*Post by Ken Pledger**Post by comp.soft-sys.matlab*....

I try to proove an equation but I faced some difficulties for prooving

for n odd values

sum i=1 to ((n-1)/2) [cos((2*i-1)*pi/n)] = 0,5

in other words,

cos(pi/n)+cos(3*pi/n)+cos(5*pi/n)+.............+cos((n-2)*pi/n) = 0,5

....

You might have been answered more promptly in a news group such as

<sci.math> which carries much more traffic than <aus.mathematics>. But

since nobody else has replied yet, here are a couple of methods to prove

your equation. (Notice "prove", not "proove". :-)

I was assuming that it was homework -- certainly it's a standard

enough kind of question. Still, as long as you've (Ken) put an

answer out there, here's another approach, which may hopefully be

instructive -- essentially, we restore some missing symmetry to the

equation.

Draw a unit circle, and mark the points corresponding to each of

the angles in the above sum. These form part of a regular n-gon;

the remaining points in this n-gon are easily seen to arise from pi

and the negations of the original angles. Now consider the sum of

the cosines of the angles in the entire n-gon. If your original

sum is S1, then since Cos(-x) = Cos(x) this new sum (call it S2)

satisfies S2 = 2*S1 - 1.

Treating the points as complex numbers, it is clear that they are

the roots of x^n + 1 = 0. By elementary symmetric function theory,

the sum of the roots is the negative of the coefficient of x^(n-1),

which is 0. S2 is the real part of this sum, thus S2 = 0, hence

S1 = 1/2.

Alternatively, one could use a vector argument (really, it's the

same approach, but may feel more natural if one is not comfortable

with the above) -- S2 is the x value of the sum of the vectors in

the n-gon. But a straightforward rearrangement turns these vectors

into the sides of a regular n-gon, thus their sum is zero and the

result follows as before.

No doubt there are other ways to proceed, but what I feel is the

key point here is that rewriting the equation in a more symmetric

form (as a function over the vertices of a regular n-gon) greatly

helps simplify matters. Exploiting symmetry is a very powerful

mathematical tool.

Cheers,

Geoff.

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Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --

***@maths.usyd.edu.au | Gameplayer by vocation.

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