No, there are no relations among variables $x_i$. Polynomial ring is freely generated commutative $R$-algebra with generators $\{x_1,\ldots,x_n\}$. This is similar to basis of vector space, however there is difference since you will not get all polynomials as linear combinations of generators, but you have to multiply them as well. Still, if you consider this analogy, you have that basis for vector space must be linear independent; here, however, you have to consider multiplication as well, so there are no polynomial relations among generators, i.e. something like $x_1 = x_2^2$ can't hold by definition.
That said, your first example still makes sense, and I will get back to it, but your second does not. It seems that you are treating variables as if they were real/complex numbers. That is not the case. Expressions like $\sin x_1$ don't even make sense because $\sin$ is not a polynomial. It's like trying to multiply general vectors, there is no structure to support the notion.
But let us get back to your second example. What you described is actually quotient ring $R[x_1,x_2]/(x_1-x_2^2)$, which is no longer polynomial ring, because it is not free, i.e. there are relations among the generators.
If you are familiar with free groups, this behaves the same way. Free groups are those that do not have relations among their generators. As soon as you impose some relation, you get corresponding quotient group, which is no longer free.