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Math Help - Trigonometric Proof (Very Challenging)

  1. #1
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    Trigonometric Proof (Very Challenging)

    Prove that cot(A) + cot(B) + cot(C) = (a^2 +b^2 +c^2) / 4(Area of ABC), where ABC is a traingle and a, b, c are opposite sides relative to the angles.
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  2. #2
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    Quote Originally Posted by strummer View Post

    Prove that cot(A) + cot(B) + cot(C) = (a^2 +b^2 +c^2) / 4(Area of ABC), where ABC is a traingle and a, b, c are opposite sides relative to the angles.
    actually it's quite easy: let S be the area of the triangle. then since S=\frac{bc\sin A}{2} and: bc\cos A = \frac{b^2 +c^2 - a^2}{2}, we'll have: \cot A= \frac{b^2 +c^2 - a^2}{4S}.

    similarly: \cot B= \frac{a^2 + c^2 - b^2}{4S}, and \cot C= \frac{a^2 + b^2 - c^2}{4S}. now add these three relations and you're done!
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  3. #3
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    Quote Originally Posted by strummer View Post
    Prove that cot(A) + cot(B) + cot(C) = (a^2 +b^2 +c^2) / 4(Area of ABC), where ABC is a traingle and a, b, c are opposite sides relative to the angles.
    hey mate,

    I'll take on the challenge!

    As I'm sure you know cot(x) = cos(x)/sin(x)

    Thus for each A,B,C we need to resolve cot(x) or moreover sin(x), cos(x)

    for cot(A)
    (c^2 - (a^2 + b^2))/(2ac*sin(B))

    Cosine Rule :
    a^2 = b^2 + c^2 - 2bc*cos(A)
    Thus cos(A) = (a^2 - (b^2 + c^2)) / (2bc)

    Sine Rule :
    a/sin(A) = b/sin(B)

    Thus,
    sin(A) =(a/b)sin(B)

    Hence,

    cot(A) = (((b^2 + c^2)-a^2) / 2bc)/ ((a/b)sin(B))
    cot(A) = (( (b^2 + c^2)-a^2) / (2ac * sin(B)))

    for cot(B)

    Cosine Rule :
    b^2 = a^2 + c^2 - 2ac*cos(B)
    cos(B) = ((a^2 + c^2)-b^2) /(2ac)

    Thus,

    cot(B) = (((a^2 + c^2)-b^2) /(2ac))/sin(B)
    cot(B) = (((b^2 + c^2)-b^2) / (2ac * sin(B)))

    for cot(C)

    Cosine Rule :
    c^2 = a^2 + b^2 - 2ab*cos(C)
    cos(C) = (c^2 - (a^2 + b^2))/(2ab)

    Sine Rule:
    b/sin(B) = c/sin(C)
    Thus,
    sin(C) = (c/b)sin(B)

    Thus,

    cot(C) = ((c^2 - (a^2 + b^2))/(2ab))/ ( (c/b)sin(B) )
    cot(C) = (c^2 - (a^2 + b^2))/(2ac*sin(B))

    Hence we now can define cot(A) + cot(B) + cot(C)

    cot(A) + cot(B) + cot(C)
    = (( (b^2 + c^2)-a^2) / (2ac * sin(B))) + cot(B) = (((a^2 + c^2)-b^2) / (2ac * sin(B))) + ((a^2 + b^2)-c^2)/(2ac*sin(B))
    = (1/(2ac * sin(B))(a^2 + b^2 + c^2)

    Area of ABC = (1/2)(ac * sin(B) --> ac*sin(B) = 2 *(Area of ABC)

    Sub in,
    cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/ (2 * (2 * (Area of ABC)))
    cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/(4 x (Area of ABC))
    = RHS

    Thus true,

    Hope this helps,

    David
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  4. #4
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    Quote Originally Posted by David24 View Post
    hey mate,

    I'll take on the challenge!

    As I'm sure you know cot(x) = cos(x)/sin(x)

    Thus for each A,B,C we need to resolve cot(x) or moreover sin(x), cos(x)

    for cot(A)
    (c^2 - (a^2 + b^2))/(2ac*sin(B))

    Cosine Rule :
    a^2 = b^2 + c^2 - 2bc*cos(A)
    Thus cos(A) = (a^2 - (b^2 + c^2)) / (2bc)

    Sine Rule :
    a/sin(A) = b/sin(B)

    Thus,
    sin(A) =(a/b)sin(B)

    Hence,

    cot(A) = (((b^2 + c^2)-a^2) / 2bc)/ ((a/b)sin(B))
    cot(A) = (( (b^2 + c^2)-a^2) / (2ac * sin(B)))

    for cot(B)

    Cosine Rule :
    b^2 = a^2 + c^2 - 2ac*cos(B)
    cos(B) = ((a^2 + c^2)-b^2) /(2ac)

    Thus,

    cot(B) = (((a^2 + c^2)-b^2) /(2ac))/sin(B)
    cot(B) = (((b^2 + c^2)-b^2) / (2ac * sin(B)))

    for cot(C)

    Cosine Rule :
    c^2 = a^2 + b^2 - 2ab*cos(C)
    cos(C) = (c^2 - (a^2 + b^2))/(2ab)

    Sine Rule:
    b/sin(B) = c/sin(C)
    Thus,
    sin(C) = (c/b)sin(B)

    Thus,

    cot(C) = ((c^2 - (a^2 + b^2))/(2ab))/ ( (c/b)sin(B) )
    cot(C) = (c^2 - (a^2 + b^2))/(2ac*sin(B))

    Hence we now can define cot(A) + cot(B) + cot(C)

    cot(A) + cot(B) + cot(C)
    = (( (b^2 + c^2)-a^2) / (2ac * sin(B))) + cot(B) = (((a^2 + c^2)-b^2) / (2ac * sin(B))) + ((a^2 + b^2)-c^2)/(2ac*sin(B))
    = (1/(2ac * sin(B))(a^2 + b^2 + c^2)

    Area of ABC = (1/2)(ac * sin(B) --> ac*sin(B) = 2 *(Area of ABC)

    Sub in,
    cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/ (2 * (2 * (Area of ABC)))
    cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/(4 x (Area of ABC))
    = RHS

    Thus true,

    Hope this helps,

    David
    I guess the OP now has a choice as to which path to follow .....
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