The question is to evaluate $$\prod_{i=1}^{60} \sin(3i-2) \sin(3i-1)$$ where $i$ is the degree for angles.
I tried writing few terms $(\sin 1 \sin2)(\sin4 \sin 5)(\sin 7 \sin 8)...(\sin 178 \sin 179)$ which can be rearranged as $(\sin 1 \sin 179)(\sin 2 \sin 178)...(\sin 89 \sin 91)$ which can also be written as $2^{-60} (1+ \cos 178)(1+\cos 176)...(1+\cos 2)$ .I couldn't proceed after this.Any ideas?Also how to evaluate it $60$ is replaced by any natural number $n$.Thanks.
In conventional notation, you are after $$A=\prod_{i=1}^{60}\sin\frac{(3i-2)\pi}{180}\sin\frac{(3i-1)\pi}{180}.$$ Then $$A=\frac{P_{180}}{P_{60}}$$ where $$P_N=\prod_{j=1}^{N-1}\sin\frac{j\pi}{N}.$$ It is well-known that $$P_N=\frac{N}{2^{N-1}}.\tag1$$ Thus $$A=\frac{3}{2^{120}}.$$
A hint for $(1)$. Consider the product $$N=\prod_{j=1}^{N-1}(1-\exp(2\pi ij/N)).$$
After you rearranged,
$$ \color{green}{ (\sin 1° \sin179°) (\sin2° \sin178°) ...(\sin44°\sin136°)}(\sin45°\sin135°)\color{red}{(\sin46°\sin134°)...(\sin88°\sin92°) (\sin89°\sin91°)} $$
Note that the green as well as red terms obey
$\left(\color{teal}{\sin\theta \sin(180°- \theta)}\right) = ( \sin\theta \cdot \sin\theta) $
$$ \left[ \ \text{Using: } \sin(180°-\theta) = \sin\theta \ \right] $$
So our product series is:
$$ (\sin^2 1°)( \sin^2 2°) ... (\sin^2 45°) ... ( \sin^2 88° )(\sin^2 89°) $$
