If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$ 
Let $0\leq\alpha\leq \pi $. $V_n (\alpha) $ denote the number of sign changes in the sequence $\cos\alpha,\cos2\alpha,\cos3\alpha,\ldots,\cos n\alpha $. Then prove that $$\lim\limits_{n\to\infty}\dfrac{V_n (\alpha)}{n}=\dfrac{\alpha}{\pi}.$$

I saw a hint where $\dfrac{V_n (\alpha)}{n}$ is considered as the probability. I mean how this expression is a probability of something. If it is, how can I progress further in this way?
Update: I have a solution to this problem
In $n\alpha$ rotation the number of times full circle rotation occures $=\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$
In one full circle rotation sign change occures 2 times. Hence in $\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$ full rotation sign change occures $=2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$
Now the rest angle is $n\alpha-\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor\times2\pi$
If we consider 0 as a change of sign in case of $\cos\left( \dfrac{\pi}{2}\right)$ and $\cos\left(\dfrac{3\pi}{2}\right)$ then:-
(1) If $0\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{\pi}{2 }$ sign changes 0 times
(2) If $\dfrac{\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{3\pi}{2 }$ sign changes 1 times
(3) If $\dfrac{3\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<2\pi$ sign changes 2 times
Let $f$ be a function such that $$f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)=\begin{cases}0,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=0\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=1\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=2\\ 2,\text{ when } \left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=3\end{cases}$$
Therefore $\dfrac{V_n(\alpha)}{n}=\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)}{n}$
Hence $$\dfrac{V_n(\alpha)}{n}\geq \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}$$ and $$\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}\leq \dfrac{V_n(\alpha)}{n}$$
$\lim\limits_{n\to \infty}\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}=\dfrac{\alpha}{\pi}$ and $\lim\limits_{n\to\infty} \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}=\dfrac{\alpha}{\pi}$
Hence by Sandwich Theorem We get $\lim\limits_{n\to \infty}\dfrac{V_n(\alpha)}{n}=\dfrac{\alpha}{\pi}$ [Proved]
Is this correct?

*

*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.$\qquad\square$
