| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then evaluate integral |
| Difficulty | Standard +0.3 This is a standard two-part question requiring compound angle formula manipulation (routine for P3) followed by a straightforward definite integral. The identity proof follows a well-established method, and the integration is direct substitution of the proven result. Slightly easier than average due to clear signposting and standard techniques. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State correct expansion of \(\cos(3x - x)\) or \(\cos(3x + x)\) | B1 | |
| Substitute expansions in \(\frac{1}{2}(\cos 2x - \cos 4x)\), or equivalent | M1 | |
| Simplify and obtain the given identity correctly | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain integral \(\frac{1}{4}\sin 2x - \frac{1}{8}\sin 4x\) | B1 | |
| Substitute limits correctly in an integral of the form \(a\sin 2x + b\sin 4x\) | M1 | |
| Obtain given answer following full, correct and exact working | A1 | [3] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State correct expansion of $\cos(3x - x)$ or $\cos(3x + x)$ | B1 | |
| Substitute expansions in $\frac{1}{2}(\cos 2x - \cos 4x)$, or equivalent | M1 | |
| Simplify and obtain the given identity correctly | A1 | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain integral $\frac{1}{4}\sin 2x - \frac{1}{8}\sin 4x$ | B1 | |
| Substitute limits correctly in an integral of the form $a\sin 2x + b\sin 4x$ | M1 | |
| Obtain given answer following full, correct and exact working | A1 | [3] |
---4 (i) Using the expansions of $\cos ( 3 x - x )$ and $\cos ( 3 x + x )$, prove that
$$\frac { 1 } { 2 } ( \cos 2 x - \cos 4 x ) \equiv \sin 3 x \sin x$$
(ii) Hence show that
$$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } \sin 3 x \sin x \mathrm {~d} x = \frac { 1 } { 8 } \sqrt { } 3$$
\hfill \mbox{\textit{CAIE P3 2010 Q4 [6]}}