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Math Help - proving sum

  1. #1
    Newbie darknight's Avatar
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    proving sum

    A+B+C = 180 (pi)
    prove that,
    cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)
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  2. #2
    Senior Member pacman's Avatar
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    cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)

    = cotA cotB cotC + (cotA cotB cotC)(secAsecBsecC)

    = cotA cotB cotC + (cos A/sin A)(cos B/sin B)(cos C/sin C)(1/(cosA cos B cos C))

    cotA + cotB + cotC = (cot A cot B cot C) + (csc A csc B csc C)

    = (cot A cot B cot (pi - A - B)) + (csc A csc B csc (pi - A - B))

    = ?
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  3. #3
    Newbie darknight's Avatar
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    Quote Originally Posted by pacman View Post
    cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)

    = cotA cotB cotC + (cotA cotB cotC)(secAsecBsecC)

    = cotA cotB cotC + (cos A/sin A)(cos B/sin B)(cos C/sin C)(1/(cosA cos B cos C))

    cotA + cotB + cotC = (cot A cot B cot C) + (csc A csc B csc C)
    after that,

    cotA + cotB + cotC
    = (cot A cot B cot C) + (csc A csc B csc C)
    = (cosAcosBcosC / sinAsinBsinC) + (1/sinAsinBsinC)

    taking 1/sinAsinBsinC outside the bracket,
    = 1/sinAsinBsinC (cosAcosBcosC + 1) ....................(1)
    ____________________________________________

    Now,
    cotA + cotB + cotC
    = cosA/sinA + cosB/sinB + cosC/sinC


    cosAsinBsinC + cosBsinCsinA + cosCsinBsinA
    = _________________________________________......... ........................(2)
    sinAsinBsinC

    Equating 1 and 2,

    cosAsinBsinC + cosBsinCsinA + cosCsinBsinA
    ________________________________________ = 1/sinAsinBsinC (cosAcosBcosC +1)
    sinAsinBsinC


    sinAsinBsinC cancel out.

    cosAsinBsinC + cosBsinCsinA + cosCsinBsinA = cosAcosBcosC +1

    cosAsinBsinC + cosBsinCsinA + cosCsinBsinA - cosAcosBcosC = 1

    (cosAsinBsinC + cosBsinCsinA) + (cosCsinBsinA - cosAcosBcosC) = 1

    sinC(sinBcosA + cosBsinA) + cosC(sinBsinA - cosAcosB) = 1

    sinC(sin(A+B)) + cosC(-(cosAcosB - sinBsinA)) = 1

    sinC (sin (pi - C)) + cosC (- (cos(A+B)) =1

    sinC (sinC) + cosC (-(-cos(pi-C)) =1

    sinC^2 + cosC (-(-cosC)) =1

    sinC^2 + cosC^2 =1

    1=1

    Hence proved

    Thanks a bunch for kick-starting my brain. Just worked on it for a minutes and I got it.
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  4. #4
    Senior Member pacman's Avatar
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    Darknight, i believe it is wrong to prove an identity working on both sides because if you do, you are treating it as an equation not as an identity. Any piece of your thought on this?
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  5. #5
    Newbie darknight's Avatar
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    Re:

    Quote Originally Posted by pacman View Post
    Darknight, i believe it is wrong to prove an identity working on both sides because if you do, you are treating it as an equation not as an identity. Any piece of your thought on this?
    That's a good point, and you are correct: the proper way to
    prove such identities is to begin on one side and algebraically
    transform it into the form shown on the other side. Working on both
    sides is technically incorrect because in doing so we are assuming
    that the equality to be proven is already true.

    However, in the sense that it is not mathematically incorrect - if
    the identity is true, then the result of working on both sides will
    eventually result in an equality that is "obviously" true.
    That is, such a method won't lead to a wrong decision
    on the truth or falseness of the identity in question. But in so far
    as mathematical rigor, it is insufficient because of the reason
    mentioned at the end of my first paragraph. Often, however, working on
    both sides can help us form a rigorous proof of an identity. I will
    illustrate this with the example.

    Identity to be proven:

    (1-TanA)/SecA + SecA/TanA = (1+TanA)/(SecATanA)

    If we work on both sides, the first step is to multiply both sides by
    Sec[a]Tan[a]:

    (1-TanA)TanA + SecASecA = 1+TanA

    Multiplying through, we find

    TanA - TanA^2 + SecA^2 = 1+TanA,
    or
    -TanA^2 + SecA^2 = 1,

    which is equivalent to the identity 1+TanA^2 = SecA^2, so we are
    done. This is not a rigorous proof, but it leads to one, in which we
    see the proper method is as follows:

    (1-TanA)/SecA + SecA/TanA

    = (1-TanA)TanA/ (SecATanA) + SecA^2/(SecATanA)

    = (TanA - TanA^2 + SecA^2)/(SecATanA)

    = (TanA - TanA^2 + 1 + TanA^2)/(SecATanA)

    = (TanA + 1)/(SecATanA).

    Therefore we have taken the left-hand side and transformed it into the
    right-hand side using algebraic manipulation and other known
    trigonometric identities. But it should not be too hard to see that
    the basic "flow" of the proof is essentially the same procedure as the
    "improper" method, just rearranged a bit.

    I believe by retracing the steps it should be easy enough to prove my earlier identity, albeit it is much tougher than this simple example.
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  6. #6
    Senior Member pacman's Avatar
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    thanks darknight for that insightful explaination . . . .
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