If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$

HINT: let $ b_n\equiv n a \pmod {2\pi}$ indicate the angle formed with the $x$- axis in the $n^{th}$ term of the sequence. Assume that $b$ is uniformly distributed in the range between $0$ and $2\pi$.

Now firstly consider the case in which $0<b_n<\pi/2$ or $3\pi/2<b_n<2\pi$. In the next step, a change of sign will occur only if $b_{n+1}>\pi/2$. What is the probability that this occurs, given that $b_{n+1}=b_n+a$?

Then repeat the same considerations for the case in which $\pi/2<b_n<3\pi/2$. A change of sign will occur only if $b_{n+1}>3\pi/2$. What is the probability that this occurs?

• I will assume that the case where $\alpha\in \pi \mathbb{Q}$ is easy, because the sequence $\big(\cos(k\alpha)\big)_{k\ge1}$ is periodic in this case, and if we consider $0$ as positive number then $V_{2q}(p\pi/q)=2p\pm 1$ and the result holds in this case.
• Now we assume that $\alpha\notin \pi\mathbb{Q}$. This implies that the sequence $\big(k\alpha \mod(2\pi)\big)_{k\geq 1}$ is equidistributed in $[0,2\pi]$. See Equidistributed sequences.

Now, let $f$ be the $2\pi$ periodic function defined by $$f(\theta)=\cases{0, & if $\cos\theta \cos(\theta+\alpha)\geq0$,\\ 1,& if $\cos\theta \cos(\theta+\alpha)<0$.}$$ With this definition, $$V_n(\alpha)=\text{card}\left\{k\in\{1,\ldots,n\}:f(k\alpha)=1\right\}$$ But if we define $$\mathcal{I}=\cases{\left(\frac{\pi}{2}-\alpha,\frac{\pi}2\right)\cup \left(\frac{3\pi}{2}-\alpha,\frac{3\pi}2\right) ,&if $0<\alpha<\pi/2$,\cr \left[0,\frac{\pi}{2}\right)\cup \left(\frac{3\pi}{2}-\alpha,\frac{3\pi}2\right)\cup\left(\frac{5\pi}{2}-\alpha,2\pi\right] ,&if $\pi/2<\alpha<\pi$.}$$ Then for $\theta\in[0,2\pi]$ we have $$f(\theta)=1\iff \theta\in\mathcal{I}$$ So, equidistribution of the sequence implies that $$\lim_{n\to\infty}\frac{V_n(\alpha)}{n}=\frac{\text{length}(\mathcal{I})}{2\pi}=\frac{\alpha}{\pi}$$ Done.$\qquad\square$