| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.2 This is a standard Further Maths question using de Moivre's theorem with well-established techniques. Parts (a) and (b) follow routine procedures (binomial expansion, algebraic manipulation). Part (c) requires recognizing that tan(5θ)=0 gives roots, then forming a quadratic from tan(π/5) and tan(2π/5). Part (d) uses Vieta's formulas. While multi-step and requiring careful algebra, it's a textbook application of known methods without novel insight, making it moderately above average difficulty. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 5\theta + i\sin 5\theta = (c+is)^5 = c^5 + 5c^4is + 10c^3i^2s^2 + 10c^2i^3s^3 + 5ci^4s^4 + i^5s^5\) | M1 | Attempts to expand \((c+is)^5\) including binomial coefficients (NB may only see real terms here) |
| \(\cos 5\theta = \text{Re}(c+is)^5 = c^5 + 10c^3i^2s^2 + 5ci^4s^4 = c^5 - 10c^3s^2 + 5cs^4\) | M1 | Extracts real terms and uses \(i^2 = -1\) to eliminate i |
| \(\cos 5\theta \equiv \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\) | A1* | Achieves the printed result with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(z+\frac{1}{z}\right)^5 = z^5 + 5z^4\left(\frac{1}{z}\right) + 10z^3\left(\frac{1}{z^2}\right) + 10z^2\left(\frac{1}{z^3}\right) + 5z\left(\frac{1}{z^4}\right) + \frac{1}{z^5}\) | M1 | Expands \(\left(z+\frac{1}{z}\right)^5\) including binomial coefficients and uses \(z^n + \frac{1}{z^n} = 2\cos n\theta\) at least once |
| \((2\cos\theta)^5 = 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\) | ||
| \(\cos 5\theta = 16\cos^5\theta - 5\cos 3\theta - 10\cos\theta\) | ||
| Uses \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\) to obtain \(\cos 5\theta\) in terms of single angles | M1 | Uses correct identity for \(\cos 3\theta\) |
| \(\cos 5\theta \equiv \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\) | A1* | Achieves printed result with no errors seen (may need careful checking) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin 5\theta \equiv 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\) | B1 | This expression (or equivalent) with no i's seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta}{\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta}\) | M1 | Uses \(\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta}\) and substitutes results from part (a) |
| \(= \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta} = \frac{t^5 - 10t^3 + 5t}{5t^4 - 10t^2 + 1}\) | A1* | Achieves the printed result with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan 5\theta = 0\) or \(\frac{t^5 - 10t^3 + 5t}{5t^4 - 10t^2 + 1} = 0\) | M1 | Considers \(\tan 5\theta = 0\); may be implied by \(t^5 - 10t^3 + 5t = 0\) or \(t^4 - 10t^2 + 5 = 0\) |
| \(\tan^5\theta - 10\tan^3\theta + 5\tan\theta = 0\) or \(t^5 - 10t^3 + 5t = 0\) | M1 | Equate numerator to 0 |
| \(\tan^4\theta - 10\tan^2\theta + 5 = 0\) or \(t^4 - 10t^2 + 5 = 0\) | A1 | Correct quartic |
| \(x^2 - 10x + 5 = 0\) | A1 | \(x^2 - 10x + 5 = 0\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Product of roots: \(\tan^2\frac{\pi}{5}\tan^2\frac{2\pi}{5} = 5\) | M1 | Must clearly state product of roots or e.g. \(\alpha\beta = 5\) or \(x_1x_2 = 5\) and uses their constant in (c) or solves quadratic and attempts product of roots |
| Or: \(x = \frac{10 \pm \sqrt{100-20}}{2} = 5 \pm 2\sqrt{5}\) and \((5+2\sqrt{5})(5-2\sqrt{5}) = \ldots\) | ||
| \(\tan^2\frac{\pi}{5}\tan^2\frac{2\pi}{5} = 5 \Rightarrow \tan\frac{\pi}{5}\tan\frac{2\pi}{5} = \sqrt{5}\) | A1 | Shows the given result with no errors |
## Question 8:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 5\theta + i\sin 5\theta = (c+is)^5 = c^5 + 5c^4is + 10c^3i^2s^2 + 10c^2i^3s^3 + 5ci^4s^4 + i^5s^5$ | M1 | Attempts to expand $(c+is)^5$ including binomial coefficients (NB may only see real terms here) |
| $\cos 5\theta = \text{Re}(c+is)^5 = c^5 + 10c^3i^2s^2 + 5ci^4s^4 = c^5 - 10c^3s^2 + 5cs^4$ | M1 | Extracts real terms and uses $i^2 = -1$ to eliminate i |
| $\cos 5\theta \equiv \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$ | A1* | Achieves the printed result with no errors seen |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(z+\frac{1}{z}\right)^5 = z^5 + 5z^4\left(\frac{1}{z}\right) + 10z^3\left(\frac{1}{z^2}\right) + 10z^2\left(\frac{1}{z^3}\right) + 5z\left(\frac{1}{z^4}\right) + \frac{1}{z^5}$ | M1 | Expands $\left(z+\frac{1}{z}\right)^5$ including binomial coefficients and uses $z^n + \frac{1}{z^n} = 2\cos n\theta$ at least once |
| $(2\cos\theta)^5 = 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta$ | | |
| $\cos 5\theta = 16\cos^5\theta - 5\cos 3\theta - 10\cos\theta$ | | |
| Uses $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ to obtain $\cos 5\theta$ in terms of single angles | M1 | Uses **correct** identity for $\cos 3\theta$ |
| $\cos 5\theta \equiv \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$ | A1* | Achieves printed result with no errors seen (may need careful checking) |
---
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin 5\theta \equiv 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$ | B1 | This expression (or equivalent) with no i's seen |
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta}{\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta}$ | M1 | Uses $\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta}$ and substitutes results from part (a) |
| $= \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta} = \frac{t^5 - 10t^3 + 5t}{5t^4 - 10t^2 + 1}$ | A1* | Achieves the printed result with no errors seen |
---
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan 5\theta = 0$ or $\frac{t^5 - 10t^3 + 5t}{5t^4 - 10t^2 + 1} = 0$ | M1 | Considers $\tan 5\theta = 0$; may be implied by $t^5 - 10t^3 + 5t = 0$ or $t^4 - 10t^2 + 5 = 0$ |
| $\tan^5\theta - 10\tan^3\theta + 5\tan\theta = 0$ or $t^5 - 10t^3 + 5t = 0$ | M1 | Equate numerator to 0 |
| $\tan^4\theta - 10\tan^2\theta + 5 = 0$ or $t^4 - 10t^2 + 5 = 0$ | A1 | Correct quartic |
| $x^2 - 10x + 5 = 0$ | A1 | $x^2 - 10x + 5 = 0$ or equivalent |
---
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Product of roots: $\tan^2\frac{\pi}{5}\tan^2\frac{2\pi}{5} = 5$ | M1 | Must clearly state product of roots or e.g. $\alpha\beta = 5$ or $x_1x_2 = 5$ and uses their constant in (c) or solves quadratic and attempts product of roots |
| Or: $x = \frac{10 \pm \sqrt{100-20}}{2} = 5 \pm 2\sqrt{5}$ and $(5+2\sqrt{5})(5-2\sqrt{5}) = \ldots$ | | |
| $\tan^2\frac{\pi}{5}\tan^2\frac{2\pi}{5} = 5 \Rightarrow \tan\frac{\pi}{5}\tan\frac{2\pi}{5} = \sqrt{5}$ | A1 | Shows the given result with no errors |\begin{enumerate}
\item (a) Use de Moivre's theorem to\\
(i) show that
\end{enumerate}
$$\cos 5 \theta \equiv \cos ^ { 5 } \theta - 10 \cos ^ { 3 } \theta \sin ^ { 2 } \theta + 5 \cos \theta \sin ^ { 4 } \theta$$
(ii) find an expression for $\sin 5 \theta$ in terms of $\cos \theta$ and $\sin \theta$\\
(b) Hence show that
$$\tan 5 \theta = \frac { t ^ { 5 } - 10 t ^ { 3 } + 5 t } { 5 t ^ { 4 } - 10 t ^ { 2 } + 1 }$$
where $t = \tan \theta$ and $\cos 5 \theta \neq 0$\\
(c) Hence find a quadratic equation whose roots $\operatorname { are } ^ { 2 } \tan ^ { 2 } \frac { \pi } { 5 }$ and $\tan ^ { 2 } \frac { 2 \pi } { 5 }$
Give your answer in the form $a x ^ { 2 } + b x + c = 0$ where $a , b$ and $c$ are integers to be found.\\
(d) Deduce that $\tan \frac { \pi } { 5 } \tan \frac { 2 \pi } { 5 } = \sqrt { 5 }$\\
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\hfill \mbox{\textit{Edexcel F2 2017 Q8 [12]}}