I have just learned the Box-Muller method for generating normal random values. My notes then consider the Polar Marsaglia method, which is more efficient than Box-Muller. In Box-Muller:
$$X=\sqrt{-2\log U_2} \cos{2\pi U_1}$$ $$Y=\sqrt{-2\log U_2} \sin{2\pi U_1}$$
I note that in the above: $-1 \leq \cos{2\pi U_1} \leq 1$ and the same for sine. Polar Marsaglia scheme is then defined as follows: $V_1 = 2U_1 -1$ and $V_2 = 2U_2-1$, also $R=V_1^2 + V_2^2$, then $X$ and $Y$ are:
$$X = \sqrt{\frac{-2\log{U}}{R}}V_1$$ and $$Y = \sqrt{\frac{-2\log{U}}{R}}V_2$$
My question is: why do we have to divide by $\sqrt{R}$ there? Because $-1 \leq V_1,V_2 \leq 1$ just like the cosine and sine in the Box-Muller. So essentially I do not understand why the cosine and sine are replaced by $\frac{V_1}{\sqrt{R}}$ and $\frac{V_2}{\sqrt{R}}$
Let us write the Marsaglia formulas under the form :
$$X = \sqrt{-2\log{U}}\frac{V_1}{\sqrt{R}},Y = \sqrt{-2\log{U}}\frac{V_2}{\sqrt{R}}$$
By comparison with Box-Muller formulas, you see that the cosine and the sine are resp. replaced by
$$C=\frac{V_1}{\sqrt{R}},S=\frac{V_2}{\sqrt{R}} \ \ \text{with} \ \ C^2+S^2=1$$
which is exactly the same because the common range of $V_1$ and $V_2$ is [-1,1]. Dividing vector $(V_1,V_2)$ by its norm $R$ provides a point on the unit circle, thus of the form $(C=\cos \alpha, S=\sin \alpha)$ for a certain $\alpha \in [0 , 2\pi)$ because the four quadrants are covered.