12 (a) A geometric sequence has first term 1 and common ratio \frac{1}{2} - AQA - A-Level Maths Pure - Question 12 - 2022 - Paper 1

We know that ( \sin 30^{\circ} = \frac{1}{2} ). Therefore, we can rewrite the series as:

$\sum_{n=1}^{\infty} (\sin 30^{\circ})^{n} = \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^{n}$

This is again a geometric series where the first term is ( \frac{1}{2} ) and ( r = \frac{1}{2} ):

$S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$

For the sum of a geometric series: $S = \frac{a}{1 - r}$ where ( a = 1 ) and ( r = \cos \theta ). Setting up the equation:

$\frac{1}{1 - \cos \theta} = 2 - \sqrt{2}$

This leads to: $1 = (2 - \sqrt{2})(1 - \cos \theta)$

Expanding and rearranging, we find: $\cos \theta = 1 - \frac{1}{2 - \sqrt{2}}$

To simplify: $\cos \theta = \frac{2 - (1 - \sqrt{2})}{2 - \sqrt{2}} = \frac{1 + \sqrt{2}}{2 - \sqrt{2}}$

Using a numerical or algebraic approach, we can find:\n Given the equation: $\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \text{ or } \theta = \frac{5\pi}{3} \text{ in the first cycle.}$

However, we need the smallest positive solution: Thus, the answer is $\theta = \frac{3\pi}{4}.$