| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Method of differences |
| Difficulty | Challenging +1.8 This is a Further Maths FP3 reduction formula question requiring trigonometric identities to establish the recurrence relation, then telescoping summation to evaluate a definite integral. While it involves multiple steps and careful algebraic manipulation, the method of differences technique is standard for this module and the trigonometric identity needed (sin A - sin B product formula) is well-known. The definite integral evaluation requires systematic application of the recurrence but follows a clear path once established. |
| Spec | 1.05l Double angle formulae: and compound angle formulae8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin nx}{\sin x} - \frac{\sin(n-2)x}{\sin x} = \frac{\sin nx - \sin nx\cos 2x + \cos nx\sin 2x}{\sin x}\) | M1 | Expands \(\sin(n-2)x\) correctly |
| \(= \frac{\sin nx - \sin nx(1-2\sin^2 x) + 2\sin x\cos x\cos nx}{\sin x}\) | M1 | Replaces \(\cos 2x\) and \(\sin 2x\) by correct trig identities |
| \(= 2\sin nx\sin x + 2\cos nx\cos x = 2\cos(n-1)x\) | ||
| \(\therefore (I_n - I_{n-2}) = \int 2\cos(n-1)x\,dx\) | A1* | Correct completion with no errors. \(I_n - I_{n-2}\) does not need to be seen explicitly but \(\int 2\cos(n-1)x\,dx\) must be seen including the integral sign |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin nx}{\sin x} - \frac{\sin(n-2)x}{\sin x} = \frac{2\cos\!\left(\frac{nx+nx-2x}{2}\right)\sin\!\left(\frac{nx-nx+2x}{2}\right)}{\sin x}\) | M1 | Use of the correct factor formula |
| \(= \frac{2\cos(nx-x)\sin x}{\sin x}\) | M1 | Attempts to replace \(nx+nx-2x\) with \(2nx-2x\) and attempts to replace \(nx-nx+2x\) with \(2x\) |
| \(= 2\cos(n-1)x\) | ||
| \((I_n - I_{n-2}) = \int 2\cos(n-1)x\,dx\) | A1* | Correct completion with no errors. \(\int 2\cos(n-1)x\,dx\) must be seen including the integral sign |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int \frac{\sin((n-1)x+x)}{\sin x}\,dx\) | M1 | Uses \(\sin nx = \sin((n-1)x+x)\) |
| \(= \int \frac{\sin(n-1)x\cos x + \sin x\cos(n-1)x}{\sin x}\,dx\) | M1 | Expands \(\sin((n-1)x+x)\) correctly |
| \(= \frac{1}{2}\int \frac{\sin nx + \sin(n-2)x}{\sin x}\,dx + \int\cos(n-1)x\,dx\) | ||
| \(= \frac{1}{2}I_n + \frac{1}{2}I_{n-2} + \int\cos(n-1)x\,dx\) | ||
| \(\therefore I_n - I_{n-2} = \int 2\cos(n-1)x\,dx\) | A1* | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin nx}{\sin x} = \frac{\sin((n-2)x+2x)}{\sin x}\) | M1 | Uses \(\sin nx = \sin((n-2)x+2x)\) |
| \(= \frac{\sin(n-2)x(1-2\sin^2 x)+2\sin x\cos x\cos(n-2)x}{\sin x}\) | M1 | Replaces \(\cos 2x\) and \(\sin 2x\) by correct trig identities |
| \(= \frac{\sin(n-2)x}{\sin x} - 2\sin x\sin(n-2)x + 2\cos x\cos(n-2)x\) | ||
| \(= \frac{\sin(n-2)x}{\sin x} + 2\cos((n-2)x+x)\) | ||
| \(I_n = I_{n-2} + 2\int\cos(n-1)x\,dx\) | ||
| \(\therefore I_n - I_{n-2} = \int 2\cos(n-1)x\,dx\) | A1* | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin nx = \sin((n-1)x+x)\) and \(\sin(n-2)x = \sin((n-1)x-x)\) | M1 | |
| \(\frac{\sin nx}{\sin x} - \frac{\sin(n-2)x}{\sin x} = \frac{\sin(n-1)x\cos x+\cos(n-1)x\sin x - (\sin(n-1)x\cos x - \sin x\cos(n-1)x)}{\sin x}\) | M1 | Replaces \(\sin((n-1)x+x)\) and \(\sin((n-1)x-x)\) with correct expansions |
| \(= \frac{2\sin x\cos(n-1)x}{\sin x}\) | ||
| \((\therefore I_n - I_{n-2}) = \int 2\cos(n-1)x\,dx\) | A1 | Correct completion with no errors. \(\int 2\cos(n-1)x\,dx\) must be seen including the integral sign |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\cos 4x\,dx = k\sin 4x\) or \(\int\cos 2x\,dx = k\sin 2x\) | M1 | \(\cos 4x\) integrated to \(\pm k\sin 4x\) or \(\cos 2x\) integrated to \(\pm k\sin 2x\) |
| \(2\int\cos 4x\,dx = \frac{1}{2}\sin 4x\) and \(2\int\cos 2x\,dx = \sin 2x\) | A1 | Both \(2\cos 4x\) and \(2\cos 2x\) integrated correctly with correct (possibly unsimplified) coefficients |
| \(\int\frac{\sin 5x}{\sin x}\,dx = \frac{2\sin(4x)}{4} + I_3\) or \(\int\frac{\sin 3x}{\sin x}\,dx = \frac{2\sin(2x)}{2} + I_1\) | M1 | One application of reduction formula. May appear in any form, no integration needed e.g. \(I_5 = \int 2\cos 4x\,dx + I_3\) or \(I_3 = \int 2\cos 2x\,dx + I_1\) |
| \(\int\frac{\sin 5x}{\sin x}\,dx = \frac{2\sin(4x)}{4} + I_3\) and \(\int\frac{\sin 3x}{\sin x}\,dx = \frac{2\sin(2x)}{2} + I_1\) | M1 | Two applications of reduction formula. Note that \(\int\frac{\sin 3x}{\sin x}\,dx\) may be attempted using trig identities and can score full marks as long as use of the reduction formula is seen at least once |
| \(I_1 = \frac{\pi}{12}\) | B1 | Could be implied by their final answer |
| \(\left[\frac{2\sin(4x)}{4} + \frac{2\sin(2x)}{2}\right]_{\frac{\pi}{12}}^{\frac{\pi}{6}} = \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{4}-\frac{1}{2}\) | M1 | Correct use of given limits at least once on an expression of the form \(\pm k\sin 4x\) or \(\pm k\sin 2x\) |
| \(\int_{\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\sin 5x}{\sin x}\,dx = \frac{1}{12}(\pi + 6\sqrt{3}-6)\) | A1 | cao |
# Question 7:
## Part (a) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin nx}{\sin x} - \frac{\sin(n-2)x}{\sin x} = \frac{\sin nx - \sin nx\cos 2x + \cos nx\sin 2x}{\sin x}$ | M1 | Expands $\sin(n-2)x$ correctly |
| $= \frac{\sin nx - \sin nx(1-2\sin^2 x) + 2\sin x\cos x\cos nx}{\sin x}$ | M1 | Replaces $\cos 2x$ and $\sin 2x$ by correct trig identities |
| $= 2\sin nx\sin x + 2\cos nx\cos x = 2\cos(n-1)x$ | | |
| $\therefore (I_n - I_{n-2}) = \int 2\cos(n-1)x\,dx$ | A1* | Correct completion with no errors. $I_n - I_{n-2}$ does not need to be seen explicitly but $\int 2\cos(n-1)x\,dx$ must be seen including the integral sign |
## Part (a) — Way 2 (factor formula):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin nx}{\sin x} - \frac{\sin(n-2)x}{\sin x} = \frac{2\cos\!\left(\frac{nx+nx-2x}{2}\right)\sin\!\left(\frac{nx-nx+2x}{2}\right)}{\sin x}$ | M1 | Use of the correct factor formula |
| $= \frac{2\cos(nx-x)\sin x}{\sin x}$ | M1 | Attempts to replace $nx+nx-2x$ with $2nx-2x$ and attempts to replace $nx-nx+2x$ with $2x$ |
| $= 2\cos(n-1)x$ | | |
| $(I_n - I_{n-2}) = \int 2\cos(n-1)x\,dx$ | A1* | Correct completion with no errors. $\int 2\cos(n-1)x\,dx$ must be seen including the integral sign |
## Part (a) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \frac{\sin((n-1)x+x)}{\sin x}\,dx$ | M1 | Uses $\sin nx = \sin((n-1)x+x)$ |
| $= \int \frac{\sin(n-1)x\cos x + \sin x\cos(n-1)x}{\sin x}\,dx$ | M1 | Expands $\sin((n-1)x+x)$ correctly |
| $= \frac{1}{2}\int \frac{\sin nx + \sin(n-2)x}{\sin x}\,dx + \int\cos(n-1)x\,dx$ | | |
| $= \frac{1}{2}I_n + \frac{1}{2}I_{n-2} + \int\cos(n-1)x\,dx$ | | |
| $\therefore I_n - I_{n-2} = \int 2\cos(n-1)x\,dx$ | A1* | Correct completion with no errors |
## Part (a) — Way 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin nx}{\sin x} = \frac{\sin((n-2)x+2x)}{\sin x}$ | M1 | Uses $\sin nx = \sin((n-2)x+2x)$ |
| $= \frac{\sin(n-2)x(1-2\sin^2 x)+2\sin x\cos x\cos(n-2)x}{\sin x}$ | M1 | Replaces $\cos 2x$ and $\sin 2x$ by correct trig identities |
| $= \frac{\sin(n-2)x}{\sin x} - 2\sin x\sin(n-2)x + 2\cos x\cos(n-2)x$ | | |
| $= \frac{\sin(n-2)x}{\sin x} + 2\cos((n-2)x+x)$ | | |
| $I_n = I_{n-2} + 2\int\cos(n-1)x\,dx$ | | |
| $\therefore I_n - I_{n-2} = \int 2\cos(n-1)x\,dx$ | A1* | Correct completion with no errors |
## Part (a) — Way 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin nx = \sin((n-1)x+x)$ and $\sin(n-2)x = \sin((n-1)x-x)$ | M1 | |
| $\frac{\sin nx}{\sin x} - \frac{\sin(n-2)x}{\sin x} = \frac{\sin(n-1)x\cos x+\cos(n-1)x\sin x - (\sin(n-1)x\cos x - \sin x\cos(n-1)x)}{\sin x}$ | M1 | Replaces $\sin((n-1)x+x)$ and $\sin((n-1)x-x)$ with correct expansions |
| $= \frac{2\sin x\cos(n-1)x}{\sin x}$ | | |
| $(\therefore I_n - I_{n-2}) = \int 2\cos(n-1)x\,dx$ | A1 | Correct completion with no errors. $\int 2\cos(n-1)x\,dx$ must be seen including the integral sign |
---
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\cos 4x\,dx = k\sin 4x$ or $\int\cos 2x\,dx = k\sin 2x$ | M1 | $\cos 4x$ integrated to $\pm k\sin 4x$ or $\cos 2x$ integrated to $\pm k\sin 2x$ |
| $2\int\cos 4x\,dx = \frac{1}{2}\sin 4x$ and $2\int\cos 2x\,dx = \sin 2x$ | A1 | Both $2\cos 4x$ and $2\cos 2x$ integrated correctly with correct (possibly unsimplified) coefficients |
| $\int\frac{\sin 5x}{\sin x}\,dx = \frac{2\sin(4x)}{4} + I_3$ or $\int\frac{\sin 3x}{\sin x}\,dx = \frac{2\sin(2x)}{2} + I_1$ | M1 | One application of reduction formula. May appear in any form, no integration needed e.g. $I_5 = \int 2\cos 4x\,dx + I_3$ or $I_3 = \int 2\cos 2x\,dx + I_1$ |
| $\int\frac{\sin 5x}{\sin x}\,dx = \frac{2\sin(4x)}{4} + I_3$ and $\int\frac{\sin 3x}{\sin x}\,dx = \frac{2\sin(2x)}{2} + I_1$ | M1 | Two applications of reduction formula. Note that $\int\frac{\sin 3x}{\sin x}\,dx$ may be attempted using trig identities and can score full marks as long as use of the reduction formula is seen at least once |
| $I_1 = \frac{\pi}{12}$ | B1 | Could be implied by their final answer |
| $\left[\frac{2\sin(4x)}{4} + \frac{2\sin(2x)}{2}\right]_{\frac{\pi}{12}}^{\frac{\pi}{6}} = \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{4}-\frac{1}{2}$ | M1 | Correct use of given limits at least once on an expression of the form $\pm k\sin 4x$ or $\pm k\sin 2x$ |
| $\int_{\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\sin 5x}{\sin x}\,dx = \frac{1}{12}(\pi + 6\sqrt{3}-6)$ | A1 | cao |
---7. Given that
$$I _ { n } = \int \frac { \sin n x } { \sin x } \mathrm {~d} x , \quad n \geqslant 1$$
\begin{enumerate}[label=(\alph*)]
\item prove that, for $n \geqslant 3$
$$I _ { n } - I _ { n - 2 } = \int 2 \cos ( n - 1 ) x \mathrm {~d} x$$
\item Hence, showing each step of your working, find the exact value of
$$\int _ { \frac { \pi } { 12 } } ^ { \frac { \pi } { 6 } } \frac { \sin 5 x } { \sin x } d x$$
giving your answer in the form $\frac { 1 } { 12 } ( a \pi + b \sqrt { 3 } + c )$, where $a$, $b$ and $c$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2016 Q7 [10]}}