Are $\sin(\alpha\beta)$ and $\sin(\alpha^{\beta})$ expressible in terms of $\sin(\alpha)$ and $\sin(\beta)$? There is a well known formula for expressing $\sin(\alpha+\beta)$ just using $\sin(\alpha)$ and $\sin(\beta)$. It is enough to replace $\cos$ in the formula $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$ by its equivalent form in terms of $\sin$ function. Thus $\sin(\alpha+\beta)$ is expressible via elementary functions (polynomials, radicals, fractions, ...) and $\sin(\alpha), \sin(\beta)$.

Question 1: What about $\sin(\alpha\cdot\beta)$ and $\sin(\alpha^{\beta})$? Can we express them just using $\sin(\alpha)$ and $\sin(\beta)$ (in any non-trivial sense)? If no, how to prove this fact?
Question 2: Also we can express $\tan(\alpha+\beta)$ in terms of $\tan(\alpha)$ and $\tan(\beta)$. What about $\tan(\alpha\cdot\beta)$ and $\tan(\alpha^{\beta})$?

 A: The underlying reason why you shouldn't expect any such formulas to exist is that $e^{ix}$
is a homomorphism with respect to addition:
$$e^{i(x+y)}=e^{ix}\cdot e^{iy}$$
but not with respect to multiplication, and certainly not with respect to exponentiation.
The angle-addition formulae for $\sin$ and $\cos$ come from the homomorphism property:
$$e^{i(x+y)}=\cos(x+y)+i\sin(x+y)$$
$$\begin{align*}
e^{ix}e^{iy}&=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))\\
&=\biggl(\cos(x)\cos(y)-\sin(x)\sin(y)\biggr)+i\biggl(\sin(x)\cos(y)+\cos(x)\sin(y)\biggr)
\end{align*}$$
(Then just identify the real and imaginary parts.)
Because $e^{ixy}$ is not going to have a relationship with $e^{ix}$ and $e^{iy}$ all the time, perhaps a natural next step would be to relax the question to only being related to one of them, say $e^{ix}$. The only time one could have any hope of a relationship with $e^{ix}$ is if $y$ is an integer, because that is precisely when 
$$e^{ixy}=(e^{ix})^y$$
is true for all $x$ (see here on complex exponentiation). And, naturally, this is precisely the situation when $\sin(xy)$ and $\cos(xy)$ have actual expressions of the sort you want (see here).
