Question 4 - A-Level Maths

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2010
Session	June
Topic	Addition & Double Angle Formulae

Using the expansions of $\cos ( 3 x - x )$ and $\cos ( 3 x + x )$, prove that $$\frac { 1 } { 2 } ( \cos 2 x - \cos 4 x ) \equiv \sin 3 x \sin x$$
Hence show that $$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } \sin 3 x \sin x \mathrm {~d} x = \frac { 1 } { 8 } \sqrt { } 3$$