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

Given that ( \sin 30^{\circ} = \frac{1}{2} ), we can substitute this into the sum formula:

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

This is also a geometric series with first term ( \frac{1}{2} ) and common ratio ( \frac{1}{2} ). Thus,

$\sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1.$

Using the formula for the sum to infinity, we have:

$\sum_{n=0}^{\infty} (\cos \theta)^n = \frac{1}{1 - \cos \theta}.$

Setting this equal to ( 2 - \sqrt{2} ), we get:

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

Rearranging, we find:

$1 = (2 - \sqrt{2})(1 - \cos \theta) \Rightarrow 1 = 2 - \sqrt{2} - 2\cos \theta + \sqrt{2}\cos \theta.$

From this, we can isolate ( \cos \theta ) and solve to find:

$\cos \theta = \frac{1 - (2 - \sqrt{2})}{1 + \sqrt{2}} = \frac{\sqrt{2} - 1}{1 + \sqrt{2}}.$

Through trigonometric identities, we can deduce possible values for ( \theta ), identifying the smallest positive result as:

$\theta = \frac{3\pi}{4}.$