The diagram shows two right-angled triangles with angles p and q as marked - Scottish Highers Maths - Question 4 - 2023

Similarly, to find ( \cos q ), we look at the triangle with angle q. Here, the adjacent side is 6 and the hypotenuse is 5. Using the cosine definition again,

$\cos q = \frac{6}{\sqrt{(4^2 + 6^2)}} = \frac{6}{\sqrt{52}} = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$

Using the sum of angles formula for cosine:

$\cos(p + q) = \cos p \cos q - \sin p \sin q$

We already found ( \cos p = \frac{4}{5} ) and ( \cos q = \frac{3}{\sqrt{13}} ). Now we need ( \sin p ) and ( \sin q ).

From the triangle for ( p ): $\sin p = \sqrt{1 - \cos^2 p} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$

For ( q ): $\sin q = \sqrt{1 - \cos^2 q} = \sqrt{1 - \left(\frac{3}{\sqrt{13}}\right)^2} = \sqrt{1 - \frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$

Substituting these into the cosine addition formula gives:

$\cos(p + q) = \left(\frac{4}{5}\right) \left(\frac{3}{\sqrt{13}}\right) - \left(\frac{3}{5}\right) \left(\frac{2}{\sqrt{13}}\right)$

This simplifies to:

$\frac{12}{5\sqrt{13}} - \frac{6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}}$