帮帮孩子求个柿子吧：
∣∑i=12022∑j=12022cos(ijπ2023)∣\displaystyle \Bigg| \sum_{i=1}^{2022} \sum_{j=1}^{2022} \cos \Big(\frac{ij\pi}{2023}\Big) \Bigg|i=1∑2022j=1∑2022cos(2023ijπ)
