I am given:
$$ X= U \cos(V) \tag{1}\\ $$ $$ Y = U \sin V \tag{2}$$
Now, I need to:
a) Give the respective ranges for $U$ and $V$ in order that the transformation defined is one to one. and
b) With the given range found in (a), find $U$ and $V$ in terms of $X$ and $Y$.
I don't know how to answer (a). I've tried sketching $X$ and $Y$ and can see that there are certain ranges for they will not be invertible since $\cos(V)$ and $\sin(V)$ are periodic. But I don't know how to translate this into specific ranges for $U$ and $V$.
If I start with $X$, then I can see that $0 \le V \le \pi$ and $U>0$ works because $X$ will be able to take on values from $-U$ to $U$ as defined in $(1)$ but then that will not work for $(2)$ ie $Y$ will not be invertible in that range. So, what is the solution and how should I approach this question?
For (b), I tried to solve by following the usual way of solving simultaneous equations and got the following:
$$\cos^{-1}\left(\frac{X}{U}\right)=V$$ $$\sin^{-1}\left(\frac{Y}{U}\right)=V $$
and it does not look like this will lead to an answer. What should I do to proceed? I suspect a trig identity may be involved.
EDIT:
The full question is:
Notice that if $X \neq 0$, we can divide the equations to obtain $$ \frac{Y}{X} = \frac{\sin V}{\cos V} = \tan V $$ If we square both equations and add them, we obtain $$ X^2 + Y^2 = U^2 \sin^2 V + U^2 \cos^2 V = U^2 $$ Solving these two equations for $U$ and $V$ looks straight-forward, but we must be careful when inverting the $\tan$ function using $\arctan$. Since the range of $\arctan$ is $(-\frac{\pi}{2},\frac{\pi}{2})$, we will need to restrict $V$ to lie in this interval. Also notice that when $X < 0$, $\cos(\arctan(\frac{Y}{X})) = -X$ and $\sin(\arctan(\frac{Y}{X})) = -Y$. We will therefore need to incorporate a switch of signs when $X$ is negative.
When $X = 0$, it is easy to see that setting $U = |Y|$ and $V = \frac{\pi}{2}$ if $Y \geq 0$ and $V = -\frac{\pi}{2}$ if $Y < 0$ will yield the correct values of $X$ and $Y$. Putting all this together, we can invert the transformation when $(U,V)$ lies in the set $(-\infty,\infty) \times [-\frac{\pi}{2},\frac{\pi}{2}]$ using the following formulas: \begin{align} U &= \begin{cases} \sqrt{X^2+Y^2} & \text{if $X \geq 0$}\\ -\sqrt{X^2+Y^2} & \text{if $X < 0$} \end{cases}\\ V &= \begin{cases} \arctan\left(\frac{Y}{X}\right) &\text{if $X \neq 0$} \\ \frac{\pi}{2} & \text{if $X = 0$ and $Y \geq 0$}\\ -\frac{\pi}{2} & \text{if $X = 0$ and $Y < 0$} \end{cases} \end{align}
To compute the joint density, all we need is the determinant of the Jacobian of the original transformation: $$ \begin{vmatrix} \frac{\partial X}{\partial U} & \frac{\partial X}{\partial V}\\ \frac{\partial Y}{\partial U} & \frac{\partial Y}{\partial V} \end{vmatrix} = \begin{vmatrix} \cos V & -U \sin V\\ \sin V & U \cos V \end{vmatrix} =U \cos^2 V + U \sin^2 V = U $$ Using the change-of-variables formula, we can compute the joint density of $(U,V)$ as \begin{align} f(u,v) &= \frac{1}{2\pi}\exp\left(-\frac{u^2 \cos^2 v + u^2 \cos^2 v}{2}\right)\cdot |u| \cdot \mathbb{1}\left\{-\frac{\pi}{2}\leq v \leq \frac{\pi}{2}\right\}\\ &= \frac{1}{2} \cdot |u| \cdot \exp\left(-\frac{u^2}{2}\right) \cdot \frac{1}{\pi}\cdot\mathbb{1}\left\{-\frac{\pi}{2}\leq v \leq \frac{\pi}{2}\right\} \end{align} where $\mathbb{1}\left\{\cdot\right\}$ is an indicator function equal to 1 if the expression in the braces is true and zero otherwise. Notice that this implies $U$ and $V$ are independently distributed, with $|U| \sim \text{Weibull}(1,2)$ and $V \sim \text{Uniform}(-\frac{\pi}{2},\frac{\pi}{2})$.