Can you tackle the sign change?

So number of sign changes indicates number of roots...isn't it so??Now cos(x)=0 has general solution x= $\frac{2m+1}{2}$ $\pi$ in interval (0,n $\alpha$ ]. Now 0< $\frac{2m+1}{2}$ $\pi$ <=n $\alpha$ . 0<2m+1<(2n $\alpha$ )/ $\pi$ . (- $\frac{1}{2}$ )<m<=n $\alpha$ / $\pi$ - $\frac{1}{2}$ .So now well $V_n$ ( $\alpha$ )=[n $\alpha$ / $\pi$ - $\frac{1}{2}$ ].Now this problem should be a cakewalk and final answer is nothing other than the mysterious $\pi$ . ([.] used in expression of Vn( $\alpha$ ) is GIF.)