How do you verify the identity #cos(x+y)+cos(x-y)=2cosxcosy#?

2 Answers
Aug 16, 2017

Remember your formula:

#cos(x + y) = (cosx * cosy) - (sinx*siny)#

Now, try this:

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

...so you can apply your formula again:

# = cosx * cos(-y) - sinx * sin(-y)#

Now here's the trick: remember that cosine is a symmetrical function about x = 0. This means that cos(-y) = cos(y) for all y.
Sine, however, is NOT symmetrical. sin(-y) = -sin(y) for all y.
(look at the graphs of these functions to verify this).

So you can rewrite #cos(x-y)# as:

#cosx * cosy - (sinx * (-siny))#
#= (cosx*cosy) + (sinx * siny)#

So therefore:
#cos(x + y) + cos(x - y) = #

#((cosx * cosy) - (sinx*siny)) + ( (cosx*cosy) + (sinx * siny))#

#= (cosx * cosy) + (cosx * cosy)#

#= 2(cosx * cosy)#

Aug 16, 2017

#"see explanation"#

Explanation:

#"using the "color(blue)"addition formulae for cosine"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(cos(A+-B)=cosAcosB∓sinAsinB)color(white)(2/2)|)))#

#"left side "#

#cos(x+y)+cos(x-y)#

#=cosxcosycancel(-sinxsiny)+cosxcosycancel(+sinxsiny)#

#=2cosxcosy=" right side "rArr" verified"#