6. (i) Without using a calculator, find the exact value of $(\sin 22.5^{\circ} + \cos 22.5^{\circ})^{2}$ You must show each stage of your working - Edexcel - A-Level Maths Pure - Question 27 - 2013 - Paper 1

Starting with the equation:

$\cos 2\theta + \sin \theta = 1$, we can rewrite it as:

$\cos 2\theta = 1 - \sin \theta$.

Using the identity for $\cos 2\theta$, we substitute:

$2\sin^{2}\theta - 1 = 1 - \sin \theta$.

Rearranging this leads to:

$2\sin^{2}\theta + \sin \theta - 2 = 0$.

This can be factored or solved using the quadratic formula, recognizing:

The expression can be rearranged to the form $k \sin^{2}\theta - \sin \theta = 0$.

Thus, we find that $k = 2$.

Using the earlier derived expression:

$2\sin^{2}\theta + \sin \theta - 2 = 0$, we can factor this as:

$(2\sin \theta - 1)(\sin \theta + 2) = 0$.

From $2\sin \theta - 1 = 0$, we solve:

$\sin \theta = \frac{1}{2}$.

This gives possible solutions $\theta = 30^{\circ}, 150^{\circ}$ in the range $0 < \theta < 360^{\circ}$.

For $\sin \theta + 2 = 0$, this is not feasible since the sine function cannot be less than -1.

Thus, the valid solutions are:

$\theta = 30^{\circ}, 150^{\circ}$.