help with trig manipulation? (cot formula, Brocard points)
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From: John Nagelson on 12 Apr 2010 13:46 Hi, I'd be grateful for some gentle help with the following bit of trig manipulation. In the context of Brocard points, it appears that.. if sin^3(x) = sin(A-x)sin(B-x)sin(C-x) and A+B+C=pi, then cot(x) = cot(A) + cot(B) + cot(C). Trying to derive the equation from the two conditions, I got horribly bogged down. What's the easiest way to do it? Best regards, John
From: spudnik on 12 Apr 2010 15:31 oh, yes; les God-am points de Brocard -- what were they, and are you using trilinears, or is it "intrinsic?" > if sin^3(x) = sin(A-x)sin(B-x)sin(C-x) and A+B+C=pi, > then cot(x) = cot(A) + cot(B) + cot(C). --Light: A History! http://21stcenturysciencetech.com
From: Robert Israel on 15 Apr 2010 18:13
John Nagelson <johnnagelson(a)yahoo.co.uk> writes: > Hi, I'd be grateful for some gentle help with the following bit of > trig manipulation. > > In the context of Brocard points, it appears that.. > > if sin^3(x) = sin(A-x)sin(B-x)sin(C-x) and A+B+C=pi, > > then cot(x) = cot(A) + cot(B) + cot(C). Using Maple: > S1:= simplify(expand(sin(x)^3 - sin(A-x)*sin(B-x)*sin(Pi-A-B-x))); S1 := sin(A) sin(B) cos(A) cos(B) sin(x) + sin(x) - sin(B) cos(B) cos(x) - sin(A) cos(A) cos(x) 2 + sin(B) cos(B) cos(x) cos(A) 2 2 2 + sin(A) cos(A) cos(x) cos(B) - cos(A) sin(x) cos(B) > S2:= simplify(expand(convert(cot(x) - cot(A) - cot(B) - cot(Pi-A-B), sincos))); S2 := - (sin(A) sin(B) cos(A) cos(B) sin(x) + sin(x) - sin(B) cos(B) cos(x) - sin(A) cos(A) cos(x) 2 + sin(B) cos(B) cos(x) cos(A) 2 2 2 + sin(A) cos(A) cos(x) cos(B) - cos(A) sin(x) cos(B) )/( sin(x) sin(A) sin(B) (sin(A) cos(B) + cos(A) sin(B))) And note that the numerator of S2 is -S1. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |