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,

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

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

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

$=cos^2y-sin^2y$

$=LHS$

$QED$