Is it possible to get a formula for this expression?$\sin 1 \sin 2 \sin 3 ...\sin n=?$ Is it possible to get a formula for this expression? 
$$\sin 1 \sin 2 \sin 3 ...\sin n=?$$
I know that I can remove half of the terms.  I use the formula 
$$(sin x)^2=\frac{1-\cos 2x}{2}$$ 
but if I do that i think I'm stuck.
 A: One approach is to use Fourier series. Define
$$f_n(x) = \sin(x) \cdots \sin(nx).$$
Its Fourier series can be written as
$$f_n(x) = \frac{1}{2}a_{n0} + \sum_{k=1}^\infty a_{nk} \cos(kx) + \sum_{k=1}^\infty b_{nk} \sin(kx).$$
Then we can compute coefficients recursively to get:
$$a_{n0} = b_{(n-1)0},$$
$$a_{nk} = \frac{1}{2}\left(b_{(n-1)(n+k)} + b_{(n-1)(n-k)}\right),$$
$$b_{nk} = \frac{1}{2}\left(a_{(n-1)(n-k)} - a_{(n-1)(n+k)}\right),$$
where we take the convention $a_{-k} = a_{k}$ and $b_{-k} = - b_{k}$ for positive $k$.
The first coefficients are $b_{11} = 1$, with all other $b_{1k}$ and $a_{1k}$ equal to zero. The next non-zero coefficients become $a_{21} = \frac{1}{2}$ and $a_{23} = - \frac{1}{2}$. After that, we have $b_{32} = \frac{1}{4}$, $b_{34} = \frac{1}{4}$ and $b_{36} = - \frac{1}{4}$. It is not difficult to write a program to compute more values. Afterwards, you just plug in the coefficients and set $x = 1$ to get a superposition of sines or cosines.
Although this is not an explicit formula, it does give some insights into the form of the superposition. For instance, notice the coefficients will be divided by two each time. Also notice that the function will alternate between a superposition of sines and cosines.
