Question #12d17

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Question #12d17

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Answer 1 · 2016-07-05T09:07:12

26) and 27)

Explanation:

26) We know that
$cos(x-y)=Cos(x) Cos(y) + Sin(x) Sin(y)$
and
$cos^2x+sin^2x=1$
then substituting
$cos(x-y) = (sqrt[1 - (3/5)^2]) (5/13) + (3/5) sqrt[1 - (5/13)^2]$

27) $cos(2x) = sin x = cos(pi/2-x)$
so
$2x=pi/2-x+2k pi$
solving for $x$
$x = pi/6+2/3k pi$ for $k = 0,pm1,pm2,...$

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Question #12d17

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