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

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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.

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

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