| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Challenging +1.2 Part (i) is a standard Further Pure technique using De Moivre's theorem and binomial expansion to derive a multiple angle formula - routine for FP3 students. Part (ii) applies this result to solve a trigonometric equation, requiring careful algebraic manipulation and consideration of the given domain. While this is a multi-step problem requiring several techniques, it follows a well-established pattern for this topic with no novel insights needed. The difficulty is elevated above average due to the algebraic complexity and being Further Maths content, but remains a standard textbook-style question. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02o Loci in Argand diagram: circles, half-lines8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2i\sin\theta = e^{i\theta} - e^{-i\theta}\) | B1 | Any equivalent form; If use \(z\), must define it |
| \((2i\sin\theta)^5 = (e^{i\theta} - e^{-i\theta})^5\) | ||
| \(= e^{5i\theta} - 5e^{3i\theta} + 10e^{i\theta} - 10e^{-i\theta} + 5e^{-3i\theta} - e^{-5i\theta}\) | M1* | Binomial expansion; can be unsimplified |
| \(32i\sin^5\theta = (e^{5i\theta}-e^{-5i\theta}) - 5(e^{3i\theta}-e^{-3i\theta}) + 10(e^{i\theta}-e^{-i\theta})\) | M1dep* | Grouping terms; Award B1 then sc M1A1 for candidates who omit this stage from otherwise complete argument |
| \(= 2i\sin 5\theta - 5(2i\sin 3\theta) + 10(2i\sin\theta)\) | ||
| \(\sin^5\theta = \frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta)\) | A1 | AG; must convince on the \(\frac{1}{16}\) and on the elimination of \(i\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(16\sin^5\theta - 10\sin\theta = \sin 5\theta - 5\sin 3\theta\) | M1* | Attempts to eliminate \(\sin 5\theta\) and \(\sin 3\theta\); Or \(16\sin^5\theta = 6\sin\theta\) |
| \(16\sin^5\theta - 6\sin\theta = 0\) | A1 | |
| \(\sin\theta = 0, \pm\sqrt[4]{\frac{3}{8}}\) | M1dep* | Must have 3 values for \(\sin\theta\) |
| \(\theta = 0, \pm 0.899\) | A1 | |
| [4] |
# Question 7:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2i\sin\theta = e^{i\theta} - e^{-i\theta}$ | B1 | Any equivalent form; If use $z$, must define it |
| $(2i\sin\theta)^5 = (e^{i\theta} - e^{-i\theta})^5$ | | |
| $= e^{5i\theta} - 5e^{3i\theta} + 10e^{i\theta} - 10e^{-i\theta} + 5e^{-3i\theta} - e^{-5i\theta}$ | M1* | Binomial expansion; can be unsimplified |
| $32i\sin^5\theta = (e^{5i\theta}-e^{-5i\theta}) - 5(e^{3i\theta}-e^{-3i\theta}) + 10(e^{i\theta}-e^{-i\theta})$ | M1dep* | Grouping terms; Award B1 then sc M1A1 for candidates who omit this stage from otherwise complete argument |
| $= 2i\sin 5\theta - 5(2i\sin 3\theta) + 10(2i\sin\theta)$ | | |
| $\sin^5\theta = \frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta)$ | A1 | AG; must convince on the $\frac{1}{16}$ and on the elimination of $i$ |
| **[4]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $16\sin^5\theta - 10\sin\theta = \sin 5\theta - 5\sin 3\theta$ | M1* | Attempts to eliminate $\sin 5\theta$ and $\sin 3\theta$; Or $16\sin^5\theta = 6\sin\theta$ |
| $16\sin^5\theta - 6\sin\theta = 0$ | A1 | |
| $\sin\theta = 0, \pm\sqrt[4]{\frac{3}{8}}$ | M1dep* | Must have 3 values for $\sin\theta$ |
| $\theta = 0, \pm 0.899$ | A1 | |
| **[4]** | | |
---7 (i) By expressing $\sin \theta$ in terms of $\mathrm { e } ^ { \mathrm { i } \theta }$ and $\mathrm { e } ^ { - \mathrm { i } \theta }$, show that
$$\sin ^ { 5 } \theta \equiv \frac { 1 } { 16 } ( \sin 5 \theta - 5 \sin 3 \theta + 10 \sin \theta ) .$$
(ii) Hence solve the equation
$$\sin 5 \theta + 4 \sin \theta = 5 \sin 3 \theta$$
for $- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.
8 consists of the set of matrices of the form $\left( \begin{array} { c c } a & - b \\ b & a \end{array} \right)$, where $a$ and $b$ are real and $a ^ { 2 } + b ^ { 2 } \neq 0$, combined under the operation of matrix multiplication.\\
(i) Prove that $G$ is a group. You may assume that matrix multiplication is associative.\\
(ii) Determine whether $G$ is commutative.\\
(iii) Find the order of $\left( \begin{array} { c c } 0 & - 1 \\ 1 & 0 \end{array} \right)$.
\hfill \mbox{\textit{OCR FP3 2014 Q7 [8]}}