Validity of a differing frame of reference than that used in Landau and Lifshitz's solution For the following problem, first problem in chapter 2 (page 16) of Landau and Lifshitz's Classical Mechanics text:

I am trying to see whether the picture I drew when originally solving the problem before looking at the solution

is valid. The answer would then be $\frac{\cos(\theta_1)}{\cos(\theta_2)}=$ ...
Based on the solution, the proper drawing should be

Can I get any input on whether the different cases are equivalent?
Edit: to address geometric confusion.
Put the solution that L & L give us out of mind and just read the problem. There is no reason NOT to draw this setup, right?

If we agree on that, my question boils down to a trigonometric one. Why use $\theta_2$ and $\phi_2$ as opposed to $\theta_1$ and $\phi_1$?

 A: 
Solution
The momentum component is to be constrained constant along the Plane$(x)$ and not the Normal$(y)$. This is because the potential energy is independent of $x$.
$$U=\begin{cases}U_1&y<0\\U_2&y>0
\end{cases}$$
So, we have the equations
\begin{align*}
v_1\sin\theta_1&=v_2\sin\phi_1\tag{1}\\
\frac12mv_1^2+U_1&=\frac12mv_2^2+U_2\tag{2}\\
\end{align*}
Putting the value of $v_2$ from equation $(1)$ into $(2)$, one gets
\begin{align*}
\frac12mv_1^2+U_1&=\frac12m\left(v_1\frac{\sin\theta_1}{\sin\phi_1}\right)^2+U_2 \\
(U_1-U_2)&=\frac12mv_1^2\left[\left(\frac{\sin\theta_1}{\sin\phi_1}\right)^2-1\right] \\
\boxed{\frac{\sin\theta_1}{\sin\phi_1}=\sqrt{1+\frac{2}{mv_1^2}(U_1-U_2)}}\tag{1}
\end{align*}
Thus, we get the relation between the angles.

Answer to your question

Why use $\theta_2$ and $\phi_2$ as opposed to $\theta_1$ and $\phi_1$?

The answer could look different by the use of trigonometric identity $\cos\left(\frac{\pi}2-\alpha\right)=\sin\alpha$ in the numerator and/or denominator of equation $(1)$ but is exactly the same in all the following appearances physically because they are all equal to the same physical quantity. $$\frac{\cos\theta_2}{\cos\phi_2}=\frac{\cos\theta_2}{\sin\phi_1}=\frac{\sin\theta_1}{\cos\phi_2}=\frac{\sin\theta_1}{\sin\phi_1}=\sqrt{1+\frac{2}{mv_1^2}(U_1-U_2)}=\frac{\frac{v_1^{\text{along plane}}}{{v_1}}}{\frac{v_2^{\text{along plane}}}{{v_2}}}$$

Note that the solution of the book has $\theta_1=\theta_1$ and $\theta_2=\phi_1$ with no other angles used in this answer or your figure.
A: OK given the diagram of the problem you present in your fourth figure, the answer is the same if you use pairs of angles relative to the normal, or pairs of angles relative to the plane.  [My notation is a little different. Subscripts 1 and 2 are regions 1 and 2 respectively.  Angle theta is relative to the normal and angle phi is relative to the plane.]  We know:
$$V_1sin(\theta_1) = V_2sin(\theta_2)$$
But from the geometry of the problem we also know that $$\theta_x = \frac{\pi}{2} - \phi_x$$
If we substitute this in the above, we get:  $$V_1sin(\frac{\pi}{2} - \phi_1) = V_2sin(\frac{\pi}{2} - \phi_2)$$
And there is a trigonometric identity that says $$sin(\frac{\pi}{2} -\theta) = cos(\theta)$$
[See wikipedia here, section "Reflections, shifts and periodicity".]
And this gives, of course $$V_1cos(\phi_1) = V_2cos(\phi_2)$$
which is how you viewed the problem.
So the ratios of sines or cosines in the problem are the same, as long as you use the correct angles.
Sameer Baheti correctly notes this too.
