Re: Simple, but a bit hard, Trigonometry problem.

In article <t8fc53tkm5er9m8hgeagjnsu2mah640gnk@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

On 24 May 2007 04:52:53 -0700, gtsavdar@xxxxxxxxx wrote:


I've managed to solve it but only after a _massive_ number of
calculations.
Is there any other simpler solution? There must be!

If:
a = Sin(5°)
b = Sin(49°)
c = Sin(87°)

then prove that: Sin(73°) = (a^2 - b^2 + a c) / ( 4 a (a^2 - b^2 +
a c) - (a-b+c) )

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If the symbol of degree inside the Sin() does not appear in some
browsers then the problem is the follow:

If:
a = Sin(5 degrees)
b = Sin(49 degrees)
c = Sin(87 degrees)

then prove that: Sin(73 degrees) = (a^2 - b^2 + a c) / ( 4 a (a^2 -
b^2 + a c) - (a-b+c) )

Of course, it's easy to verify it numerically, at least up to an
acceptable accuracy, but I don't see an easy way to prove it
algebraically.

I can set it up as a horrible system of simultaneous nonlinear
polynomial equations which could be solved, in theory, using Grobner
bases. However it's clear that the degrees are too high for such a
method to be practical, at least for the simple way I'm
conceptualizing it.

So as of now, I don't see any practical way of doing it algebraically,
even with the help of Maple.

Maybe this works. Let Q = exp( pi i / 180). Then all these sines
can be expressed in terms of Q, e.g.,
sin(5 degrees) = (Q^5 - Q^{-5}) / (2i)
so the problem is to prove that Q satisfies some polynomial,
knowing that it satisfies Q^360 - 1 = 0. Well the minimal
polynomial for Q is no doubt available somewhere, Maple can
surely check whether some other polynomial is divisible by it.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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