| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Mixed trigonometric products |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple advanced techniques: integration by substitution, deriving a reduction formula using integration by parts, applying de Moivre's theorem with binomial expansion, and synthesizing all parts to evaluate a definite integral. While each component is a standard Further Maths technique, the multi-stage synthesis and the need to connect parts (b), (c), and (d) elevates this above routine exercises. However, it follows well-established patterns for reduction formula questions. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.05l Double angle formulae: and compound angle formulae1.08h Integration by substitution4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \sin\theta\cos^n\theta\,d\theta = -\dfrac{\cos^{n+1}\theta}{n+1}(+C)\) | M1 A1 | Allow with "\(+C\)" missing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\frac{1}{2}\pi}\sin^{m-1}\theta\sin\theta\cos^n\theta\,d\theta = \left[-\dfrac{\sin^{m-1}\theta\cos^{n+1}\theta}{n+1}\right]_0^{\frac{1}{2}\pi} + \dfrac{m-1}{n+1}\int_0^{\frac{1}{2}\pi}\cos^{n+2}\theta\sin^{m-2}\theta\,d\theta\) | M1 A1 | Integrates by parts using part (a). Allow with limits missing |
| \(= \dfrac{m-1}{n+1}\int_0^{\frac{1}{2}\pi}\cos^n\theta\sin^{m-2}\theta(1-\sin^2\theta)\,d\theta\) | M1 | Applies \(\cos^2\theta = 1 - \sin^2\theta\) |
| \(I_{m,n} = \dfrac{m-1}{n+1}I_{m-2,n} - \dfrac{m-1}{n+1}I_{m,n}\) | M1 | Combines with LHS |
| \(\dfrac{n+m}{n+1}I_{m,n} = \dfrac{m-1}{n+1}I_{m-2,n}\) leading to \(I_{m,n} = \dfrac{m-1}{m+n}I_{m-2,n}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z + z^{-1} = 2\cos\theta\) | B1 | |
| \((z+z^{-1})^5 = (z^5+z^{-5}) + 5(z^3+z^{-3}) + 10(z+z^{-1})\) | M1 A1 | Expands and groups. A0 if no grouping shown |
| \(2^5\cos^5\theta = 2\cos 5\theta + 5(2\cos 3\theta) + 10(2\cos\theta)\) | M1 | Substitutes \(z^n + z^{-n} = 2\cos n\theta\) |
| \(\cos^5\theta = \dfrac{1}{16}\cos 5\theta + \dfrac{5}{16}\cos 3\theta + \dfrac{5}{8}\cos\theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_{2,5} = \dfrac{1}{7}I_{0,5}\) | M1 | Applies reduction formula in part (b) |
| \(\dfrac{1}{7}I_{0,5} = \dfrac{1}{7}\int_0^{\frac{1}{2}\pi}\cos^5\theta\,d\theta = \dfrac{1}{7}\int_0^{\frac{1}{2}\pi}\dfrac{1}{16}\cos 5\theta + \dfrac{5}{16}\cos 3\theta + \dfrac{5}{8}\cos\theta\,d\theta\) | M1 | Applies part (c) |
| \(= \dfrac{1}{7}\left[\dfrac{1}{80}\sin 5\theta + \dfrac{5}{48}\sin 3\theta + \dfrac{5}{8}\sin\theta\right]_0^{\frac{1}{2}\pi}\) | A1 | |
| \(= \dfrac{8}{105}\) | A1 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \sin\theta\cos^n\theta\,d\theta = -\dfrac{\cos^{n+1}\theta}{n+1}(+C)$ | M1 A1 | Allow with "$+C$" missing |
**Total: 2 marks**
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\frac{1}{2}\pi}\sin^{m-1}\theta\sin\theta\cos^n\theta\,d\theta = \left[-\dfrac{\sin^{m-1}\theta\cos^{n+1}\theta}{n+1}\right]_0^{\frac{1}{2}\pi} + \dfrac{m-1}{n+1}\int_0^{\frac{1}{2}\pi}\cos^{n+2}\theta\sin^{m-2}\theta\,d\theta$ | M1 A1 | Integrates by parts using part (a). Allow with limits missing |
| $= \dfrac{m-1}{n+1}\int_0^{\frac{1}{2}\pi}\cos^n\theta\sin^{m-2}\theta(1-\sin^2\theta)\,d\theta$ | M1 | Applies $\cos^2\theta = 1 - \sin^2\theta$ |
| $I_{m,n} = \dfrac{m-1}{n+1}I_{m-2,n} - \dfrac{m-1}{n+1}I_{m,n}$ | M1 | Combines with LHS |
| $\dfrac{n+m}{n+1}I_{m,n} = \dfrac{m-1}{n+1}I_{m-2,n}$ leading to $I_{m,n} = \dfrac{m-1}{m+n}I_{m-2,n}$ | A1 | AG |
**Total: 5 marks**
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z + z^{-1} = 2\cos\theta$ | B1 | |
| $(z+z^{-1})^5 = (z^5+z^{-5}) + 5(z^3+z^{-3}) + 10(z+z^{-1})$ | M1 A1 | Expands and groups. A0 if no grouping shown |
| $2^5\cos^5\theta = 2\cos 5\theta + 5(2\cos 3\theta) + 10(2\cos\theta)$ | M1 | Substitutes $z^n + z^{-n} = 2\cos n\theta$ |
| $\cos^5\theta = \dfrac{1}{16}\cos 5\theta + \dfrac{5}{16}\cos 3\theta + \dfrac{5}{8}\cos\theta$ | A1 | |
**Total: 5 marks**
---
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_{2,5} = \dfrac{1}{7}I_{0,5}$ | M1 | Applies reduction formula in part (b) |
| $\dfrac{1}{7}I_{0,5} = \dfrac{1}{7}\int_0^{\frac{1}{2}\pi}\cos^5\theta\,d\theta = \dfrac{1}{7}\int_0^{\frac{1}{2}\pi}\dfrac{1}{16}\cos 5\theta + \dfrac{5}{16}\cos 3\theta + \dfrac{5}{8}\cos\theta\,d\theta$ | M1 | Applies part (c) |
| $= \dfrac{1}{7}\left[\dfrac{1}{80}\sin 5\theta + \dfrac{5}{48}\sin 3\theta + \dfrac{5}{8}\sin\theta\right]_0^{\frac{1}{2}\pi}$ | A1 | |
| $= \dfrac{8}{105}$ | A1 | |
**Total: 4 marks**8
\begin{enumerate}[label=(\alph*)]
\item Find $\int \sin \theta \cos ^ { n } \theta d \theta$, where $n \neq - 1$.\\
Let $I _ { m , n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { m } \theta \cos ^ { n } \theta d \theta$.
\item Show that, for $m \geqslant 2$ and $n \geqslant 0$,
$$I _ { m , n } = \frac { m - 1 } { m + n } I _ { m - 2 , n }$$
\item By considering the binomial expansion of $\left( z + \frac { 1 } { z } \right) ^ { 5 }$, where $z = \cos \theta + i \sin \theta$, use de Moivre's theorem to show that
$$\cos ^ { 5 } \theta = a \cos 5 \theta + b \cos 3 \theta + c \cos \theta$$
where $a$, $b$ and $c$ are constants to be determined.
\item Using the results given in parts (b) and (c), find the exact value of $I _ { 2,5 }$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q8 [16]}}