How do you use $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$ to find cos(-pi/12)?

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How do you use $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$ to find cos(-pi/12)?

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Answer 1 · 2016-03-02T16:50:08

$cos (- \frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $cos (-pi/12) = sqrt(2 + sqrt3)/2$

Explanation:

$cos (- \frac{π}{12}) = cos (\frac{π}{12})$ $cos (-pi/12) = cos (pi/12)$
$sin (\frac{π}{6}) = \frac{1}{2}$ $sin (pi/6) = 1/2$ -->
${cos}^{2} (\frac{π}{6}) = 1 - {sin}^{2} (\frac{π}{6}) = 1 - \frac{1}{4} = \frac{3}{4}$ $cos^2 (pi/6) = 1 - sin^2 (pi/6) = 1 - 1/4 = 3/4$ -->
$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} .$ $cos (pi/6) = sqrt3/2.$ (cos $(\frac{π}{6})$ $(pi/6)$ is positive)
Use the identity: $cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} = 2 {cos}^{2} (\frac{π}{12}) - 1$ $cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1$
$2 {cos}^{2} (\frac{π}{12}) = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$ $2cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/2$
${cos}^{2} (\frac{π}{12}) = \frac{2 + \sqrt{3}}{4}$ $cos^2 (pi/12) = (2 + sqrt3)/4$
$cos (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $cos (pi/12) = sqrt(2 + sqrt3)/2$ (since cos (pi/12) is positive)

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How do you use $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$ to find cos(-pi/12)?

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How do you use $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$ to find cos(-pi/12)?