Question

prove the identity (cosx+cosy)^2+(sinx-siny)^2=2+ 2 cos (x+y)

Answer 1

If you know that \( \cos(x+y)=\cos x \cos y - \sin x \sin y\) then you can expand the LHS and use the identity \(\cos^2x+\sin^2x =1\)

Answer 2

LHS:
\[ (\cos x + \cos y)^2+ (\sin x - \sin y)^2\\
=(\cos^2 x + 2\cos x \cos y + \cos^2 y) + (\sin^2 - 2\sin x \sin y + \sin^2y)\\
=\underbrace{\cos^2x+\sin^2x}+\underbrace{\cos^2y+\sin^2y}+2\cos x \cos y - 2\sin x \sin y\\
=\;\;\;\;\;\;\;\;\;\;1 \;\;\;\;\;\;\;\;\;\;+\;\;\;\;\;\;\;\;\ 1\;\;\;\;\;\;\;\;\;\;+ 2(\cos x\cos y - \sin x \sin y)\\
=2 + 2(\cos x\cos y - \sin x \sin y)\\
=2 + 2[\cos(x+y)]\]

Answer 3

thank you so much!

Answer 4

yw =]