Is there a formula for $\sin(xy)$ Can we express a trigonometric function for the product of two angles as a function of trigonometric functions of its factors?
For example: Is there a formula for $\sin(xy)$ as a function of $\sin x$ and $\sin y$ or other trigonometric functions of $x$ and $y$.
 A: Actually, there is a sort of formula. Let
$$T_d^n=\cos\left(\frac{n-d}{2}\pi\right)\frac{2^dn}{n+d}{{\frac{n+d}2}\choose d}$$
then we have 
$$\cos(n \arccos(x))=\sum_{d=0}^\infty T_d^nx^d \,\,\,\,\,\,\,\,\,\forall |x|<1$$
as both satisfy the differential equation (with a unique solution)
$$(1-x^2)f''=xf'-n^2f, f(0)=\cos\left(\frac{n\pi}2\right),f'(0)=n\sin\left(\frac{n\pi}2\right)$$
Then note that $\cos(nx)=\cos(n \arccos(\cos x))$ for $|x|\leq\pi$ (for integer $n$ however the statement is true for all $x$). Therefore we have
$$\cos(n x)=\sum_{d=0}^\infty T_d^n\cos^d(x) \,\,\,\,\,\,\,\,\,\forall |x|\leq\pi$$
and by differentiating once and doing some multiplying, we get
$$\sin(n x)=\sum_{d=0}^\infty \frac{d\cdot T_d^n}{n}\cos^{d-1}(x)\sin(x) \,\,\,\,\,\,\,\,\,\forall |x|\leq\pi$$
Note: When $n$ is an integer, $T_d^n=0$ for $d>n$ (the binomial coefficient is $0$), so the series will actually be finite. The $\cos\left({\frac{n-d}{2}\pi}\right)$ will also have a repeating pattern of $\{1,0,-1,0\}$, and the formula will be valid for all $x$. These are the so-called Chebyshev Polynomials.
A: AFAIK, no, not for the general case $x,y \in \mathbb{R}$. But if $y=n \in \mathbb{N}$, we have
$$\sum_{k=0}^n \binom{n}{k}\cos^kx\sin^{n-k}x\sin \left[ \frac{1}{2}(n-k)\pi \right]$$
There's a couple of other formulations of the linked page, but as said, I doubt that this has been extended to cover all reals. But I have been wrong before and would myself also be very interested in finding such a function.
A: If, say, $-\pi/2 \le x \le \pi/2$, and $-\pi/2 \le y \le \pi/2$, then
$$\sin(xy) = \sin(\arcsin(\sin(x)) \arcsin(\sin(y)))$$
expresses $\sin(xy)$ as a function of $\sin(x)$ and $\sin(y)$.
Other formulas can be used for other intervals.  
On the other hand, 
there is no way to write $\sin(xy)$  as a continuous function of a finite
number of functions of the form $\sin(\alpha x)$, $\cos(\alpha x)$, 
$\sin(\alpha y)$, $\cos(\alpha y)$
, valid for all real $x,y$ (i.e. no continuous function $F$ on $[-1,1]^n$, choices $v_j$ of 
$\sin(\alpha_j x)$, $\cos(\alpha_j x)$, $\sin(\alpha_j y)$, $\cos(\alpha_j y)$, and constants $\alpha_1 \ldots \alpha_n$, 
such that $F(v_1,\ldots,v_n) = \sin(xy)$ for all $x,y \in \mathbb R$).
Such a function $F(v_1, \ldots,v_n)$ would be uniformly
continuous on $\mathbb R^2$, and $\sin(xy)$ is not.
A: One can use binomial expansion in combination with the complex extension of trig functions:
$$\cos(xy)=\frac{e^{xyi}+e^{-xyi}}{2}=\frac{a^{xy}+a^{-xy}}2$$
Using $a=e^i$ for simplicity.
We also have:
$$(a+a^{-1})^n=\sum_{i=0}^{\infty}\frac{n!a^{n-i}a^{-i}}{i!(n-i)!}=\sum_{i=0}^{\infty}\frac{n!a^{n-i}}{i!(n-2i)!}$$
Which is obtained by binomial expansion.
We also have:
$$(a+a^{-1})^n=(a^{-1}+a)^n=\sum_{j=0}^{\infty}\frac{n!a^{2j-n}}{j!(n-j)!}$$
And, combining the two, we get:
$$(a+a^{-1})^n=\frac{\sum_{i=0}^{\infty}\frac{n!a^{n-2i}}{i!(n-i)!}+\sum_{j=0}^{\infty}\frac{n!a^{2j-n}}{j!(n-j)!}}2=\frac12\sum_{i=0}^{\infty}\frac{n!}{i!(n-i)!}(a^{n-2i}+a^{-(n-2i)})$$
If we have $\cos(n)=\frac{a^n+a^{-n}}2$, then we have
$$(2\cos(n))^k=(a^n+a^{-n})^k=\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}\frac{a^{n(k-2i)}+a^{-n(k-2i)}}2$$
Furthermore, the far right of the last equation can be simplified back into the form of cosine:
$$\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}\frac{a^{n(k-i)}+a^{-n(k-i)}}2=\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$
Thus, we can see that for $\cos(ny)$, it simply the first of the many terms in $\cos^n(y)$ and we may rewrite the summation formula as:
$$(2\cos(n))^k=\cos(nk)+\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$
And rearranging terms, we get:
$$\cos(nk)=2^k\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$
This becomes explicit formulas for $n=0,1,2,3,\dots$
I note that there is no way by which you may reduce the above formula without the knowledge that $n,k\in\mathbb{Z}$.
Also, it is quite difficult to produce the formulas for, per say, $\cos(10x)$ because as you proceed to do so, you will notice that it requires knowledge of $\cos(8x),\cos(6x),\cos(4x),\dots$, which you can eventually solve, starting with $\cos(2x)$ (it comes out to be the well known double angle formula), using this to find, $\cos(4x)$, use that to find $\cos(6x)$, etc. all the way to $\cos(10x)$.
Notably, this can be easier than Chebyshev Polynomials because it only requires that you know the odd/even formulas less than the one you are trying to solve. (due to $-2i$)
But this is the closest I may give to you for the formula of $\cos(xy)$, $x,y\in\mathbb{R}$.
It is also true for $x,y\in\mathbb{C}$.
As others have noted, this can also be solved in terms of the Chebyshev Polynomial:
$$T_n(\cos(x))=\cos(nx)$$
Trivially,
$$T_0(\cos(x))=1$$
$$T_1(\cos(x))=\cos(x)$$
Through the sum of angles formula, it is derivable that we have:
$$T_n(\cos(x))=2T_{n-1}(\cos(x))-T_{n-2}(\cos(x))$$
A much easier recursive formula for $n\in\mathbb{Z}$.
The formula for $\sin(nk)$ is easily derivable with binomial expansion:
$$\sin(nk)=\frac{e^{-nki}-e^{nki}}{2i}=\frac{c^{nk}-c^{-nk}}2$$
The solution is very similar to the cosine, with the exception that complex numbers will appear more than one may like.
Also, there is no Chebyshev polynomial for sine as far as I have seen.  Probably easier to use $\sin(nk)=\cos(nk+\frac12\pi)$
Addendum
I shall proceed to attempt to explain how to further use my recursive definition.
Start with
$$\cos(nk)=2\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$
We also have:
$$\cos(n(k-2j))=2\cos^{k-2j}(n)-\sum_{i=1}^{\infty}\frac{(k-2j)!}{i!(k-2j-i)!}(\cos(n(k-2j-2i)))$$
Combine the above two to get:
$$\cos(nk)=2\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(2\cos^{k-2j}(n)-\sum_{j=1}^{\infty}\frac{(k-2i)!}{j!(k-2i-j)!}(\cos(n(k-2j-2i))))$$
I'm going to call all of the numbers with the factorials $\beta_i$:
$$\cos(nk)=2\cos^k(n)-\sum_{i=1}^{\infty}\beta_i(2\cos^{k-2j}(n)-\sum_{j=1}^{\infty}\beta_j\cos(n(k-2j-2i))))$$
Through a very painful process, you may factor out each and every term this way, I just imagine it isn't so much of a beautiful process.
You will most likely also run into the problem of divergence, which may be fixed using $\cos(k-2i)=\cos(2i-k)$, allowing the terms in the binomial expansion to always have positive exponents so you don't run into $0^{-1}$ or divergence problems.
A: Unlikely to be explicitly/purely in terms of $\sin x$, $\sin y$, etc.
But 
$\, \sin xy=\sin \left[ \frac{(x+y)^{2}}{4}-\frac{(x-y)^{2}}{4} \right]
           =\sin \frac{(x+y)^{2}}{4} \cos \frac{(x-y)^{2}}{4}-
            \cos \frac{(x+y)^{2}}{4} \sin \frac{(x-y)^{2}}{4}$
or approximately for small $\, x$ and $y$,
$\sin xy \approx \sin x \sin y
 \left( 1+\frac{\sin^{2} x+\sin^{2} y}{6}+\frac{3}{40}\sin^{4} x-\frac{5}{36}\sin^{2} x \sin^{2} y+\frac{3}{40} \sin^{4} y \right)$
