How do you evaluate $tan (\frac{7 π}{12})$ $tan ((7pi)/12)$ ?

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How do you evaluate $tan (\frac{7 π}{12})$ $tan ((7pi)/12)$ ?

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Answer 1 · 2016-03-17T13:49:14

$- \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2} - \sqrt{3}}$ $- sqrt(2 + sqrt3)/(sqrt2 - sqrt3)$

Explanation:

On the trig unit circle,
$tan (\frac{7 π}{12}) = tan (\frac{π}{12} + \frac{6 π}{12}) = tan (\frac{π}{12} + \frac{π}{2}) = - cot (\frac{π}{12})$ $tan ((7pi)/12) = tan (pi/12 + (6pi)/12) = tan (pi/12 + pi/2) = -cot (pi/12)$ .
Find $cot (\frac{π}{12}) = \frac{cos (\frac{π}{12})}{sin (\frac{π}{12})}$ $cot (pi/12) = cos (pi/12)/sin (pi/12)$
Apply the trig identity: $cos 2 a = 2 {cos}^{2} a - 1 = 1 - 2 {sin}^{2} a .$ $cos 2a = 2cos^2 a - 1 = 1 - 2sin^2 a.$
$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} = 2 {cos}^{2} (\frac{π}{12}) - 1 .$ $cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1.$
${cos}^{2} (\frac{π}{12}) = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{4}$ $cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/4$
$cos (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $cos (pi/12) = sqrt(2 + sqrt3)/2$ --> (since (pi/12) is in Quadrant I)
By the same way:
$sin (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin (pi/12) = sqrt(2 - sqrt3)/2$
Finally:
$tan (\frac{7 π}{12}) = - \frac{cos}{sin} = - \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}$ $tan ((7pi)/12) = - cos/(sin) = - sqrt(2 + sqrt3)/(sqrt(2 - sqrt3))$

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How do you evaluate $tan (\frac{7 π}{12})$ $tan ((7pi)/12)$ ?

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