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三角函數問題 煩請解惑
版主: thepiano
Re: 三角函數問題 煩請解惑
1 / (2n^2) = 2 / (4n^2) = [(2n + 1) - (2n - 1)] / [1 + (2n + 1)(2n - 1)]
arctan(1 / (2n^2)) = arctan([(2n + 1) - (2n - 1)] / [1 + (2n + 1)(2n - 1)]) = arctan(2n + 1) - arctan(2n - 1)
所求 = π/2 - π/4 = π/4
arctan(1 / (2n^2)) = arctan([(2n + 1) - (2n - 1)] / [1 + (2n + 1)(2n - 1)]) = arctan(2n + 1) - arctan(2n - 1)
所求 = π/2 - π/4 = π/4
Re: 三角函數問題 煩請解惑
鋼琴老師 謝謝
但這一段跳太快了
為什麼
arctan([(2n + 1) - (2n - 1)] / [1 + (2n + 1)(2n - 1)])
可以 = arctan(2n + 1) - arctan(2n - 1)??
謝謝你喔
但這一段跳太快了
為什麼
arctan([(2n + 1) - (2n - 1)] / [1 + (2n + 1)(2n - 1)])
可以 = arctan(2n + 1) - arctan(2n - 1)??
謝謝你喔