🚀 go-pugleaf

On Mon, 10 May 1999, Daniel Lichtman wrote:

:Date: Mon, 10 May 1999 22:17:44 +0200
:From: Daniel Lichtman <dlichtma@cybercable.fr>
:Newsgroups: can.schoolnet.math, can.schoolnet.math.jr, fj.sci.math,
:    fr.sci.maths, sci.math, uk.education.maths
:Subject: Re: Please Help problem
:
:The two sides of the equation are equal!!
:So it is always true!!
:Daniel
:Dimitris Sourmelidis <xman@mypad.com> a écrit dans le message :
:373727b7@news3.us.ibm.net...
:> Guys here is a math prob!
:>
:> if there is a triangle where : sin(Á) * cos^3(Â/2) = sin(Â) * cos^3(Á/2)

sin(A)=2sin(A/2)cos(A/2). Similarly for sin(B). Cancel cos(A/2) and
cos(B/2) and 2. We can do this as A,B can't be pi radians (180 degrees),
as we're in a triangle.
Hence sin(A/2)cos^2(B/2)=sin(B/2)cos^2(A/2).
Thus sin(A/2)-sin(B/2)=sin(A/2)sin^2(B/2)-sin(B/2)sin^2(A/2).
Hence either sin(A/2)=sin(B/2) or else 1=-sin(A/2)sin(B/2).
In the latter case we must have |sin(A/2)|=|sin(B/2)|=1, so A, B are odd
multiples of pi, a contradiction.
Thus the former situtation occurs.
Hence, A/2=B/2 or A/2+B/2=pi, whence A+B=2pi, a contradiction since
A+B<pi.
Thus A/2=B/2 and A=B. Thus the triangle has two angles (i.e. A,B) equal.
From this one can show the corresponding sides (i.e. a,b) are equal- see,
for example, Euclid Book 1, right near the start.

:> then we want to proof
:> that this triangle has two equal sides!
:>
:>
:> Please post a reply to xman@mypad.com
:>
:>
:> Thanks to any reply..
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