9. (a) Express $2 \, ext{sin} \theta - 4 \, \text{cos} \theta$ in the form $R \text{sin}(\theta - \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 5

To find the maximum value of
$H(\theta) = 4 + 5(2 \text{sin} 3\theta - 4 \text{cos} 3\theta)^2$
First, we need to evaluate the maximum of $2 \text{sin} 3\theta - 4 \text{cos} 3\theta$.
The maximum occurs when
$\sqrt{2^2 + (-4)^2} = \sqrt{20}$
The maximum value for the square is 20. Thus,
$H(\theta) = 4 + 5(\sqrt{20})^2 = 4 + 5(20) = 104$.

To find the value of $\theta$ where
$H(\theta) = 104$, we set
$2 \text{sin} 3\theta - 4 \text{cos} 3\theta = \sqrt{20}$.
We can use
$\tan(3\theta) = \frac{-4}{2}$, which gives us specific values for $3\theta$.
Solving gives us:
$3\theta = \frac{\pi}{2} + n\pi, \text{ for } n \in \mathbb{Z}$
Therefore, the least value of $\theta$ is
$\theta = \frac{\pi}{6} + \frac{n\pi}{3}$ where we take $n = 0$.

The minimum value occurs when the expression
$2 \text{sin} 3\theta - 4 \text{cos} 3\theta$ is minimized.
Its minimum value would be $-\sqrt{20}$.
Evaluating gives us:
$H(\theta) = 4 + 5(-\sqrt{20})^2 = 4 + 5(20) = 104$.

To find when this minimum occurs, we work with:
$3\theta = \frac{3\pi}{2} + n\pi$
Solving gives us $\theta = \frac{\pi}{2}$, thus the maximum point reflects the angular occurrence for the minimum value.