2023 H2 Maths Paper 1, Question 6

We use $\cos 2\theta = 2\cos^{2}\theta - 1$, i.e. $\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}$, and square it.

Applying the same identity again to $\cos^{2}(2\theta) = \frac{1 + \cos 4\theta}{2}$,

Since $R$ lies in the first quadrant, $y > 0$, so $y^{2} = \left( 9 - x^{2} \right)^{3/2}$. By the disc method,

Using $x = 3\sin\theta$, so $dx = 3\cos\theta\, d\theta$ and $9 - x^{2} = 9\cos^{2}\theta$,

When $x = 1.5$: $\sin\theta = 0.5 \Rightarrow \theta = \frac{\pi}{6}$. When $x = 3$: $\sin\theta = 1 \Rightarrow \theta = \frac{\pi}{2}$.

Using the result of (a),

Let $G(\theta) = \frac{\sin 4\theta}{32} + \frac{\sin 2\theta}{4} + \frac{3\theta}{8}$. At $\theta = \frac{\pi}{2}$, $\sin 4\theta = \sin 2\pi = 0$ and $\sin 2\theta = \sin\pi = 0$, so $G\left( \frac{\pi}{2} \right) = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$. At $\theta = \frac{\pi}{6}$, $\sin 4\theta = \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ and $\sin 2\theta = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, so