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
$cos (x - y) = cos (x) cos (y) + sin (x) sin (y)$ $cos(x-y)=Cos(x) Cos(y) + Sin(x) Sin(y)$
and
${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$
then substituting
$cos (x - y) = ⎛ ⎝ \sqrt{1 - {(\frac{3}{5})}^{2}} ⎞ ⎠ (\frac{5}{13}) + (\frac{3}{5}) \sqrt{1 - {(\frac{5}{13})}^{2}}$ $cos(x-y) = (sqrt[1 - (3/5)^2]) (5/13) + (3/5) sqrt[1 - (5/13)^2]$

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

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Question #12d17

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