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Comments

vlorbik

thanks for bringing this to my attention.
i immediately sat down and proved it, of course:
put a+b+c = \pi and use the standard textbook formula
for tan(x+y) -- twice -- out it comes.

but the derivation of the stated formula
is slightly messy ... and so it seems unlikely
that this proof would yield any insight
as to why this theorem is true "really"
(as opposed to, say, the law of sines).

caitmcq

You know, you really are a nerd.

David Harmon

My trig is too rusty for me to grapple with this directly, but:

Are there similar correspondences for any other polygon? Could it be related to the "accident" that the three angles of a triangle sum to a half circle?

Foxy

Using the law of cosines, I was able to solve for the tan values of each angle in terms of the lengths of the sides. What's really cool about that, then, is you get this really big term under a radical that, according to Heron's formula, is the area of the triangle. Cool.

Another value of interest, and it seems as though it would be significant more than in just this problem, is (A^2 + B^2 - C^2)(B^2 + C^2 - A^2)(C^2 + A^2 - B^2).

Good stuff.

I like math

There is demonstration for this fact using complex numbers say A+B+C=180

(cosA+isinA)*(cosB+isinB)*(cosC+isinC)=-1

So this product is real number dividing each factor by the cos we have

(1+itanA)*(1+itanB)*(1+itanC) is real

by foiling and since the imaginary part is 0 we get that identity.

That was the nicest way I could show that identity

My interest in it was that it can be used to show Hieron's formula. As I was trying to find nicer ways to show it I got the way above.

BlGene

A+B+C = 180
C = 180-(A+B)
tan( 180 - (A+B) ) = (tan(180)-tan(A+B))/(1+tan(180)tan(A+B))

tan(180) = 0

tan(C) = -tan(A+B)
tan(C) = - (tan(A)+tan(B))/(1 - tan(A)tan(B))
tan(C) - tan(A)tan(B)tan(C) = -tan(A) - tan(B)

tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C)

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