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Homework Help: Trig proof

  1. Dec 5, 2005 #1
    a teacher wrote on the board

    tan-1(a) + tan-1(b) + tan-1(c) = pi
    [they are inverse trig functions btw, not the recipricol 1/tanx = cotx]

    hence prove that

    a + b + c = abc

    wow. do you have any idea where i can start? thanks. i've been uttered clueless.
     
  2. jcsd
  3. Dec 5, 2005 #2

    siddharth

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    You should first find what
    [tex] \tan^{-1} x + \tan^{-1} y [/tex] is.


    That is, can you find the (??) in the equation below?
    [tex] \tan^{-1} x + \tan^{-1} y = \tan^{-1} (??) [/tex]
    Finding the above is very interesting.
    You have to consider the cases when

    xy<1
    xy>1 and x,y >0
    xy>1 and x,y <0
     
    Last edited: Dec 5, 2005
  4. Dec 5, 2005 #3
    where did x and y come from agn?
     
  5. Dec 5, 2005 #4

    siddharth

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    x and y are just two variables. You can replace them with a and b if you wish.
     
  6. Dec 5, 2005 #5
    so you mean:
    [tex]
    arctan(a) + arctan(b) + arctan(c) = \pi
    [/tex]
     
  7. Dec 5, 2005 #6

    Tide

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    If the first statement is true then a, b and c can be represented by the interior angles of a triangle. There is no triangle for which [itex]a + b + c = a \cdot b \cdot c[/itex] because the maximum possible value of [itex]a \cdot b \cdot c[/itex] is [itex](\pi/3)^3[/itex] which is less than [itex]\pi[/itex].
     
  8. Dec 5, 2005 #7

    NateTG

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    An example solution is:
    [tex]a=b=c=\sqrt{3}[/tex]
    so
    [tex]\tan^{-1}(a)=\tan^{-1}(b)=\tan^{-1}(c)=\frac{\pi}{3}[/itex]
    Then
    [tex]abc=3 \sqrt{3}[/tex]
    and
    [tex]a+b+c= 3 \sqrt{3}[/tex]
     
  9. Dec 5, 2005 #8

    Tide

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    Yikes! What was I thinking!

    Thanks for catching that, Nate!
     
  10. Dec 6, 2005 #9
    er, i guess i got it

    hehe i got it that makes more sort of. systematic sense.

    using the double angle for tan thing

    tan [a + b] = tan (a) + tan (b)
    ---------------
    1 - tan(a)tan(b)

    state that tan (a) = A
    tan (b) = B you'll see why later.

    anyways take the arctan of both sides of the double identity for tan and you get

    a + b = arctan [tan (a) + tab (b) / 1 -tan(a)tan(b)]

    now becuase tan (a) = A
    a = arctan A and vice vresa for B

    hence you end up wiht the arctan identity

    arctan (A) + arctan (B) = arctan [(A+B)/(1-AB)]

    and then you use that for a, b, and c you end up with the simple equation that tan pi = 0, hence

    a + b + c - abc = 0
    hence
    a + b + c = abc

    try that method. just for those of you who awanted a more systematic proof and that made more induction sort of sense. thanks again guys.
     
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