Can sine be made into a homomorphism? Consider the usual sine function $\mathbb{R}\rightarrow \mathbb{R}$. Is there some (single) group structure we can put on $\mathbb{R}$ with respect to which sine becomes a homomorphism?
I suspect the answer is either no for a trivial reason, or yes by a simple set-theoretic argument (probably providing a great many such group structures of no interest). 
This latter seems plausible if I can "replace" the reals (and the sine function) by some arbitrary equal-sized set (and sufficiently similar function). Indeed, if I only asked that we have a pair of group structures $*_1$ and $*_2$ so that sine is a homomorphism from one to the other, then such an argument does go through (if I'm not mistaken). 
So I'm putting this question forward mostly in case there's a pleasant surprise. Depending on the answer, one could of course ask for further restrictions (abelian, continuous (probably impossible), torsion, torsion-free etc), but for now I'll leave as is. 
 A: If you require the group structure to be continuous, this is impossible.  Indeed, in that case, the image $[-1,1]$ would have to be a group as well.  But $[-1,1]$ is not homogeneous, so it cannot be a topological group.
Without requiring continuity, it is possible.  Let me first give a construction where you restrict from $\mathbb{R}$ to $X=[-\pi/2,\pi/2]$; I will write $f:X\to X$ for the sine function.  In that case, the (set-theoretic) dynamical system given by $f$ has a particularly simple structure: $f$ is injective with a unique fixed point 0, $X_n=f^n(X)\setminus f^{n+1}(X)$ has cardinality $2^{\aleph_0}$ for all $n$, and $\bigcap f^n(X)=\{0\}$.  Such a dynamical system is completely determined by giving the sequence of sets $X_n$ for $n\geq 0$ with bijections $X_n\to X_{n+1}$.  Let $Y$ be the free $\mathbb{Q}$-vector space on $X$; then it is easy to see that the induced homomorphism $\tilde{f}:Y\to Y$ will have the same properties.  Choosing compatible bijections $X_n\to Y_n$, we get a group structure on $X$ for which $f$ is a homomorphism.
Extending this to all of $\mathbb{R}$ is now easy: identify $\mathbb{R}$ with $X\times \mathbb{Z}$ in such a way that the projection $X\times\mathbb{Z}\to X$ sends each $x\in \mathbb{R}$ to the unique $y\in [-\pi/2,\pi/2]$ such that $\sin x=\sin y$.  Then the sine map $\mathbb{R}\to \mathbb{R}$ can be identified with the projection to $X$ followed by $f:X\to X$.
This construction is quite flexible if you want the group structure to have various properties.  For instance, you could replace $Y$ with the free group on $X$, or the free abelian group on $X$, or the free $\mathbb{F}_p$-vector space on $X$, or many other constructions; you could also replace $X\times\mathbb{Z}$ with any semidirect product of $X$ with a countable group.
