# Question #52e7f

##### 1 Answer

- See explanation

#### Explanation:

Using the appropriate

#color(blue)"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)

#sinx=3/5 "allows opposite and hypotenuse to be completed"# Use

#color(blue)"Pythagoras' theorem"# to find adjacent = 4 or recognise the well known '3,4,5' triangle.

#sinx=3/5rArrcosx=4/5#

#"------------------------------------------------------"# Similarly

#cosy=5/13# allows adjacent and hypotenuse to be completed. Using Pythagoras or recognising '5,12,13' triangle to obtain opposite = 12.

#cosy=5/13rArrsiny=12/13#

#"-----------------------------------------------------------"# We now have all the required ratios to complete the expansion.

#cos(x-y)=cosxcosy+sinxsiny#

#=(4/5xx5/13)+(3/5xx12/13)#

#=20/65+36/65=56/65" thus shown"#