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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.

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    $\sin x=\sqrt{\frac{1-\cos 2x}{2}}$. You need the square root. – 2017-01-18
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    See this for reference https://in.answers.yahoo.com/question/index?qid=20140728071736AAA6igp – 2017-01-18
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    @juniven don't change incorrect stuff in the question, unless the OP indicates that is what the OP meant. It changes the meaning of the question if it states, "I used X" when X is wrong, as compared to "I used Y" when Y is the correct formula. – 2017-01-18
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    $n<\pi$ I assume? – 2017-01-18
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    no,n is whatever positive value i desire – 2017-01-18
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    Wolfram Alpha does not give [any satisfiable answer](http://www.wolframalpha.com/input/?i=product(sin(k),+k%3D1..n)), so I guess in the general case there is no simple formula. I didn't find anything either. – 2017-01-18
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    Your question is equivalent to $\prod\limits_{k=1}^n (1-q^k)=??? (q,n)$ . – 2017-01-18
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    @ Thomas Andrews I will remember it next time. Thanks for the advice – 2017-01-18
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    All questions are legitimate, this one in particular. But there is something that should be done almost always : give a context. Otherwise, we can spend all our life asking questions which are not motivated by anything. My experience is that it is completely different when one has to focus on really deep (abstract or applied) issues. – 2017-01-18

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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.

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    Another observation: the resulting trig polynomial of $x$ has about $6n^2/\pi^2$ zeros in $[0,2\pi)$. So I believe that implies that a positive proportion of the coefficients $a_0, \ldots, a_{n(n+1)/2}, b_1, \ldots, b_{n(n+1)/2}$ must be non-zero, so the expression can't have massive cancellation. – 2017-01-18