Cosine Summation Lemma

Question

Just to give some background, I am attempting to solve part (b) of this Fourier matrix - multiplicity of eigenvalues? problem from Artin's Algebra textbook. If I can prove the following lemma, I should be able to finish my work.

Lemma: Let $m$ and $n$ be positive integers such that n does not divide m. Then $\sum_{l=0}^{n-1} cos(2πml/n) = 0$

I was able to prove this for even n. I was also able to prove it for the special case n=3, but I am stumped on how to prove the general case. Please help me!

Answer 1

Observe the angle-sum identities \begin{align*} \sin(lx + \frac 12 x) &= \sin lx \cos \frac x2 + \cos lx \sin \frac x2 \\ \sin(lx - \frac 12 x) &= \sin lx \cos \frac x2 - \cos lx \sin \frac x2 \end{align*} lead to $$ \cos lx \sin \frac x2 = \frac 12 \left[ \sin(lx + \frac 12 x) - \sin(lx - \frac 12 x) \right].$$ You can sum both sides from $l=0$ to $l=n-1$ and use the telescoping nature of the sum to obtain $$ \sum_{l=0}^{n-1} \cos lx \sin \frac x2 = \frac 12 \sum_{l=0}^{n-1} \left[ \sin(lx + \frac 12 x) - \sin(lx - \frac 12 x) \right] = \frac 12 \left[ \sin (n-\frac 12) x - \sin( -\frac 12 x) \right].$$ The difference-of-sines formula gives you $$ \sin (n-\frac 12)x - \sin(-\frac 12 x) = 2 \cos \left(\frac{n-1}{2}\right) x \sin nx.$$ Thus $$\sum_{l=0}^{n-1} \cos lx = \frac 1{ \sin \frac x2} \left[ \cos \left(\frac{n-1}{2}\right) x \sin nx \right].$$ With $x = \dfrac{2\pi m}{n}$ you have $$\sin nx = \sin 2\pi m = 0$$ and $$\sin \frac x2 = \sin \frac{\pi m}{n} \not= 0$$ because $n$ does not divide $m$. Thus the sum is zero.

Answer 2

We have the following identity $$\sum^{n-1}_{k=1} cos(ak+c) = \frac{sen(an+c-\frac{a}{2}) - sen(c+\frac{a}{2}) }{2sen (\frac{a}{2})}.$$

With $c=0$, $a=2\pi m/n$, so $an=2\pi m$. And the difference in the numerator is zero.

To prove the formula

Use

$$sen(p) -sen(q) = 2sen(\frac{p-q}{2}) cos(\frac{p+q}{2}) $$

with $p=a(k+1)+b=ak+b+a$ and $q= ak+b$, then

$$ \frac{p-q}{2} =\frac{a}{2}$$

$$\frac{p+q}{2}= \frac{2(ak+b) +a}{2} =ak+b+\frac{a}{2}$$

so

$$sen(a(k+1)+b)-sen(ak+b)= 2sen (\frac{a}{2}) cos(ak+b+\frac{a}{2}) $$

applying $\sum^{n-1}_{k=1}$ we have by the telescopic sum

$$sen(an+b) - sen(a+b)= 2sen (\frac{a}{2}) \sum^{n-1}_{k=1} cos(ak+b+\frac{a}{2}) $$ then

$$\sum^{n-1}_{k=1} cos(ak+b+\frac{a}{2}) = \frac{sen(an+b) - sen(a+b) }{2sen (\frac{a}{2})} $$ take $b+\frac{a}{2} =c$ so $$\sum^{n-1}_{k=1} cos(ak+c) = \frac{sen(an+c-\frac{a}{2}) - sen(c+\frac{a}{2}) }{2sen (\frac{a}{2})}. $$