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.
 A: The identity $$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
means that the point $\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}}\right)$ is on the unit circle $x^2+y^2=1$. By definition, there exists an angle $y$ (actually, an infinity of angles) such that $\cos(y)=\frac{u}{\sqrt{u^2+v^2}}$ and $\sin(y)=\frac{v}{\sqrt{u^2+v^2}}$.
Assume for example that $u>0$ and $v>0$ (*), so that $y$ is in the first quadrant. Then, $y=\arcsin(\frac{v}{\sqrt{u^2+v^2}})+2k\pi$ and $x=\arccos(\frac{u}{\sqrt{u^2+v^2}})+2k\pi$, where $k$ is an integer.
(*) For other possible signs of $u$ and $v$, we obtain similar formulas.
A: 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.
