Is this a correct formula for sin(a b)? First of all, recall the complex definitions of the two trigonometric functions:
$$ \sin(x) = \frac{e^{ix}-e^{-ix}}{2i}, \space \cos(x) = \frac{e^{ix}+e^{-ix}}{2}. $$
Now, the formula for $\sin(2x)$ can be derived from these two. But so can it be extended by multipying each new result by $ 2\cos(2^{b}x) $:
$$ 2\sin x \cos x = \sin(2x), \space 4\sin x \cos x \cos(2x) = \sin(4x), \space ...
 $$
and so on. Therefore
$$ 2^{b}\sin x \prod_{n=0}^{b-1}\cos(2^{n}x) = \sin(2^{b}x). $$
However, it might be possible - and this is the step that could be a fallacy - to rewrite the product so as to generalize the outcome to some real numbers.
$$ 2^{b}\sin\ x \prod_{n=0}^{\infty}\frac{\cos(2^{n}x)}{\cos(2^{n}2^{b}x)} = \sin(2^{b}x) =>   \prod_{n=0}^{\infty}\frac{\cos(2^{n}x)}{\cos(2^{n}bx)} = \frac{\sin(bx)}{b\sin x}.$$
So far, is this true?
Pick x = 1 and notice this is false for b = $\frac{\pi}{4}$. However, for x = 1 and b = $\pi$ or x = 1 and b = 1 the product goes to appropriate values.
Can I now deduce from this the formula for the reciprocal of the product?
$$ \prod_{n=0}^{\infty}\frac{\cos(2^{n}bx)}{\cos(2^{n}x)} = \frac{b\sin x}{\sin(bx)} $$
Instead of starting the derivation with sinx, I could have started it with $\frac{1}{\sin x}$ and then continued multiplying the denominator. Hence the invalidity of the formula for $\frac{\pi}{2^{n}}, \space n > 1 \space and \space n \in \mathbb{N}, \space$ cannot be explained by the first product's singularities, right?
Is the idea anyhow correct, at least for some numbers?
 A: As I mentioned in comments, I think the conjecture you intended was actually
$$ \prod_{n=1}^{\infty}\frac{\cos(2^{-n}bx)}{\cos(2^{-n}x)} = \frac{\sin(bx)} {b\sin x}.$$
This follows immediately from the following identity:
$$ \prod_{n=1}^{\infty}{\cos(2^{-n}t)} = \frac{\sin t}t,\qquad (1) $$
as we may simply combine the cases $t=bx$ and $t=x$ to get:
\begin{eqnarray*}
\prod_{n=1}^{\infty}\frac{\cos(2^{-n}bx)}{\cos(2^{-n}x)} &=& 
\frac{\prod_{n=1}^{\infty}{\cos(2^{-n}bx)}}{\prod_{n=1}^{\infty}{\cos(2^{-n}x)}}\\
\\&=&\frac{\frac{\sin(bx)}{bx}}{\frac{\sin x}x}
\\&=&\frac{\sin(bx)} {b\sin x}
\end{eqnarray*}
The proof of $(1)$  is similar to your original argument:
For $m\in \mathbb{N}$ we have: $$\frac{\sin(2^{-m}t)}{2^{-m}t}\prod_{n=1}^m \cos(2^{-n}t)=\frac{\sin t}t.$$
Letting $m\to \infty$ gives us $(1)$.
This argument seems to have been given many times on this site: see for example here or here or here or here.
I thought the way you arrived at the conjecture was very clever (+1) and reminded me of Euler's construction of the Gamma function as an infinite product.
A: note that:
$$\sin(2x)=\frac{e^{2ix}-e^{-2ix}}{2i}=\frac{(e^{ix}+e^{-ix})(e^{ix}-e^{-ix})}{2i}=2\frac{e^{ix}-e^{-ix}}{2i}\frac{e^{ix}+e^{-ix}}{2}=2\sin(x)\cos(x)$$
Try this expansion with your hypothesis and see what you get
