| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2004 |
| Session | June |
| 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 Part (i) is a standard derivation using the addition formula cos(A+B) followed by the double angle formula—a routine exercise in formula manipulation. Part (ii) requires recognizing how to rearrange the identity to integrate cos³x, then evaluating a straightforward definite integral. This is typical P2/C3 material with clear signposting ('hence') and no novel insight required, making it slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Make relevant use of the \(\cos(A+B)\) formula | M1* | |
| Make relevant use of \(\cos 2A\) and \(\sin 2A\) formulae | M1* | |
| Obtain a correct expression in terms of \(\cos A\) and \(\sin A\) | A1 | |
| Use \(\sin^2 A = 1 - \cos^2 A\) to obtain an expression in terms of \(\cos A\) | M1(dep*) | |
| Obtain given answer correctly | A1 | Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Replace integrand by \(\frac{1}{4}\cos 3x + \frac{3}{4}\cos x\), or equivalent | B1 | |
| Integrate, obtaining \(\frac{1}{12}\sin 3x + \frac{3}{4}\sin x\), or equivalent | \(\text{B1} + \text{B1}\sqrt{}\) | |
| Use limits correctly | M1 | |
| Obtain given answer | A1 | Total: 5 |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Make relevant use of the $\cos(A+B)$ formula | M1* | |
| Make relevant use of $\cos 2A$ and $\sin 2A$ formulae | M1* | |
| Obtain a correct expression in terms of $\cos A$ and $\sin A$ | A1 | |
| Use $\sin^2 A = 1 - \cos^2 A$ to obtain an expression in terms of $\cos A$ | M1(dep*) | |
| Obtain given answer correctly | A1 | **Total: 5** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Replace integrand by $\frac{1}{4}\cos 3x + \frac{3}{4}\cos x$, or equivalent | B1 | |
| Integrate, obtaining $\frac{1}{12}\sin 3x + \frac{3}{4}\sin x$, or equivalent | $\text{B1} + \text{B1}\sqrt{}$ | |
| Use limits correctly | M1 | |
| Obtain given answer | A1 | **Total: 5** |7 (i) By expanding $\cos ( 2 x + x )$, show that
$$\cos 3 x \equiv 4 \cos ^ { 3 } x - 3 \cos x$$
(ii) Hence, or otherwise, show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { 3 } x \mathrm {~d} x = \frac { 2 } { 3 }$$
\hfill \mbox{\textit{CAIE P2 2004 Q7 [10]}}