Discussion:
Help:proof on cosine
(too old to reply)
comp.soft-sys.matlab
2007-04-22 22:08:08 UTC
Hello everybody,
I try to proove an equation but I faced some difficulties for prooving
the following equation:
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
Best Regards
Ken Pledger
2007-04-30 22:13:42 UTC
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". :-)

The obvious trick is to use complex numbers. If we write

C = (sum i=1 to (n-1)/2) (cos((2i-1)pi/n)) and also

S = (sum i=1 to (n-1)/2) (sin((2i-1)pi/n))

then C + iS is a geometric series by de Moivre's Theorem, so it can
easily be summed. However, the sum then needs some slightly tricky
cancellation and simplification before you can find its real part C.

A shorter method uses the less well-known trick of multiplying
both sides by 2.sin(pi/n) and using the prosthaphaeresis formula

2.cosA.sinB = sin(A + B) - sin(A - B).

That method gives 2C.sin(pi/n)

= (sum i=1 to (n-1)/2) 2.(cos((2i-1)pi/n)).sin(pi/n)

= (sum i=1 to (n-1)/2) (sin(2i.pi/n) - sin(2(i-1)pi/n) from above

= sin((n-1)pi/n) - sin0 as the other terms cancel in pairs

= sin(pi - (pi/n)) - 0

= sin(pi/n)

so C = 1/2 as required.

Ken Pledger.
Fred the Wonder Worm
2007-04-30 23:37:20 UTC
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.

-----------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
***@maths.usyd.edu.au | Gameplayer by vocation.
-----------------------------------------------------------------------------
Alex. Lupas
2007-05-08 15:03:48 UTC
Post by comp.soft-sys.matlab
Hello everybody,
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
Thanks in advance, Best Regards
============
Let n=2m+1 and

(1) T_m(x)=cos(m*arccos(x)) =(Chebychev-polynomial)
and

(2) Q_m(x)=(T_m(x)+T_{m+1}(x))/(1+x) .

Because T_m(-1)=(-1)^m observe that Q_m(x) is a polynomial of degree
m.
Post by comp.soft-sys.matlab
From (1)-(2) it follows that the roots of Q_m(x) are the numbers
x_j=cos{(2j-1)*pi/n} .

If Q_m(x)= Ax^m+ Bx^{m-1}+... , then

(*) x_1+x_2+...+ x_m = -B/A (Vieta).

The question is to find A and B.

If C(m,k):=m!/(k!(m-k)!) ,

(a)_0=1 , (a)_k:=a(a+1)...(a+k-1) for k in {1,2,...}
and

F(-m,a;b;z):= SUM_{0=<k=<m}(-1)^kC(m,k)(a)_kz^k/(b)_k

use the fact that

(3) Q_m(x)=F(-m,m+1;1/2; (1-x)/2 ).
Post by comp.soft-sys.matlab
From (3) yo find A and B .Then use (*) .