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Math Help - Simplifying trig expression

  1. #1
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    Simplifying trig expression

    I have to simplify (or get it in terms of tan I guess?) \cot (\frac{2\pi }{3} - x)



    I'm not sure how to get the reference angle and subtract the angle 'x' from it to get an expressional value...how would I do this?
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  2. #2
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    Re: Simplifying trig expression

    Quote Originally Posted by daigo View Post
    I have to simplify (or get it in terms of tan I guess?) \cot (\frac{2\pi }{3} - x)



    I'm not sure how to get the reference angle and subtract the angle 'x' from it to get an expressional value...how would I do this?
    tan2pi/3==-root3 I'm writing t for tanx

    tan(2pi/3-x) = (-root3-t)/(1+root3*t) So cot (2pi/3-x) = (1+root3*t)/(-root3-t) = -(1+root3*t)/root3+t)
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  3. #3
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    Re: Simplifying trig expression

    note the cofunction identity ...

    \cot{t} = \tan\left(\frac{\pi}{2} - t\right)

    so ...

    \cot\left(\frac{2\pi}{3} - x\right) = \tan\left[\frac{\pi}{2} - \left(\frac{2\pi}{3} - x\right)\right] = \tan\left(x -\frac{\pi}{6}\right)
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  4. #4
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    Re: Simplifying trig expression

    Got a sign wrong in my last reply. Should read
    tan(2pi/3-x)=(-root3-t)/(1-root3*t) So cot(2pi/3-x)=(1-root3*t)/(-root3-t)=-(1-root3*t)/(root3+t)= (root3*t-1)/(root3+t)
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