(i) To show that \(\cos 2θ = \frac{2 \sin 2θ - π}{4θ}\), we start by considering the geometry of the circle and the shaded area. The area of the circle is \(πr^2\), so half the area is \(\frac{1}{2}πr^2\).

The area of sector OAB is \(\frac{1}{2}r^2θ\), and the area of triangle OAB is \(\frac{1}{2}r^2 \sin θ\).

The area of the segment ABC is \(\frac{1}{2}r^2(θ - \sin θ)\).

Equating the shaded area to half the circle's area, we have:

\(\frac{1}{2}r^2θ + \frac{1}{2}r^2(θ - \sin θ) = \frac{1}{2}πr^2\)

Simplifying gives:

\(r^2θ - \frac{1}{2}r^2 \sin θ = \frac{1}{2}πr^2\)

Dividing through by \(r^2\) and rearranging gives:

\(θ - \frac{1}{2} \sin θ = \frac{1}{2}π\)

Using the double angle identity \(\cos 2θ = 1 - 2 \sin^2 θ\), we can express \(\sin θ\) in terms of \(θ\) and solve for \(\cos 2θ\).

(ii) Using the iterative formula \(θ_{n+1} = \frac{1}{2} \cos^{-1} \left( \frac{2 \sin 2θ_n - π}{4θ_n} \right)\), start with \(θ_1 = 1\).

Calculate successive iterations:

\(θ_2 = \frac{1}{2} \cos^{-1} \left( \frac{2 \sin 2(1) - π}{4(1)} \right)\)

Continue iterating until the value stabilizes to 4 decimal places.

The final value of \(θ\) correct to 2 decimal places is 0.95.