In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2

Using the quadratic formula for the equation:

$4 \sin^2 \theta - 52 \sin \theta + 25 = 0$

We set ( a = 4, b = -52, c = 25 ). The formula is given by:

$\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Calculating the discriminant:

$b^2 - 4ac = (-52)^2 - 4 \cdot 4 \cdot 25 = 2704 - 400 = 2304$

Taking the square root:

$\sqrt{2304} = 48$

Now substituting into the quadratic formula:

$\sin \theta = \frac{52 \pm 48}{8}$

Calculating the two potential solutions:

1. $\sin \theta = \frac{100}{8} = 12.5$ (not valid for sin)
2. $\sin \theta = \frac{4}{8} = 0.5$

Thus, ( \theta = 180° - 30° = 150° \text{ or } rac{5\pi}{6} \text{ radians} $$, since θ is obtuse.

The formula for the sum to infinity of a geometric series is:

$S = \frac{a}{1 - r}$

where ( a ) is the first term and ( r ) is the common ratio. We already established the terms:

1. First term, ( a = 12 \cos \theta = 12 \cos(150°) = 12 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -6\sqrt{3} $$
2. Now, using the established common ratio from part (a):

$r = \frac{5 + 2 \sin(150°)}{12 \cos(150°)}$

Calculating ( \sin(150°) = \frac{1}{2} ) and ( \cos(150°) = -\frac{\sqrt{3}}{2} ):

$r = \frac{5 + 2 \cdot \frac{1}{2}}{12 \cdot -\frac{\sqrt{3}}{2}} = \frac{6}{-6\sqrt{3}} = -\frac{1}{\sqrt{3}}$

Substituting into the sum formula:

$S = \frac{-6\sqrt{3}}{1 - \left(-\frac{1}{\sqrt{3}}\right)} = \frac{-6\sqrt{3}}{1 + \frac{1}{\sqrt{3}}} = \frac{-6\sqrt{3} \cdot \sqrt{3}}{\sqrt{3} + 1} = \frac{-18}{\sqrt{3} + 1}$

Thus, we express this in the required form:

$S = k(1 - \sqrt{3})$

Finding ( k):

This implies that k can be calculated to find the constant that satisfies the equation.