f(x) = 5 ext{cos}x + 12 ext{sin}x Given that f(x) = R ext{cos}(x - eta), where R > 0 and 0 < eta < \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 5

We start with:

$\cos(x - \alpha) = \frac{6}{13}.$

This implies:

$x - \alpha = \arccos\left(\frac{6}{13}\right) \approx 1.091.$

Hence, we find:

$x = 1.091 + 1.176 = 2.267.$

We must also consider the cosine symmetry, thus:

$x - \alpha = -\arccos\left(\frac{6}{13}\right) \Rightarrow x = -1.091 + 1.176 = 0.085.$

Both solutions must be within the interval [0, 2\pi). Therefore, valid solutions are:

• $x \approx 2.267$
• $x \approx 0.084.$

The maximum value occurs when:

$R_{max} = R = 13.$

The maximum occurs when:

$\cos(x - \alpha) = 1 \Rightarrow x - \alpha = 0 \Rightarrow x = \alpha \approx 1.176.$