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

In article <1v1n53pmpevei4k2rmvqaf862bpsp9vn0d@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

On Tue, 29 May 2007 00:15:20 GMT, Gerry Myerson
<gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

In article <mnim53dkv2fieoo3t0a5prf7nm0spq7gc3@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

On Mon, 28 May 2007 21:21:02 +0100, David Hartley <me9@xxxxxxxxxxx>
wrote:

In message <qtfm53l2nlpf05dq7r5b07moq1upe20cnv@xxxxxxx>, quasi
<quasi@xxxxxxxx> writes
On Mon, 28 May 2007 15:46:16 -0500, quasi <quasi@xxxxxxxx> wrote:

On Mon, 28 May 2007 15:30:21 -0500, quasi <quasi@xxxxxxxx> wrote:

On Mon, 28 May 2007 14:42:53 -0500, quasi <quasi@xxxxxxxx> wrote:

On Mon, 28 May 2007 14:27:57 -0500, quasi <quasi@xxxxxxxx> wrote:

Finding angles whose sines have a given relation is more problematic.

As a test question, do there exist 3 acute angles with integer degree
measures n1,n2,n3 such that

sin(n1)=sin(n2)*sin(n3)

Ok, that was too easy.

Let n1=30, n2=n3=45.

I'll try to think of a better one.

Ok, here's a "trig problem" ...

Consider the multiplicative group G generated by

{sin(1),...,sin(44)} union {sin(46),...,sin(89)}

where the angles above are in degrees.

Is G the free abelian group with 88 generators?

No, they're not independent.

sin (30) sin(2n) = sin(n) sin(90-n) for each n =1 to 44

Nice.

Ok, let's restrict the angles to less than 45 degrees.

Here's the revised problem.

Let G be the multiplicative group G generated by

{sin(1),...,sin(44)}

where the angles above are in degrees.

Is G the free abelian group with 44 generators?

quasi

You might be interested in my paper,
Rational products of sines of rational angles,
Aequationes Math. 45 (1993), no. 1, 70--82.
MR 93m:11140

Thanks, I'll take a look when I get a chance.

But given that you haven't replied with a counterexample, I'll take
that as suggesting that the answer to the question as to whether G is
free abelian on 44 generators is "yes".

No such suggestion was intended.

I don't think there's a couterexample in the paper,
though sin 12 sin 48 = sin 18 sin 30 comes close.

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

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