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

1. May 15, 2004

### man_fire

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. May 15, 2004

### arildno

Have you used Heron's formula in deriving 2)?

3. May 15, 2004

### Vance

Is this the way you brought that form up ?

$$S=\frac{1}{2}absinC$$

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

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

4. May 16, 2004

### man_fire

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.

5. May 16, 2004

### man_fire

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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