Express 2sinθ − 1.5cosθ in the form R sin(θ − α), where R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 5

The maximum value of 2sinθ − 1.5cosθ occurs when the sine function equals its maximum value, which is 1. Substituting this value gives:

$ext{Max Value} = 2(1) - 1.5 imes 0 = 2$

The maximum occurs at θ when sinθ is at its peak. Therefore:

heta = rac{π}{2}

To find the maximum value of H, we assess the equation:

$H = 6 + 2 imes 1 - 1.5 imes (-1) = 6 + 2 + 1.5 = 9.5$

To find t when this maximum occurs, we solve for t:

t = rac{ ext{expression where sine and cosine reach max}}{4} t/25 = k

where k needs to be determined based on max values of sine and cosine surfacing when both are maximized. Solving, the value lies around t = 5.5 hours.

To determine when H = 7:

Starting from the equation:

$7 = 6 + 2 imes 0.5 - 1.5 imes cos(4πt / 25)$

We isolate cos(4πt / 25), yielding:

$0.5 = -1.5cos(4πt / 25)$

$cos(4πt / 25) = -\frac{1}{3}$

Using the cos inverse function, we can resolve:

$4πt / 25 = ext{cos}^{-1}(-1/3)$

This gives approximate solutions that can be converted into time, ultimately yielding the minute callback needed in real terms (_rescue of energy formula needed).