Existence problem for a polynomial with complex coefficients

Let $n$ be a nonnegative integer and $a_{0}, a_{1}, ..., a_{n}$ real numbers. For any real number $t$ let $f(t)= \sum_{k=0}^{n}a_{k}\cos(kt)$. Could you help me with the following two questions ?

a) Prove that there exists a polynomial $P \in \mathbb{C}[X]$ such that for all real $t$ : $f(t)=e^{-int}P(e^{it})$.

b)Suppose that for all real $t$ $f(t)=0$. Prove that all $a_{k}$s are equal to zero.

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Hint: $\cos(kt)=\frac12(\mathrm e^{\mathrm ikt}+\mathrm e^{-\mathrm ikt})$. – Did Jan 10 '12 at 22:49
Please try to find a better title. – Mariano Suárez-Alvarez Jan 10 '12 at 22:50
As for b). Try to evaluate $\int\limits_{0,2\pi} f(t)\cos(k t)dt$ using two approaches – Norbert Jan 11 '12 at 0:49

(a): $\cos(kt) = \frac{1}{2} (e^{ikt} +e^{-ikt})$, so take $P(X):=a_0 X^n + \sum\limits_{k=1}^n \frac{1}{2} a_k (X^{n+k} + X^{n-k})$.
(b): If $f(t)=0$ for all real $t$, then by (a), the polynomial $P(X)$ vanishes on the unit circle, so it must also vanish as an element of ${\Bbb C}[X]$.