If $sin θ = \frac{\sqrt{403}}{22}$ $sintheta=sqrt403/22$ and $\frac{π}{2} < θ < π$ $pi/2<theta<pi$ , how do you find $tan 2 θ$ $tan2theta$ ?

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If $sin θ = \frac{\sqrt{403}}{22}$ $sintheta=sqrt403/22$ and $\frac{π}{2} < θ < π$ $pi/2<theta<pi$ , how do you find $tan 2 θ$ $tan2theta$ ?

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Answer 1 · 2016-11-20T21:18:15

$tan 2 θ = \frac{9 \sqrt{403}}{161}$ $tan2theta=(9sqrt403)/161$

Explanation:

If $θ \in [\frac{π}{2}, π]$ $theta in [pi/2,pi]$ , $cos θ < 0$ $costheta<0$ , so:
$cos θ = - \sqrt{1 - {sin}^{2} θ}$ $costheta=-sqrt{1-sin^2theta}$
$cos θ = - \sqrt{1 - \frac{403}{484}} = - \sqrt{\frac{81}{484}} = - \frac{9}{22}$ $costheta=-sqrt{1-403/484}=-sqrt{81/484}=-9/22$

$tan θ = \frac{sin θ}{cos θ} = \frac{\frac{\sqrt{403}}{22}}{- \frac{9}{22}} = - \frac{\sqrt{403}}{9}$ $tantheta=sintheta/costheta=(sqrt{403}/22)/(-9/22)=-sqrt403/9$
$tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ} = \frac{- \frac{2 \sqrt{403}}{9}}{1 - \frac{403}{81}} = \frac{- \frac{2 \sqrt{403}}{9}}{- \frac{322}{81}} = \frac{9 \sqrt{403}}{161}$ $tan2theta=(2tantheta)/(1-tan^2theta)=(-(2sqrt403)/9)/(1-403/81)=(-(2sqrt403)/9)/(-322/81)=(9sqrt403)/161$

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If $sin θ = \frac{\sqrt{403}}{22}$ $sintheta=sqrt403/22$ and $\frac{π}{2} < θ < π$ $pi/2<theta<pi$ , how do you find $tan 2 θ$ $tan2theta$ ?

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If $sin θ = \frac{\sqrt{403}}{22}$ $sintheta=sqrt403/22$ and $\frac{π}{2} < θ < π$ $pi/2<theta<pi$ , how do you find $tan 2 θ$ $tan2theta$ ?