Roots of trigonometric polynomial let $a=(a_0,a_1,...,a_n)$ and $P_a(x)= \sum_{k=0}^{n}a_k \cos (kx) $
define $b=(a_n,a_{n-1},...,a_0)$
If $Z_a$ is the number of roots of $P_a$ on $[0,2\pi[$ 
then $$Z_a+Z_b \geq 2n$$
I have failed to find a clever algebra trick, I've tried multiplying $P_a$ and $P_b$ and tried induction on $deg (P)$ but to no avail.
 A: Here's a way to do it using complex analysis. I'll suppose first that $a_0 \neq 0$ and $a_n \neq 0$. Let $Q_a(z) = a_0 + a_1 z + ... + a_nz^n$, and similarly for $Q_b$. Let $C$ denote the unit circle in the plane. Now, if $Q_a$ has $m$ zeroes in the unit disk, then the image $f(C)$ winds $m$ times around 0; see the wiki page on "Argument principle" if you are unfamiliar with this fact. Each times it winds around zero it must strike the imaginary axis at least twice, giving your trig polynomial at least two zeroes. So you get 2 times the number of zeroes of $Q_a$ in the unit disk as a lower bound for $Z_a$. Similarly for $Z_b$ and $Q_b$, however $Q_b(z) = z^n Q_a(1/z)$, which means $Q_b$ has a zero inside the disk whenever $Q_a$ has a zero outside the disk. This gives a lower bound for $Z_b$, and basically $Z_a+Z_b$ is at least two times the number of zeroes of $Q_a$, which is $n$, as long as $Q_a$ has no zeroes on the unit circle $C$. However, if $Q_a(z) = 0$ with $z \in C$ then $Q_b(1/z)=0$ as well, giving you a zero of both $P_a$ and $P_b$, so that zero is essentially counted twice as well, and the result follows.
When $a_0=0$ or $a_n=0$, you can write $Q_a(z) = H_a(z)z^r$ and $Q_b(z) = H_b(z)z^s$ for $H_a$, $H_b$ with nonzero leading coefficients, and some $r$ and $s$. Then just apply the above reasoning to $H_a$, $H_b$, together with the extra zeroes from $z^r$ and $z^s$ gives you what you want.
Hope that helps.
Greg 
