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[函数] 求证三角恒等式

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hbghlyj posted 2020-1-20 18:45 |Read mode
Last edited by hbghlyj 2020-1-20 18:53${(\sin a + \sin b + \sin c + \sin d - \sin (a + b) - \sin (c + d))^2} + {( - 2 + \cos a + \cos b + \cos c + \cos d - \cos (a + b) - \cos (c + d))^2} = {(\sin a + \sin b - 2\sin (a + b) + \sin (a + b - c) + \sin (a + b - d) - \sin (a + b - c - d))^2} + {( - 1 + \cos a + \cos b - 2\cos (a + b) + \cos (a + b - c) + \cos (a + b - d) - \cos (a + b - c - d))^2}$

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original poster hbghlyj posted 2020-1-20 18:48
这是这道题的副产品。我本想通过模长均为1来导致矛盾,后来发现这是个恒等式

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