How do you verify the identity #(cosx-cosy)/(sinx+siny)+(sinx-siny)/(cosx+cosy)=0#?

Question

22959 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T15:32:21

Put on a common denominator.

#((cosx - cosy)(cosx + cosy) +(sinx - siny)(sinx + siny))/((sinx + siny)(cosx + cosy)) = 0#

#(cos^2x - cos^2y + sin^2x - sin^2y)/((sinx + siny)(cosx + cosy)) = 0#

Use the pythagorean identity #sin^2x + cos^2x = 1#:

#(1 - cos^2y - sin^2y)/((sinx + siny)(cosx + cosy)) = 0#

Use the pythagorean identity mentioned above again, except this time in the form #sin^2x = 1 - cos^2x#.

#(sin^2y - sin^2y)/((sinx + siny)(cosx + cosy)) = 0#

#0/((sinx + siny)(cosx + cosy)) = 0#

#0 = 0#

#LHS = RHS#

Identity proved!

Hopefully this helps!

Answer 2 · 2017-01-04T15:32:26

See below.

#(cosx-cosy)/(sinx+siny)+(sinx-siny)/(cosx+cosy)=#
#=(cos^2x-cos^2y+sin^2x-sin^2y)/((sinx+siny)(cosx + cosy))=#
#=0/((sinx+siny)(cosx + cosy))=0#