| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Method of differences |
| Difficulty | Challenging +1.8 This is a Further Maths reduction formula question requiring application of a given trigonometric identity to establish a recurrence relation, then using method of differences to find a specific value. While the identity is provided, students must recognize how to manipulate sin(2nθ) + sin(2(n-1)θ) to create the required form, integrate the resulting expression, and then telescope the series. This requires sophisticated technique and multi-step reasoning beyond standard A-level, though the structure is guided by the question parts. |
| Spec | 1.05l Double angle formulae: and compound angle formulae8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(I_n + I_{n-1} = \int_0^{\pi/2} \frac{\sin(2n\theta)+\sin(2n-2)\theta}{\cos\theta}\,d\theta = \int_0^{\pi/2} \frac{2\sin(2n-1)\theta\cos\theta}{\cos\theta}\,d\theta\) | M1A1 | LR |
| \(= \int_0^{\pi/2} 2\sin(2n-1)\theta\,d\theta = \left[-\frac{2\cos(2n-1)\theta}{2n-1}\right]_0^{\pi/2}\) | dM1A1 | LR |
| \(= [0] - \left[-\frac{2}{2n-1}\right] = \frac{2}{2n-1}\) | A1 | AG |
| \(I_1 = \int_0^{\pi/2}\frac{\sin 2\theta}{\cos\theta}\,d\theta = \int_0^{\pi/2} 2\sin\theta\,d\theta = [-2\cos\theta]_0^{\pi/2} = [0]-[-2] = 2\) | B1 | |
| \(I_2 = \frac{2}{3} - 2 = -\frac{4}{3}\) | M1A1 | |
| \(I_3 = \frac{2}{5} + \frac{4}{3} = \frac{26}{15}\); \(I_4 = \frac{2}{7} - \frac{26}{15} = -\frac{152}{105}\) | A1 |
## Question 5:
| Working/Answer | Marks | Notes |
|---|---|---|
| $I_n + I_{n-1} = \int_0^{\pi/2} \frac{\sin(2n\theta)+\sin(2n-2)\theta}{\cos\theta}\,d\theta = \int_0^{\pi/2} \frac{2\sin(2n-1)\theta\cos\theta}{\cos\theta}\,d\theta$ | M1A1 | LR |
| $= \int_0^{\pi/2} 2\sin(2n-1)\theta\,d\theta = \left[-\frac{2\cos(2n-1)\theta}{2n-1}\right]_0^{\pi/2}$ | dM1A1 | LR |
| $= [0] - \left[-\frac{2}{2n-1}\right] = \frac{2}{2n-1}$ | A1 | AG |
| $I_1 = \int_0^{\pi/2}\frac{\sin 2\theta}{\cos\theta}\,d\theta = \int_0^{\pi/2} 2\sin\theta\,d\theta = [-2\cos\theta]_0^{\pi/2} = [0]-[-2] = 2$ | B1 | |
| $I_2 = \frac{2}{3} - 2 = -\frac{4}{3}$ | M1A1 | |
| $I_3 = \frac{2}{5} + \frac{4}{3} = \frac{26}{15}$; $I_4 = \frac{2}{7} - \frac{26}{15} = -\frac{152}{105}$ | A1 | |
---5 Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 n \theta } { \cos \theta } \mathrm {~d} \theta$, where $n$ is a non-negative integer.\\
(i) Use the identity $\sin P + \sin Q \equiv 2 \sin \frac { 1 } { 2 } ( P + Q ) \cos \frac { 1 } { 2 } ( P - Q )$ to show that
$$I _ { n } + I _ { n - 1 } = \frac { 2 } { 2 n - 1 } , \text { for all positive integers } n$$
(ii) Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 8 \theta } { \cos \theta } d \theta$.
\hfill \mbox{\textit{CAIE FP1 2015 Q5 [9]}}