Question 15 - A-Level Maths

Show that \(\left( 1 - \frac { 1 } { 4 } \mathrm { e } ^ { 2 \mathrm { i } \theta } \right) \left( 1 - \frac { 1 } { 4 } \mathrm { e } ^ { - 2 \mathrm { i } \theta } \right) = \frac { 1 } { 16 } ( 17 - 8 \cos 2 \theta )\)
[0pt] [3 marks]
15
Given that the series \(\mathrm { e } ^ { 2 \mathrm { i } \theta } + \frac { 1 } { 4 } \mathrm { e } ^ { 4 \mathrm { i } \theta } + \frac { 1 } { 16 } \mathrm { e } ^ { 6 \mathrm { i } \theta } + \frac { 1 } { 64 } \mathrm { e } ^ { 8 \mathrm { i } \theta } + \ldots\). has a sum to infinity, express this sum to infinity in terms of \(\mathrm { e } ^ { 2 \mathrm { i } \theta }\)
15
Hence show that \(\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 4 ^ { n - 1 } } \cos 2 n \theta = \frac { 16 \cos 2 \theta - 4 } { 17 - 8 \cos 2 \theta }\)
[0pt] [4 marks]
15
Deduce a similar expression for \(\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 4 ^ { n - 1 } } \sin 2 n \theta\)
[0pt] [1 mark]