How do you verify the following identity?

#(cos(x) - cos(y))/(sin(x)+sin(y)) + (sin(x)-sin(y))/(cos(x)+cos(y)) = 0#

1 Answer
Ken C.
Jul 4, 2016

See below.

Explanation:

A helpful starting point when verifying identities involving fractions is to add the fractions. That's what we'll start with here:
#(cosx - cosy)/(sinx+siny) + (sinx-siny)/(cosx+cosy) = 0#

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

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

As you can see, the binomials in the numerator multiplied very easily because they were differences of squares. Now we rearrange the terms a bit:
#(cos^2x-cos^2y+sin^2x-sin^2y)/((sinx+siny)(cosx+cosy))=0#

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

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

Recall the Pythagorean Identity #sin^2x+cos^2x=1#, or equivalently, #sin^2y+cos^2y=1#. Therefore:
#(1-1)/((sinx+siny)(cosx+cosy))=0#

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

#0=0#

Q.E.D.

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