| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Derive triple angle then evaluate integral |
| Difficulty | Standard +0.3 Part (i) is a standard derivation using addition formulae and double angle identities—routine A-level technique. Part (ii) requires rearranging the identity and integrating, which is straightforward once the substitution is made. This is a typical textbook exercise testing formula manipulation and integration, slightly easier than average due to its structured guidance. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State correct expansion of \(\sin(2x+x)\) | B1 | |
| Use trig formulae and Pythagoras to express \(\sin 3x\) in terms of \(\sin x\) | M1 | |
| Obtain a correct expression in any form | A1 | e.g. \(2\sin x(1-\sin^2 x) + \sin x(1-2\sin^2 x)\) |
| Obtain \(\sin 3x \equiv 3\sin x - 4\sin^3 x\) correctly AG | A1 | Accept \(=\) for \(\equiv\) |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use identity, integrate and obtain \(-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x\) | B1 B1 | One mark for each term correct |
| Use limits correctly in an integral of the form \(a\cos x + b\cos 3x\), where \(ab\neq 0\) | M1 | \(\left(-\frac{3}{8}-\frac{1}{12}+\frac{3}{4}-\frac{1}{12} = -\frac{11}{24}+\frac{2}{3}\right)\) |
| Obtain answer \(\frac{5}{24}\) | A1 | Must be exact. Accept simplified equivalent e.g. \(\frac{15}{72}\). Answer only with no working is 0/4 |
| Total | 4 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State correct expansion of $\sin(2x+x)$ | B1 | |
| Use trig formulae and Pythagoras to express $\sin 3x$ in terms of $\sin x$ | M1 | |
| Obtain a correct expression in any form | A1 | e.g. $2\sin x(1-\sin^2 x) + \sin x(1-2\sin^2 x)$ |
| Obtain $\sin 3x \equiv 3\sin x - 4\sin^3 x$ correctly **AG** | A1 | Accept $=$ for $\equiv$ |
| **Total** | **4** | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use identity, integrate and obtain $-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x$ | B1 B1 | One mark for each term correct |
| Use limits correctly in an integral of the form $a\cos x + b\cos 3x$, where $ab\neq 0$ | M1 | $\left(-\frac{3}{8}-\frac{1}{12}+\frac{3}{4}-\frac{1}{12} = -\frac{11}{24}+\frac{2}{3}\right)$ |
| Obtain answer $\frac{5}{24}$ | A1 | Must be exact. Accept simplified equivalent e.g. $\frac{15}{72}$. Answer only with no working is 0/4 |
| **Total** | **4** | |6 (i) By first expanding $\sin ( 2 x + x )$, show that $\sin 3 x \equiv 3 \sin x - 4 \sin ^ { 3 } x$.\\
(ii) Hence, showing all necessary working, find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sin ^ { 3 } x \mathrm {~d} x$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q6 [8]}}