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I've created some new formulas in trigonometry! (perhaps)

  1. May 15, 2004 #1
    Hi all,
    I'm a newbie here and sorry for my bad English.
    I wanna share with you guys 2 formulas I've created luckily on a boring day:
    1) Let R be circumscribed circle of triangle ABC, let S be it's area.
    We have: S=2R^2*sinAsinBsinC
    2)Let S be area of triangle ABC, a=BC, b=CA, c=AB.
    We have:(a+b+c)^2 >= S*12*(3^0.5)
    Did anyone know these before? If didn't, please reply to me.
     
  2. jcsd
  3. May 15, 2004 #2

    arildno

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    Have you used Heron's formula in deriving 2)?
     
  4. May 15, 2004 #3
    Is this the way you brought that form up ?

    [tex]S=\frac{1}{2}absinC[/tex]

    just multiply till we get [tex]S^3=\frac{1}{8}a^2b^2c^2sinAsinBsinC[/tex]

    finally use the relationship between R and a, sinA; b, sinB ; c and sinC, substitute them all in the above S to get that result...

    if thats da way you did, you did made a success, you found it out before me...lol

    Good luck on your next try..lol
     
  5. May 16, 2004 #4
    Hello Vance,
    You're a genius because you've found out a better way to get the result than I have. I brought that form up from:
    S=abc/4R
    and: S=0.5*absinC. It's so messy.
    Thanks for your consideration.
     
  6. May 16, 2004 #5
    Hello Arildno,
    I didn't use Heron's formula in deriving 2).
    Here is my solvation:
    cotang(A/2) + cotang(B/2) + cotang(C/2) >= 3*cotang((A+B+C)/6)=3*3^(1/2)
    and:
    cotang(A/2) = cos(A/2) / sin(A/2). We multiply both numerator and denominator with 2cos(A/2). Simply numerator = 1+cosA, and denominator = sinA. Now we multiply both of them with bc --->numerator=bc+bccosA; denominator=bcsinA=2S.
    Let's take care of the numerator. From cos theorem: a^2=b^2 + c^2 - 2bccosA ---->bccosA = (b^2 + c^2 - a^2)/2.
    With all of this, we go to:
    cotang(A/2) = ((b+c)^2 - a^2)/4S.
    It's the same with cotang(B/2) and cotang(C/2).
     
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