Choose $y\in\mathbb{R}$ so that $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. I know the identity $\sin^2y+\cos^2y=1$. Also, I notice that
$$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
without $u,v$ vanishing simultaneously. But I am not sure whether $\exists y\in\mathbb{R}$ s.t. $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. Does anyone have an idea? Thanks a lot.

It is clear that $\frac v{\sqrt{u^2+v^2}}\in[-1,1]$. Therefore, since $\cos0=1$ and $\cos\pi=-1$, the intermediate value theorem implies that there's a $y\in[0,\pi]$ such that $\cos y=\frac v{\sqrt{u^2+v^2}}$. Then\begin{align}\sin^2y&=1-\cos^2y\\&=1-\frac{v^2}{u^2+v^2}\\&=\left(\frac u{\sqrt{u^2+v^2}}\right)^2,\end{align}and therefore $\sin y=\pm\frac u{\sqrt{u^2+v^2}}$. If you got the $-$ sign, then replace $y$ by $-y$ and you're done.
