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Math Help - Trig identities!!!!!!!!!

  1. #1
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    Trig identities!!!!!!!!!

    if cotx= -2/3 and pi/2<x<pi

    solve for tan(x/2)

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  2. #2
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    \frac{\pi}{2}<x<\pi\;\implies\;\frac{\pi}{4}<\frac  {x}{2}<\frac{\pi}{2}
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  3. #3
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    Hello, andymac!

    If \cot x\:=\: -\frac{2}{3}\:\text{ and }\;\frac{\pi}{2} < x < \pi,\:\text{ find: }\tan \frac{x}{2}

    \cot x \:=\:-\frac{2}{3}\quad\Rightarrow\quad \tan x \:=\:-\frac{3}{2} \:=\:\frac{opp}{adj}

    Since x is in Quadrant 2, opp = 3,\:adj = -2 \quad\Rightarrow\quad hyp = \sqrt{13}
    . . Hence: . \cos x \:=\:-\frac{2}{\sqrt{13}}


    Half-angle identity: . \tan\frac{x}{2} \;=\;\pm\sqrt{\frac{1-\cos x}{1 + \cos x}}

    Since x is in Quadrant 2, \frac{x}{2} is in Quadrant 1.

    We have: . \tan\frac{x}{2} \;=\;\sqrt{\frac{1-(-\frac{2}{\sqrt{13}})}{1 + (-\frac{2}{\sqrt{13}})}} \;=\;\sqrt{\frac{\sqrt{13} + 2}{\sqrt{13} - 2}}

    Rationalize: . \tan\frac{x}{2} \;= \;\sqrt{\frac{\sqrt{13}+2}{\sqrt{13}-2}\cdot{\color{blue}\frac{\sqrt{13}+2}{\sqrt{13}+2  }}} \;=\;\sqrt{\frac{(\sqrt{13} + 2)^2}{9}}

    Therefore: . \boxed{\tan\frac{x}{2} \;=\;\frac{\sqrt{13}+2}{3}}

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