| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Derive triple angle then evaluate integral |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard application of addition formulae to derive a triple angle identity, followed by a routine definite integral using the derived result. Both parts follow predictable patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Make relevant use of the \(\cos(A+B)\) formula | M1* | |
| Make relevant use of the \(\cos 2A\) and \(\sin 2A\) formulae | M1* | |
| Obtain a correct expression in terms of \(\cos x\) and \(\sin x\) | A1 | |
| Use \(\sin^2 x = 1 - \cos^2 x\) to obtain an expression in terms of \(\cos x\) | M1(dep*) | |
| Obtain given answer correctly | A1 | [5] |
| (ii) Replace integrand by \(\frac{1}{2}\cos 3x + \frac{1}{2}\cos x\), or equivalent | B1 | |
| Integrate, obtaining \(\frac{1}{6}\sin 3x + \frac{1}{2}\sin x\), or equivalent | B1 + B1√ | |
| Use limits correctly | M1 | |
| Obtain given answer | A1 | [5] |
**(i)** Make relevant use of the $\cos(A+B)$ formula | M1* |
Make relevant use of the $\cos 2A$ and $\sin 2A$ formulae | M1* |
Obtain a correct expression in terms of $\cos x$ and $\sin x$ | A1 |
Use $\sin^2 x = 1 - \cos^2 x$ to obtain an expression in terms of $\cos x$ | M1(dep*) |
Obtain given answer correctly | A1 | [5]
**(ii)** Replace integrand by $\frac{1}{2}\cos 3x + \frac{1}{2}\cos x$, or equivalent | B1 |
Integrate, obtaining $\frac{1}{6}\sin 3x + \frac{1}{2}\sin x$, or equivalent | B1 + B1√ |
Use limits correctly | M1 |
Obtain given answer | A1 | [5]8 (i) By first expanding $\cos ( 2 x + x )$, show that
$$\cos 3 x \equiv 4 \cos ^ { 3 } x - 3 \cos x$$
(ii) Hence show that
$$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \left( 2 \cos ^ { 3 } x - \cos x \right) d x = \frac { 5 } { 12 }$$
\hfill \mbox{\textit{CAIE P2 2011 Q8 [10]}}