Question #d49bf

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Question #d49bf

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Answer 1 · 2017-02-28T04:30:42

$= (cos (a + b) (cos (a - b)$ $= (cos (a + b)(cos (a - b)$

Explanation:

Use these trig identities:
cos (a + b) = cos a.cos b - sin a.sin b
cos (a - b) = cos a.cos b + sin a.sin b)
In this case:
${(cos a . cos b)}^{2} - {(sin a . sin b)}^{2} =$ $(cos a.cos b)^2 - (sin a.sin b)^2 =$
$= (cos a . cos b - sin a . sin b) (cos a . cos b + sin a . sin b) =$ $= (cos a.cos b - sin a.sin b)(cos a.cos b + sin a.sin b) =$
$= cos (a + b) (cos (a - b)$ $= cos (a + b)(cos (a - b)$

Answer 2 · 2017-02-28T05:39:25

${(cos a cos b)}^{2} - {(sin a sin b)}^{2}$ $(cosacosb)^2-(sinasinb)^2$

$= {cos}^{2} a {cos}^{2} b - {sin}^{2} a {sin}^{2} b$ $=cos^2acos^2b-sin^2asin^2b$

$= {cos}^{2} a (1 - {sin}^{2} b) - (1 - {cos}^{2} a) {sin}^{2} b$ $=cos^2a(1-sin^2b)-(1-cos^2a)sin^2b$

$= {cos}^{2} a - {cos}^{2} a {sin}^{2} b - {sin}^{2} b + {cos}^{2} a {sin}^{2} b$ $=cos^2a-cos^2asin^2b-sin^2b+cos^2asin^2b$

$= {cos}^{2} a - {sin}^{2} b$ $=cos^2a-sin^2b$

$= {cos}^{2} a - (1 - {cos}^{2} b)$ $=cos^2a-(1-cos^2b)$

$= {cos}^{2} a + {cos}^{2} b - 1$ $=cos^2a+cos^2b-1$

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Question #d49bf

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