Given $tan θ = - \frac{5}{12}$ $tantheta=-5/12$ and $\frac{3 π}{2} < θ < 2 π$ $(3pi)/2<theta<2pi$ , how do you find $tan (\frac{θ}{2})$ $tan(theta/2)$ ?

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Given $tan θ = - \frac{5}{12}$ $tantheta=-5/12$ and $\frac{3 π}{2} < θ < 2 π$ $(3pi)/2<theta<2pi$ , how do you find $tan (\frac{θ}{2})$ $tan(theta/2)$ ?

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Answer 1 · 2016-11-25T12:09:05

You use the formula of the tangent of the double angle and you solve the resulting equation choosing the right solution based on the interval specified

Explanation:

$tan (θ) = \frac{2 tan (\frac{θ}{2})}{1 - {tan}^{2} (\frac{θ}{2})}$ $tan(theta) = (2tan(theta/2))/(1-tan^2(theta/2)$

or

$- \frac{5}{12} = \frac{2 tan (\frac{θ}{2})}{1 - {tan}^{2} (\frac{θ}{2})}$ $-5/12 = (2tan(theta/2))/(1-tan^2(theta/2)$

Pose $t = tan (\frac{θ}{2})$ $t = tan(theta/2)$

You have:

$5 t^{2} - 24 t - 5 = 0$ $5t^2-24t-5=0$

$t = \frac{12 \pm \sqrt{144 + 25}}{5}$ $t=(12+-sqrt(144+25))/5$

If $\frac{3 π}{2} < θ < 2 π$ $(3pi)/2 < theta < 2pi$ then $\frac{3 π}{4} < \frac{θ}{2} < π$ $(3pi)/4 < theta/2 < pi$ , and the tangent is negative in this interval, so:

$tan (\frac{θ}{2}) = \frac{12 - \sqrt{169}}{5} = - \frac{1}{5}$ $tan(theta/2) = (12 -sqrt(169))/5 = -1/5$

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Given $tan θ = - \frac{5}{12}$ $tantheta=-5/12$ and $\frac{3 π}{2} < θ < 2 π$ $(3pi)/2<theta<2pi$ , how do you find $tan (\frac{θ}{2})$ $tan(theta/2)$ ?

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Explanation:

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Given tantheta=-5/12tanθ=−512tantheta=-5/12 and (3pi)/2<theta<2pi3π2<θ<2π(3pi)/2<theta<2pi, how do you find tan(theta/2)tan(θ2)tan(theta/2)?

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Given $tan θ = - \frac{5}{12}$ $tantheta=-5/12$ and $\frac{3 π}{2} < θ < 2 π$ $(3pi)/2<theta<2pi$ , how do you find $tan (\frac{θ}{2})$ $tan(theta/2)$ ?