Triangles ABC and ADE are both right angled - Scottish Highers Maths - Question 13 - 2019

Now we find cos q using triangle ADE.

Using the same definition of cosine:

$\cos q = \frac{\text{adjacent side}}{\text{hypotenuse}}$

The length of DE (adjacent side) is 1, and the hypotenuse AD is given as (\sqrt{10}). Thus,

$\cos q = \frac{1}{\sqrt{10}}$

Using the angle addition formula for sine:

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

We need to find (\sin p) and (\sin q). From the cosine values computed:

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

Similarly,

$\sin q = \sqrt{1 - \left(\cos q\right)^2} = \sqrt{1 - \left(\frac{1}{\sqrt{10}}\right)^2} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$

Substituting the values into the sine addition formula:

$\sin(p + q) = \left(\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{10}}\right) + \left(\frac{1}{\sqrt{5}}\right) \left(\frac{3}{\sqrt{10}}\right)$

$= \frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{2 + 3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$