# Question 52e7f

Jun 17, 2016
1. See explanation

#### Explanation:

Using the appropriate $\textcolor{b l u e}{\text{Addition Formulae}}$

color(red)(|bar(ul(color(white)(a/a)color(black)(cos(A±B)=cosAcosB∓sinAsinB)color(white)(a/a)|)))#

hence cos(x-y)=cosxcosy+sinxsiny

We are given sinx and cosy but require to find cosx and siny to evaluate the right side of the expansion.

Using the following sketches (not to scale)

$\sin x = \frac{3}{5} \text{allows opposite and hypotenuse to be completed}$

Use $\textcolor{b l u e}{\text{Pythagoras' theorem}}$ to find adjacent = 4 or recognise the well known '3,4,5' triangle.

$\sin x = \frac{3}{5} \Rightarrow \cos x = \frac{4}{5}$
$\text{------------------------------------------------------}$

Similarly $\cos y = \frac{5}{13}$ allows adjacent and hypotenuse to be completed. Using Pythagoras or recognising '5,12,13' triangle to obtain opposite = 12.

$\cos y = \frac{5}{13} \Rightarrow \sin y = \frac{12}{13}$
$\text{-----------------------------------------------------------}$

We now have all the required ratios to complete the expansion.

$\cos \left(x - y\right) = \cos x \cos y + \sin x \sin y$

$= \left(\frac{4}{5} \times \frac{5}{13}\right) + \left(\frac{3}{5} \times \frac{12}{13}\right)$

$= \frac{20}{65} + \frac{36}{65} = \frac{56}{65} \text{ thus shown}$