If sina =-5÷13 and a lies in quadrant 3 then the value of cos (a÷2)?

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If sina =-5÷13 and a lies in quadrant 3 then the value of cos (a÷2)?

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Answer 1 · 2018-05-04T03:40:27

$cos (\frac{a}{2}) = - \frac{\sqrt{26}}{26}$ $cos (a/2) = - sqrt26/26$

Explanation:

$sin a = - \frac{5}{13}$ $sin a = - 5/13$ .
First find cos a.
${cos}^{2} a = 1 - {sin}^{2} a = 1 - \frac{25}{169} = \frac{144}{169}$ $cos^2 a = 1 - sin^2 a = 1 - 25/169 = 144/169$
$cos a = \pm \frac{12}{13}$ $cos a = +- 12/13$ .
Since a lies in Quadrant 3, cos a is negative.
$cos a = - \frac{12}{13}$ $cos a = - 12/13$ .
To find cos (a/2), use trig identity:
$2 {cos}^{2} (\frac{a}{2}) = 1 + cos a$ $2cos^2 (a/2) = 1 + cos a$
$2 {cos}^{2} (\frac{a}{2}) = 1 - \frac{12}{13} = \frac{1}{13}$ $2cos^2 (a/2) = 1 - 12/13 = 1/13$
${cos}^{2} (\frac{a}{2}) = \frac{1}{26}$ $cos^2 (a/2) = 1/26$
$cos (\frac{a}{2}) = \pm \frac{1}{\sqrt{26}}$ $cos (a/2) = +- 1/sqrt26$ .
Since a lies in Quadrant 3, then, $\frac{a}{2}$ $a/2$ lies in Quadrant 2, and
$cos (\frac{a}{2})$ $cos (a/2)$ is negative.
$cos (\frac{a}{2}) = - \frac{1}{\sqrt{26}} = \frac{\sqrt{26}}{26}$ $cos (a/2) = - 1/sqrt26 = sqrt26/26$

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If sina =-5÷13 and a lies in quadrant 3 then the value of cos (a÷2)?

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