| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | October |
| 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 This is a standard two-part question requiring derivation of a triple angle formula using addition formulae (routine manipulation) followed by algebraic integration using the derived identity. Both parts are textbook exercises with clear pathways: (a) expand sin(2x+x) and apply double angle formulae, (b) rearrange to isolate sin³x and integrate. The techniques are well-practiced at P3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
(a) $\sin 3x \equiv \sin(2x+x) \equiv \sin 2x \cos x + \cos 2x \sin x$ M1
$\equiv 2\sin x \cos x \cos x + (1 - 2\sin^2 x)\sin x$ M1
$\equiv 2\sin x(1 - \sin^2 x) + (1 - 2\sin^2 x)\sin x$ ddM1
$\equiv 3\sin x - 4\sin^3 x$ A1*
(4 marks)
(b) $\int_0^{\pi/3} \sin^3 x \, dx = \int_0^{\pi/3} \left(3\sin x - \frac{1}{4}\sin 3x\right) dx$ M1
$= \left[-3\cos x + \frac{1}{12}\cos 3x\right]_0^{\pi/3}$ dM1 A1
$= \frac{5}{24}$ A1
(4 marks)
(8 marks)
## Alternative 1
$\int_0^{\pi/3} \sin^3 x \, dx = \int_0^{\pi/3} \sin x(1 - \cos^2 x) \, dx = \int_0^{\pi/3} (\sin x - \sin x \cos^2 x) \, dx$ M1
$= \left[-\cos x - \left(-\frac{\cos^3 x}{3}\right)\right]_0^{\pi/3}$ dM1 A1
$= \frac{5}{24}$ A1
(4 marks)
## Alternative 2
$u = \cos x \Rightarrow du = -\sin x \, dx \Rightarrow \int_0^{\pi/3} \sin^3 x \, dx = \int_1^{1/2} (u^2 - 1) \, du$ M1
$= \left[\frac{u^3}{3} - u\right]_1^{1/2}$ dM1 A1
$= \frac{5}{24}$ A1
(4 marks)
(8 marks)5. (a) Show that
$$\sin 3 x \equiv 3 \sin x - 4 \sin ^ { 3 } x$$
(b) Hence find, using algebraic integration,
$$\int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 3 } x d x$$
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\hfill \mbox{\textit{Edexcel P3 2020 Q5 [8]}}