 x - 12\sin^2 x + 1)$ | A1 | Correct expression. Must be in terms of $x$. "$\cos 5x =$" not required |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 5\theta = \sin 2\theta \sin\theta - \cos\theta \Rightarrow \cos\theta(16\sin^4\theta - 12\sin^2\theta+1) = 2\sin^2\theta\cos\theta - \cos\theta \Rightarrow \cos\theta(16\sin^4\theta-14\sin^2\theta+2)=0$ | M1 | Uses result from (a) with $\sin 2\theta = 2\sin\theta\cos\theta$ and collects terms |
| $16\sin^4\theta - 14\sin^2\theta+2=0 \Rightarrow \sin^2\theta = \frac{7\pm\sqrt{17}}{16} \Rightarrow \sin\theta = \ldots$ | dM1 | Solves for $\sin^2\theta$ and takes square root; depends on first mark |
| $\sin\theta = \sqrt{\frac{7\pm\sqrt{17}}{16}} \Rightarrow \theta = \ldots$ | ddM1 | Full method to reach at least one value of $\theta$; depends on previous mark |
| $\theta = 0.986,\ 0.438$ | A1 | Correct values and no others in range |
# Question 7(b) - Alternative Method (cos θ substitution):
| Working | Mark | Guidance |
|---------|------|----------|
| Uses $\sin^2\theta = 1-\cos^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$ and collects terms | M1 | Uses part (a) |
| Solves for $\cos^2\theta$ by any method including calculator, takes square root to obtain at least one value for $\cos\theta$. $\frac{9\pm\sqrt{17}}{16} = 0.82019...,\ 0.30480...$ | dM1 | Depends on first mark. May be implied by values of $\cos\theta$ or $\theta$ |
| Full method to reach at least one value for $\theta$. $\sqrt{\frac{9\pm\sqrt{17}}{16}} = 0.905645...,\ 0.552092...$ | dM1 | Depends on previous mark. May be implied by values of $\theta$ |
| $(\theta =)\ 0.986,\ 0.438$ | A1 | |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
(a) Use de Moivre's theorem to show that
$$\cos 5 x \equiv \cos x \left( a \sin ^ { 4 } x + b \sin ^ { 2 } x + c \right)$$
where $a$, $b$ and $c$ are integers to be determined.\\
(b) Hence solve, for $0 < \theta < \frac { \pi } { 2 }$
$$\cos 5 \theta = \sin 2 \theta \sin \theta - \cos \theta$$
giving your answers to 3 decimal places.
\hfill \mbox{\textit{Edexcel F2 2023 Q7 [8]}}