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

For $\cos q$, we use the other right-angled triangle, where the adjacent side is 6 and the hypotenuse is the diagonal which can be calculated using the Pythagorean theorem:

$\text{hypotenuse} = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}.$
Thus, $\cos q = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}.$

Using the cosine addition formula:

$\cos(p + q) = \cos p \cdot \cos q - \sin p \cdot \sin q.$
We already have $\cos p = \frac{4}{5}$ and $\cos q = \frac{3}{\sqrt{13}}$. Next, we need $\sin p$ and $\sin q$.
We can find them as follows:

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

For $\sin q$:

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

Substituting these values back into the addition formula:

$\cos(p + q) = \left(\frac{4}{5}\right) \left(\frac{3}{\sqrt{13}}\right) - \left(\frac{3}{5}\right) \left(\frac{2}{\sqrt{13}}\right) = \frac{12}{5\sqrt{13}} - \frac{6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}}.$