# How can I show that $z_i =\cos(iw)$ where $w$ is uniform on $[0,2\pi]$ is a white noise process?

How can I show that $z_i =\cos(iw)$, where $w$ is uniform on $[0,2\pi]$ is a white noise process?

So far, I have shown $E(z_i)=0$ by integrating. However, I need to show $\operatorname{Cov}(z_i,z_{i-j})=0$ for any $i$ and any nonzero $j$.

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Using $\mathbb{E}z_i=0$ we obtain for $0<j \leq i$:
$$cov(z_i,z_{i-j}) = \int \cos(i \cdot w) \cdot \cos((i-j) \cdot w) \, d\mathbb{P} = \frac{1}{2\pi} \int_0^{2\pi} \cos(i \cdot x) \cdot \cos ((i-j) \cdot x) \, dx \\ \stackrel{\ast}{=} \frac{1}{4\pi} \cdot \int_0^{2\pi} \cos((2i-j) \cdot x) + \cos(j \cdot x) \, dx \\ = \frac{1}{4 \pi} \cdot \bigg( \bigg[\frac{\sin((2i-j) \cdot x)}{2i-j}\bigg]_0^{2\pi} + \bigg[\frac{\sin(j \cdot x)}{j}\bigg]_0^{2\pi}\bigg) =0$$
$$\cos \alpha + \cos \beta = 2 \cos \bigg(\frac{\alpha+\beta}{2}\bigg) \cdot \cos \bigg( \frac{\alpha-\beta}{2} \bigg)$$
(here: $\alpha :=2i-j$, $\beta:=j$)