The first three terms of a geometric sequence are $\sin x,{\text{ }}\sin 2x$ and $4\sin x{\cos ^2}x,{\text{ }} - \frac{\pi }{2} < x < \frac{\pi }{2}$.

(a)     Find the common ratio r.

(b)     Find the set of values of x for which the geometric series $\sin x + \sin 2x + 4\sin x{\cos ^2}x +  \ldots$ converges.

Consider $x = \arccos \left( {\frac{1}{4}} \right),{\text{ }}x > 0$.

(c)     Show that the sum to infinity of this series is $\frac{{\sqrt {15} }}{2}$.

(a)     $\sin x,{\text{ }}\sin 2x{\text{ and }}4\sin x{\cos ^2}x$

$r = \frac{{2\sin x\cos x}}{{\sin x}} = 2\cos x$     A1

Note:     Accept $\frac{{\sin 2x}}{{\sin x}}$.

[1 mark]

(b)     EITHER

$\left| r \right| < 1 \Rightarrow \left| {2\cos x} \right| < 1$     M1

OR

$- 1 < r < 1 \Rightarrow  - 1 < 2\cos x < 1$     M1

THEN

$0 < \cos x < \frac{1}{2}{\text{ for }} - \frac{\pi }{2} < x < \frac{\pi }{2}$

$- \frac{\pi }{2} < x <  - \frac{\pi }{3}{\text{ or }}\frac{\pi }{3} < x < \frac{\pi }{2}$     A1A1

[3 marks]

(c)     ${S_\infty } = \frac{{\sin x}}{{1 - 2\cos x}}$     M1

${S_\infty } = \frac{{\sin \left( {\arccos \left( {\frac{1}{4}} \right)} \right)}}{{1 - 2\cos \left( {\arccos \left( {\frac{1}{4}} \right)} \right)}}$

$= \frac{{\frac{{\sqrt {15} }}{4}}}{{\frac{1}{2}}}$     A1A1

Note: Award A1 for correct numerator and A1 for correct denominator.

$= \frac{{\sqrt {15} }}{2}$     AG

[3 marks]

Total [7 marks]