Continued product in $\sin$ series Find the value of the product $$(\sin 1°)(\sin 3°)(\sin 5°)\ldots(\sin 89°)$$ I tried multiplying and dividing by $2$ and then combining and then converting into cosine, but doesn't work out. 
 A: Notice for any integer $N > 0$, we have
$$\begin{align}
z^{2N} + 1 
&= \prod_{k=-N}^{N-1} \left( z - e^{\frac{2k+1}{2N}\pi i} \right)
= \prod_{k=0}^{N-1}\left(z - e^{\frac{2k+1}{2N}\pi i}\right)\left(z - e^{-\frac{2k+1}{2N}\pi i}\right)\\
&= \prod_{k=0}^{N-1}\left[ z^2+1 - 2z\cos\left(\frac{2k+1}{2N}\pi\right)\right]
\end{align}
$$
Set $z = 1$, we find
$$2 
= 2^N \prod_{k=0}^{N-1} \left[1 - \cos\left(\frac{2k+1}{2N}\pi\right)\right]
= 2^{2N} \prod_{k=0}^{N-1} \sin^2\left(\frac{2k+1}{4N}\pi\right)\\
$$
Since all the $\sin(\cdots)$ involved are positive, this leads to
$$\prod_{k=0}^{N-1} \sin\left(\frac{2k+1}{4N}\pi\right) = 2^{\frac12 - N}$$
Substitute $N = 45$, we get
$$\sin(1^\circ)\sin(3^\circ)\cdots\sin(89^\circ)
= \prod_{k=0}^{44}\sin\left(\frac{2k+1}{180}\pi\right) = 
2^{-89/2}$$
A: $$\prod_{i=1}^{45}\sin(2i-1)^\circ=\prod_{j=1}^{45}\cos(2j-1)^\circ$$
Let $\cos45x=\cos45^\circ$
$\implies45x=360^\circ n\pm45^\circ\iff x=8^\circ n\pm1^\circ$ where $n$ is any integer
Considering the  '+' sign,
$1\le8n+1\le90\iff0\le n\le11 \ \ \ \ (1)$
$91\le8n+1\le180\iff12\le n\le22, \ \ \ \ (2)$
$n=12\implies\cos(97^\circ)=\cos(180^\circ-83^\circ)=-\cos83^\circ$
$n=22\implies\cos(177^\circ)=\cos(180^\circ-3^\circ)=-\cos3^\circ$
$181\le8n+1\le270\iff23\le n\le33, \ \ \ \ (3)$
$n=23\implies\cos(185^\circ)=\cos(180^\circ+5^\circ)=-\cos5^\circ$
$n=33\implies\cos(265^\circ)=\cos(180^\circ+85^\circ)=-\cos85^\circ$
$271\le8n+1<360\iff34\le n<45, \ \ \ \ (4)$
$n=34\implies\cos(273^\circ)=\cos(360^\circ-87^\circ)=+\cos87^\circ$
$n=44\implies\cos(353^\circ)=\cos(360^\circ-7^\circ)=+\cos7^\circ$
There are $11+11$ negative values
Using this, the coefficient of $\cos^nx$ in $\cos(nx),$ is $1+\binom n2+\binom n4+\cdots=(1+1)^{n-1}$  which is validated by this
$\implies\cos45x=2^{45-1}\cos^{45}x+\cdots$
So, the roots $2^{n-1}\cos^{45}x+\cdots=\cos45x=\cos45^\circ$ are $(1),(2),(3),(4)$
$\implies(-1)^{22}\prod_{j=1}^{45}\cos(2j-1)^\circ=\dfrac{\dfrac1{\sqrt2}}{2^{44}}$
