| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Express cos²x or sin²x in terms of cos 2x |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard double angle formula manipulation and integration. Part (i) requires direct recall of cos²x = (1+cos2x)/2, part (ii) is routine integration with simple substitution, and part (iii) uses the identity sin²x + cos²x = 1 to deduce the answer without calculation. All steps are textbook exercises with no problem-solving or novel insight required, making it easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| State correct expression \(\frac{1}{2} + \frac{1}{2}\cos 2x\), or equivalent | B1 | [1 mark] |
| Answer | Marks | Guidance |
|---|---|---|
| Integrate an expression of the form \(a + b \cos 2x\), where \(ab \ne 0\), correctly | M1 | |
| State correct integral \(\frac{1}{2}x + \frac{1}{4}\sin 2x\), or equivalent | A1 | |
| Use correct limits correctly | M1 | |
| Obtain given answer correctly | A1 | [4 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Use identity \(\sin^2 x = 1 - \cos^2 x\) and attempt indefinite integration | M1 | |
| Obtain integral \(x - \left(\frac{1}{2}x - \frac{1}{4}\sin 2x\right)\), or equivalent | A1 | |
| Use limits and obtain answer \(\frac{1}{6}\pi - \frac{\sqrt{3}}{8}\) | A1 | [3 marks] |
**(i)**
State correct expression $\frac{1}{2} + \frac{1}{2}\cos 2x$, or equivalent | B1 | [1 mark] |
**(ii)**
Integrate an expression of the form $a + b \cos 2x$, where $ab \ne 0$, correctly | M1 | |
State correct integral $\frac{1}{2}x + \frac{1}{4}\sin 2x$, or equivalent | A1 | |
Use correct limits correctly | M1 | |
Obtain given answer correctly | A1 | [4 marks] |
**(iii)**
Use identity $\sin^2 x = 1 - \cos^2 x$ and attempt indefinite integration | M1 | |
Obtain integral $x - \left(\frac{1}{2}x - \frac{1}{4}\sin 2x\right)$, or equivalent | A1 | |
Use limits and obtain answer $\frac{1}{6}\pi - \frac{\sqrt{3}}{8}$ | A1 | [3 marks] |
*Note: Solutions that use the result of part (ii), score M1A1 for integrating 1 and A1 for the final answer.*6 (i) Express $\cos ^ { 2 } x$ in terms of $\cos 2 x$.\\
(ii) Hence show that
$$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \cos ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 6 } \pi + \frac { 1 } { 8 } \sqrt { } 3$$
(iii) By using an appropriate trigonometrical identity, deduce the exact value of
$$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sin ^ { 2 } x \mathrm {~d} x .$$
\hfill \mbox{\textit{CAIE P2 2007 Q6 [8]}}