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  1. #1
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    Question

    is arctan(-pi/3) same thing as arctan(2pi/3)
    and arctan(-4pi/3) same thing as arctan(5pi/3)?


    if not, why is it that one?

    hence solve tanx=-\sqrt{3}

    thanks
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  2. #2
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    Quote Originally Posted by BabyMilo View Post

    hence solve tanx=-\sqrt{3}

    thanks
    In the case of

     \tan(x) =-\sqrt{3}

    In the first snd and 4th quadrants x = \pi - \frac{\pi}{3} , 2\pi - \frac{\pi}{3}= \frac{2\pi}{3} , \frac{5\pi}{3}

    As \tan(x) has the period \pi in general you can say

     x = \frac{2\pi}{3} \pm n\pi , n = 0,1,2,\dots
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  3. #3
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    Quote Originally Posted by BabyMilo View Post
    is arctan(-pi/3) same thing as arctan(2pi/3)
    and arctan(-4pi/3) same thing as arctan(5pi/3)?
    if not, why is it that one?
    The correct answer is: no they are not the same.
    The \arctan is injection (one-to-one function) from \mathbb{R} to \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right).
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  4. #4
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    but they do give the same solution on my calculator?
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  5. #5
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    Quote Originally Posted by BabyMilo View Post
    but they do give the same solution on my calculator?
    In that case, I would get a better calculator if I were you.
    \arctan \left( {\frac{{ - \pi }}{3}} \right) = -0.80844879263
     \arctan \left( {\frac{{2\pi }}{3}} \right) = 1.125338832884
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  6. #6
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    Quote Originally Posted by BabyMilo View Post
    is arctan(-pi/3) same thing as arctan(2pi/3)
    and arctan(-4pi/3) same thing as arctan(5pi/3)?

    These may have been meant to be written using tan instead of arctan,
    as the brackets contain common angles.


    if not, why is it that one?

    hence solve tanx=-\sqrt{3}

    thanks
    tan\left(-\frac{\pi}{3}\right)=tan\left(\frac{2{\pi}}{3}\rig  ht)

    tan\left(-\frac{4{\pi}}{3}\right)=tan\left(\frac{5{\pi}}{3}\  right)

    tan\left(\frac{\pi}{3}\right)=\sqrt{3}

    tan(-A)=-tanA

    tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}

    tan\left(-\frac{\pi}{3}\right)=tan\left(\frac{2{\pi}}{3}\rig  ht)

    Therefore, if

    tanx=-\sqrt{3}

    then

    x=arctan\left(\frac{2{\pi}}{3}\right)
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  7. #7
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    I just realize what i was asking.
    I was supposed to ask tan(2pi/3)
    same thing as tan(-pi/3)

    thanks
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