| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question requiring systematic application of De Moivre's theorem and binomial expansion to prove a trigonometric identity, followed by routine equation solving. While it involves multiple steps and FP3 content, the techniques are well-practiced and the path is clear, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin^3\theta\cos^2\theta = \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^3\!\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^2\) | B1 | For \(\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)\) OR \(\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)\) soi. \(z\) may be used for \(e^{i\theta}\) throughout |
| \(= -\frac{1}{32i}(z^3-3z+3z^{-1}-z^{-3})(z^2+2+z^{-2})\) | M1 | For expanding brackets (binomial theorem or otherwise) |
| Full expansion with 12 terms | M1 | Two brackets expanded soi by alternate method |
| \(-\frac{1}{32i}\) | B1 | |
| \(= -\frac{1}{32i}\!\left((z^5-z^{-5})-(z^3-z^{-3})-2(z-z^{-1})\right)\) | M1 | For grouping terms. Can be seen at any stage |
| \(= -\frac{1}{16}\!\left(\frac{z^5-z^{-5}}{2i}-\frac{z^3-z^{-3}}{2i}-2\frac{z-z^{-1}}{2i}\right)\) | oe includes replacing \(z^5-z^{-5}\) with \(2i\sin 5\theta\) etc | |
| \(= -\frac{1}{16}(\sin 5\theta - \sin 3\theta - 2\sin\theta)\) | A1 | For simplification to AG www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2i\sin\theta = z - \frac{1}{z}\) | B1 | |
| \(-8i\sin^3\theta = z^3-3z+\frac{3}{z}-\frac{1}{z^3} = (z^3-\frac{1}{z^3})-(3z-\frac{3}{z}) = 2i\sin 3\theta - 6i\sin\theta\) | M1 | For RHS |
| \(32i\sin^5\theta = z^5-5z^3+10z-\frac{10}{z}+\frac{5}{z^3}-\frac{1}{z^5}\) \(= (z^5-\frac{1}{z^5})-(5z^3-\frac{5}{z^3})+(10z-\frac{10}{z})\) \(= 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta\) | M1 | For grouping terms |
| B1 | For RHS of this line and line * above | |
| \(\sin^3\theta\cos^2\theta = -\frac{1}{32i}(4(2is3\theta-6is\theta)+(2is5\theta-10is3\theta+20is\theta))\) \(= -\frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 4\sin 3\theta + 10\sin\theta - 12\sin\theta)\) | B1 | For \(-\frac{1}{32i}\) |
| \(= -\frac{1}{16}(\sin 5\theta - \sin 3\theta - 2\sin\theta)\) | A1 | For ag www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin^3\theta\cos^2\theta = 0 \Rightarrow \sin\theta = 0\) OR \(\cos\theta = 0\) | M1 | For either equation. Accept also \(\sin\theta = \pm1\). Can be implied by the A mark plus at least \(\sin^3\theta=0\) or similar |
| \(\Rightarrow \theta = r\pi\) OR \(\theta = (2r+1)\frac{1}{2}\pi\) | A1 | For either solution, AEF including a list of the first few. At least 2 in list (and no wrong solution) |
| \(\Rightarrow \theta = \frac{n\pi}{2}\) | A1 | For both of above solutions leading to general solution in form of AG where \(k=2\) |
# Question 5(i):
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin^3\theta\cos^2\theta = \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^3\!\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^2$ | B1 | For $\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)$ OR $\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)$ soi. $z$ may be used for $e^{i\theta}$ throughout |
| $= -\frac{1}{32i}(z^3-3z+3z^{-1}-z^{-3})(z^2+2+z^{-2})$ | M1 | For expanding brackets (binomial theorem or otherwise) |
| Full expansion with 12 terms | M1 | Two brackets expanded soi by alternate method |
| $-\frac{1}{32i}$ | B1 | |
| $= -\frac{1}{32i}\!\left((z^5-z^{-5})-(z^3-z^{-3})-2(z-z^{-1})\right)$ | M1 | For grouping terms. Can be seen at any stage |
| $= -\frac{1}{16}\!\left(\frac{z^5-z^{-5}}{2i}-\frac{z^3-z^{-3}}{2i}-2\frac{z-z^{-1}}{2i}\right)$ | | oe includes replacing $z^5-z^{-5}$ with $2i\sin 5\theta$ etc |
| $= -\frac{1}{16}(\sin 5\theta - \sin 3\theta - 2\sin\theta)$ | A1 | For simplification to AG www |
**Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2i\sin\theta = z - \frac{1}{z}$ | B1 | |
| $-8i\sin^3\theta = z^3-3z+\frac{3}{z}-\frac{1}{z^3} = (z^3-\frac{1}{z^3})-(3z-\frac{3}{z}) = 2i\sin 3\theta - 6i\sin\theta$ | M1 | For RHS |
| $32i\sin^5\theta = z^5-5z^3+10z-\frac{10}{z}+\frac{5}{z^3}-\frac{1}{z^5}$ $= (z^5-\frac{1}{z^5})-(5z^3-\frac{5}{z^3})+(10z-\frac{10}{z})$ $= 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta$ | M1 | For grouping terms |
| | B1 | For RHS of this line and line * above |
| $\sin^3\theta\cos^2\theta = -\frac{1}{32i}(4(2is3\theta-6is\theta)+(2is5\theta-10is3\theta+20is\theta))$ $= -\frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 4\sin 3\theta + 10\sin\theta - 12\sin\theta)$ | B1 | For $-\frac{1}{32i}$ |
| $= -\frac{1}{16}(\sin 5\theta - \sin 3\theta - 2\sin\theta)$ | A1 | For ag www |
---
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin^3\theta\cos^2\theta = 0 \Rightarrow \sin\theta = 0$ OR $\cos\theta = 0$ | M1 | For either equation. Accept also $\sin\theta = \pm1$. Can be implied by the A mark plus at least $\sin^3\theta=0$ or similar |
| $\Rightarrow \theta = r\pi$ OR $\theta = (2r+1)\frac{1}{2}\pi$ | A1 | For either solution, AEF including a list of the first few. At least 2 in list (and no wrong solution) |
| $\Rightarrow \theta = \frac{n\pi}{2}$ | A1 | For both of above solutions leading to general solution in form of AG where $k=2$ |
---5 (i) By expressing $\sin \theta$ and $\cos \theta$ in terms of $\mathrm { e } ^ { \mathrm { i } \theta }$ and $\mathrm { e } ^ { - \mathrm { i } \theta }$, prove that
$$\sin ^ { 3 } \theta \cos ^ { 2 } \theta \equiv - \frac { 1 } { 16 } ( \sin 5 \theta - \sin 3 \theta - 2 \sin \theta )$$
(ii) Hence show that all the roots of the equation
$$\sin 5 \theta = \sin 3 \theta + 2 \sin \theta$$
are of the form $\theta = \frac { n \pi } { k }$, where $n$ is any integer and $k$ is to be determined.
\hfill \mbox{\textit{OCR FP3 2012 Q5 [9]}}