How do you prove $sin(x+y)*sin(x-y)=(sinx)^2-(siny)^2$ ?

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Answer 1 · 2015-04-16T16:46:45

Using the angle sum and difference formula of the function sinus on the first member:

$(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=$

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

$=sin^2x(1-sin^2y)-(1-sin^2x)sin^2y=$

$=sin^2x-sin^2xsin^2y-sin^2y+sin^2xsin^2y=$

$=sin^2x-sin^2y$

that is the second member.

How do you prove sin(x+y)*sin(x-y)=(sinx)^2-(siny)^2sin(x+y)*sin(x-y)=(sinx)^2-(siny)^2?