How do you prove ${sin}^{2} x - {sin}^{2} y = sin (x + y) sin (x - y)$ $Sin^2x-sin^2y=sin(x+y)sin(x-y)$ ?

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Answer 1 · 2016-05-24T06:19:12

Please see below.

$sin (x + y) sin (x - y)$ $sin(x+y)sin(x-y)$

= $(sin x cos y + cos x sin y) (sin x cos y - cos x sin y)$ $(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)$

= ${sin}^{2} x {cos}^{2} y - {cos}^{2} x {sin}^{2} y$ $sin^2xcos^2y-cos^2xsin^2y$

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

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

= $sin^2x-cancel(sin^2xsin^2y)-sin^2y+cancel(sin^2xsin^2y)$

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

How do you prove Sin^2x-sin^2y=sin(x+y)sin(x-y)sin2x−sin2y=sin(x+y)sin(x−y)Sin^2x-sin^2y=sin(x+y)sin(x-y)?