* $(1):$
$$A = \sin \dfrac{2\pi}{2019} \cdot \sin \dfrac{4\pi}{2019} \cdot \sin \dfrac{6\pi}{2019} \cdot ... \cdot \sin \dfrac{1008\pi}{2019} \cdot \sin \dfrac{1010\pi}{2019} \cdot ... \cdot \sin \dfrac{2018\pi}{2019}$$
Áp dụng $\sin x = \sin (\pi - x)$ (bắt đầu từ $\sin \dfrac{1010\pi}{2019}$) ta được:
$$A = \sin \dfrac{2\pi}{2019} \cdot \sin \dfrac{4\pi}{2019} \cdot \sin \dfrac{6\pi}{2019} \cdot ... \cdot \sin \dfrac{1008\pi}{2019} \cdot \sin \dfrac{1009\pi}{2019} \cdot \sin \dfrac{1007\pi}{2019} \cdot ... \cdot \sin \dfrac{\pi}{2019}$$
$$\implies A = \sin \dfrac{\pi}{2019} \cdot \sin \dfrac{2\pi}{2019} \cdot \sin \dfrac{3\pi}{2019} \cdot ... \cdot \sin \dfrac{1009\pi}{2019}$$
* $(2):$
Áp dụng công thức sau:
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
với $n = 2019$, ta được:
$$\sin \dfrac{\pi}{2019} \cdot \sin \dfrac{2\pi}{2019} \cdot \sin \dfrac{3\pi}{2019} \cdot ... \cdot \sin \dfrac{2017\pi}{2019} \cdot \sin \dfrac{2018\pi}{2019} = \dfrac{2019}{2^{2018}}$$
Lại áp dụng $\sin x = \sin (\pi - x)$ (bắt đầu từ $\sin \dfrac{1010\pi}{2019}$), ta được:
$$\left(\sin \dfrac{\pi}{2019} \cdot \sin \dfrac{2\pi}{2019} \cdot \sin \dfrac{3\pi}{2019} \cdot ... \cdot \sin \dfrac{1008\pi}{2019} \cdot \sin \dfrac{1009\pi}{2019}\right)^2 = \dfrac{2019}{2^{2018}}$$
$$\implies A^2 = \dfrac{2019}{2^{2018}}$$
$$\implies A = \dfrac{\sqrt{2019}}{2^{1009}}$$
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