Long Question 5

Question 5 (10 marks):

\begin{array}{l} (a) Prove \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) = \cos 3 x \\ (b) Hence, \\ (i) solve the equation s i n (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2} for 0 \leq x \leq 2 π \\ and give your answer in the simplest fraction form in terms of π radian . \\ (i i) sketch the graph of y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \end{array}

Solution:

\begin{array}{l} (a) Left hand side, \\ \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) \\ = [\sin 3 x \cos \frac{π}{6} + \cos 3 x \sin \frac{π}{6}] - [\sin 3 x \cos \frac{π}{6} - \cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x (\frac{1}{2})] \\ = \cos 3 x (right hand side) \end{array}

\begin{array}{l} (b) (i) \\ \sin (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2}, 0 \leq x \leq 2 π \\ \cos \frac{3 x}{2} = \frac{1}{2} \\ \frac{3 x}{2} = \frac{π}{3}, (2 π - \frac{π}{3}), (2 π + \frac{π}{3}) \\ \frac{3 x}{2} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3} \\ x = \frac{2 π}{9}, \frac{10 π}{9}, \frac{14 π}{9} \end{array}

\begin{array}{l} (b) (i i) \\ y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \\ y = \cos 3 x - \frac{1}{2} \end{array}