How do you use half angle formula to find $\frac{tan (5 π)}{12}$ $tan (5pi)/12$ ?

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How do you use half angle formula to find $\frac{tan (5 π)}{12}$ $tan (5pi)/12$ ?

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Answer 1 · 2015-04-28T01:13:36

$tan (5 π) \equiv tan (π) = 0$ $tan(5pi) -= tan(pi) = 0$
and there is no point in using the half angle formula to evaluate
$\frac{tan (5 π)}{12}$ $(tan(5pi))/12$

Therefore, let's assume that you really want to evaluate
$tan (\frac{5 π}{12})$ $tan((5pi)/12)$

Half angle formula (for tan):
${tan}^{2} (\frac{θ}{2}) = \frac{1 - cos (θ)}{1 + cos (θ)}$ $tan^2(theta/2) = (1-cos(theta))/(1+cos(theta))$

${tan}^{2} (\frac{5 π}{12}) = tan (\frac{\frac{5 π}{6}}{2})$ $tan^2((5pi)/12) = tan(((5pi)/6)/2)$

$= \frac{1 - cos (\frac{π}{6})}{1 + cos (\frac{π}{6})}$ $=(1-cos(pi/6))/(1+cos(pi/6))$

$= \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}$ $= (1+sqrt(3)/2)/(1-sqrt(3)/2)$

$= \frac{{(1 + \frac{\sqrt{3}}{2})}^{2}}{1 - \frac{3}{4}}$ $= (1+sqrt(3)/2)^2/(1-3/4)$

$= 4 {(1 + \frac{\sqrt{3}}{2})}^{2}$ $= 4(1+sqrt(3)/2)^2$

That is
${tan}^{2} (\frac{5 π}{12}) = {(2 (1 + \frac{\sqrt{3}}{2}))}^{2}$ $tan^2((5pi)/12) = (2(1+sqrt(3)/2))^2$

Since $\frac{5 π}{12}$ $(5pi)/12$ is in Quadrant 1, the tan is positive
and
$tan (\frac{5 π}{12}) = 2 + \sqrt{3}$ $tan((5pi)/12) = 2+sqrt(3)$

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How do you use half angle formula to find $\frac{tan (5 π)}{12}$ $tan (5pi)/12$ ?

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