Question #73fd2

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Answer 1 · 2017-04-27T10:11:01

See proof below

We need

#(a+b)(a-b)=a^2-b^2#

#sin^2x+cos^2x=1#

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

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

#sin(x+y)=sinxcosy+sinycosx#

#sin(x-y)=sinxcosy-sinycosx#

Therefore,

#cos(x+y)*cos(x-y)=(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)#

#=cos^2xcos^2y-sin^2xsin^2y#

#sin(x+y)*sin(x-y)=(sinxcosy+sinycosx)(sinxcosy-sinycosx)#

#=sin^2xcos^2y-sin^2ycos^2x#

So,

#LHS=cos(x+y)*cos(x-y)+sin(x+y)*sin(x-y)#

#=cos^2xcos^2y-sin^2xsin^2y+sin^2xcos^2y-sin^2ycos^2x#

#=cos^2y(cos^2x+sin^2x)-sin^2y(sin^2x+cos^2x)#

#=cos^2y-sin^2y#

#=LHS#

#QED#