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    Prove that
    4 arccot2+arctan (24/7)=\pi

    so
    let
     arccot2=x, =>cotx=2=>tanx=1/2
    let
    arctan(24/7)=y =>tany=(24/7)

    therefore 4x+y=\pi
    \displaystyle\tan(4x+y)=\frac{ta  n(4x)+tan(y)}{1-tan(4x)(tan(y))}
    now to find tan(4x) is immensely long does anyone know of a quicker way to do this?

    Thanks
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    Double-angle forumla twice can't be that much effort?
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    Forgive me if I'm being stupid but surely tan(4x) = tan (pi -y) = -tan y

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    So this means tan(4x+y)=0 => 4x+y = npi

    and y<0.5pi and 4x<pi
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    Alternatively this could be done with complex numbers; after rewriting arccot 2, we wish to prove that:

     4\mathrm{arctan} \frac{1}{2} + \arctan \frac{24}{7} = \pi

    Noting that  \mathrm{arctan} \frac{a}{b} \equiv \mathrm{arg} (b + ai) (taking a principal argument), what if we consider the argument of the expression  (2+i)^4(7+24i) in two different ways? This is in fact almost a Machin-Like Formula.
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    (Original post by rbnphlp)
    ...
    (Original post by electriic_ink)
    Forgive me if I'm being stupid but surely tan(4x) = tan (pi -y) = -tan y
    Perhaps I am misreading this, but it seems to me that this result is assuming the very thing that you need to prove.
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    (Original post by ghostwalker)
    Perhaps I am misreading this, but it seems to me that this result is assuming the very thing that you need to prove.
    No, you're not. I've just realised the OP wrote 4x + y = pi. I misread that as fact but actually that is what he's supposed to be proving.
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    (Original post by ghostwalker)
    Perhaps I am misreading this, but it seems to me that this result is assuming the very thing that you need to prove.
    :o:

    @ unbounded:thanks unbounded seems like you had the alternative solution after all
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    (Original post by rbnphlp)
    :o:

    @ unbounded:thanks unbounded seems like you had the alternative solution after all
    Actually it really isn't that much work to use the double angle formula twice on tan(x)=1/2. You can almost work out that tan(2x)=4/3 in your head.
 
 
 
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