If there's no relation between $c_1$ and $c_2$, can I say that $c_1= r\sin \theta,c_2=r\cos \theta$?

Question

If there's no relation between $c_1$ and $c_2$, can I say that $c_1= r\sin \theta,c_2=r\cos \theta$?

In other words, does the following system in the variables $r,\theta$, always have solution?

$$ r\sin\theta = c \\ r\cos \theta = d $$

This question comes from some notes I'm reading on differential equations, where they use this to simplify these expressions:

$$ y_1=c_1\cos\beta t+c_2\sin\beta t = r\cos(\theta-\beta t)\\ y_2=-c_1\sin\beta t+c_2\cos \beta t= r\sin (\theta - \beta t) $$

Answer 1

Yes you can since the only consequence of such a system is that

$c^{2} + d^{2} =r^{2}$ and $\tan(\theta) = c/d$

for some arbitrary number $r$ and some angle $\theta$, assuming $c$ and $d$ have been given.

Such a number $r$ and such an angle $\theta$ always exist.

Answer 2

If $c_1=c_2=0$ then $r=0$ and $\theta$ can be anything.

For other values of $c_1,c_2$ there will be many solutions, but it is often convenient to assume $$r>0\ ,\qquad -\pi<\theta\le\pi\ ,$$ in which case there will be exactly one solution. We have $$r=\sqrt{c_1^2+c_2^2}$$ and $$\tan\theta=\frac{c_1}{c_2}$$ provided $c_2\ne0$. By finding the sign of $\cos\theta$ and $\sin\theta$ you can determine which quadrant $\theta$ is in, then this last equation gives you one definite value for $\theta$. Finally, in the case $c_2=0$ we have $$\theta=\begin{cases}\frac\pi2&\hbox{if}\ c_1>0\\ -\frac\pi2&\hbox{if}\ c_1<0.\end{cases}$$