Express $2 \sin \theta - 1.5 \cos \theta$ in the form $R \sin(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 5

The maximum value occurs when $\sin(\theta - \alpha)$ equals 1. Therefore, the maximum value of $2 \sin \theta - 1.5 \cos \theta$ is:

$R = 2.5$

Thus, the maximum value is $2.5$.

The maximum occurs when $\sin(\theta - \alpha) = 1$, which means:

$\theta - \alpha = \frac{\pi}{2}$

Therefore, solving for $\theta$ gives:

$\theta = \frac{\pi}{2} + \alpha \approx 1.2140$

This is the value of $\theta$ when the maximum occurs.

To find the maximum value of $H$, we need to evaluate:

$H = 6 + 2 \times 2.5 = 11$

To find $t$ when $H$ is maximum, we set the derivative with respect to $t$ to zero and solve for $t$:

$H = 6 + 2 \sin(\left(\frac{4\pi}{25}t\right)) - 1.5 \cos(\left(\frac{4\pi}{25}t\right))$

Setting $\frac{dH}{dt} = 0$ gives the critical points within the interval. Substitute to find: $\frac{dH}{dt} = 0$ leads to $t$ approximately equal to the point where the sine function is maximized, yielding:

$t \approx 3.67$

To determine when $H = 7$, we solve:

$7 = 6 + 2 \sin(\left(\frac{4\pi}{25}t\right)) - 1.5 \cos(\left(\frac{4\pi}{25}t\right))$

This simplifies to:

$\sin(\left(\frac{4\pi}{25}t\right)) - \frac{1.5}{2} \cos(\left(\frac{4\pi}{25}t\right)) = 0.5$

This gives us the values of $t$ within the interval $0 < t < 12$. Solving this numerically or graphically, we can find the times to the nearest minute.