Question #73fd2

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Question #73fd2

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Answer 1 · 2017-04-27T10:11:01

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Explanation:

We need

$(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b)=a^2-b^2$

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

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

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

$sin (x + y) = sin x cos y + sin y cos x$ $sin(x+y)=sinxcosy+sinycosx$

$sin (x - y) = sin x cos y - sin y cos x$ $sin(x-y)=sinxcosy-sinycosx$

Therefore,

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

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

$sin (x + y) \cdot sin (x - y) = (sin x cos y + sin y cos x) (sin x cos y - sin y cos x)$ $sin(x+y)*sin(x-y)=(sinxcosy+sinycosx)(sinxcosy-sinycosx)$

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

So,

$L H S = cos (x + y) \cdot cos (x - y) + sin (x + y) \cdot sin (x - y)$ $LHS=cos(x+y)*cos(x-y)+sin(x+y)*sin(x-y)$

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

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

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

$= L H S$ $=LHS$

$Q E D$ $QED$

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Question #73fd2

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