f(x) = 7 \, \cos \, 2x - 24 \, \sin \, 2x Given that f(x) = R \cos(2x + \alpha), \text{ where } R > 0 \text{ and } 0 < \alpha < 90^\circ; (a) find the value of R and the value of \alpha - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 6

Using the values of R and α, we will re-write the equation:

$R \cos(2x + \alpha) = 12.5$

Substituting R and α, we have:

$25 \cos(2x + 73.7^\circ) = 12.5$

Dividing both sides by 25:

$\cos(2x + 73.7^\circ) = 0.5$

This leads us to:

$2x + 73.7^\circ = 60^\circ \, \text{ or } \, 2x + 73.7^\circ = 300^\circ$
A secondary angle yields:

1. For the first equation:

$2x = 60^\circ - 73.7^\circ = -13.7^\circ \implies x = -6.85^\circ \, (\text{not in range})$

2. For the second equation:

$2x = 300^\circ - 73.7^\circ = 226.3^\circ \implies x = 113.15^\circ$

Thus, the only valid solution in the range is:

$x = 113.2^\circ$ (to 1 decimal place).

Using the double angle identities:

$\cos 2x = 2 \cos^2 x - 1$
$\sin 2x = 2 \sin x \cos x$

We rewrite the expression:

$14 \cos^{2} x - 48 \sin x \cos x$
= $14 \left(\frac{\cos 2x + 1}{2}\right) - 48 \left(\frac{\sin 2x}{2}\right)$

= $7 \cos 2x + 7 - 24 \sin 2x$

Thus, comparing with the form a cos 2x + b sin 2x + c:
a = 7, b = -24, c = 7.

To find the maximum value of:

$14 \cos^{2} x - 48 \sin x \cos x$, we use the derived expression:

$= 7 \cos 2x - 24 \sin 2x + 7$

The maximum value of 7 cos 2x - 24 sin 2x can be determined by using the maximum modulus:

$\sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = 25$, so

The total maximum value becomes:

$25 + 7 = 32$.