f(x)=7 \, cos \, 2x - 24 \, sin \, 2x Given that f(x)=R \, cos(2x + \alpha), where R > 0 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 5

Starting from the equation:
[ 7 \cos 2x - 24 \sin 2x = 12.5 ]
Divide through by R (which is 25):
[ \frac{7}{25} \cos 2x - \frac{24}{25} \sin 2x = \frac{12.5}{25} ]
This simplifies to:
[ \cos(2x + \alpha) = 0.5 ]
From this, we can find
[ 2x + \alpha = 60^{\circ} ; \text{or} ; 300^{\circ} ]
Thus:
[ 2x = 60^{\circ} - \alpha ; \text{or} ; 300^{\circ} - \alpha ]
Calculating for each:
For 60:
[ x = \frac{60^{\circ} - 73.7^{\circ}}{2} ]
For 300:
[ x = \frac{300^{\circ} - 73.7^{\circ}}{2} ]
The values of x fall within the range:
[ x \approx 13.2^{\circ} ]
[ x \approx 113.2^{\circ} ]
These are the solutions between 0 and 180 degrees.

Using the double angle identities, we can rewrite:

[ \cos 2x = 2 \cos^{2} x - 1 ]
[ \sin 2x = 2 \sin x \cos x ]
Thus,
[ 14 \cos^{2} x - 48 \sin x \cos x = 7 \cdot 2 \cos^{2} x - 24 \cdot 2 \sin x \cos x ]
This rearranges to:
[ 14 \cos^{2} x - 48 \sin x \cos x = 7 \cos 2x - 24 \sin 2x + 0 ]
where a = 7, b = -24, and c = 0.

The maximum value of an expression of the form a \cos 2x + b \sin 2x is given by:
[ R = \sqrt{a^2 + b^2} ]
Substituting the values:
[ R = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 ]
Therefore, the maximum value is 25.