Proof cos(x+y) cos(x-y)= cos2x + cos2y - 1 ?

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Answer 1 · 2017-03-13T13:28:37

$cos (x + y) cos (x - y) = \frac{1}{2} (cos 2 x + cos 2 y)$ $cos(x+y)cos(x-y)=1/2(cos2x+cos2y)$

not $cos 2 x + cos 2 y - 1$ $cos2x+cos2y-1$

$cos (x + y) cos (x - y)$ $cos(x+y)cos(x-y)$

= $(cos x cos y - sin x sin y) (cos x cos y + sin x sin y)$ $(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)$

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

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

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

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

= $\frac{1}{2} (2 {cos}^{2} x) - \frac{1}{2} (2 {sin}^{2} y)$ $1/2(2cos^2x)-1/2(2sin^2y)$

= $\frac{1}{2} (1 + cos 2 x) - \frac{1}{2} (1 - cos 2 y)$ $1/2(1+cos2x)-1/2(1-cos2y)$

= $\frac{1}{2} (cos 2 x + cos 2 y)$ $1/2(cos2x+cos2y)$