~tan 連乘式一題~

(tan1° + 1) (tan44° + 1) = 2
(tan2° + 1) (tan43° + 1) = 2
(tan3° + 1) (tan42° + 1) = 2
......
(tan22° + 1) (tan23° + 1) = 2
∴ (tan1° + 1) (tan2° + 1) (tan3° + 1)......(tan44° + 1)
= 2^22
至於為何有：
(tan1° + 1) (tan44° + 1) = 2
......
(tan22° + 1) (tan23° + 1) = 2
就是關鍵。

2011-02-22 19:30:49 補充：
根據tan的和角公式：
tan (a＋b) = (tan a＋tan b)÷(1－tan a × tan b)
設 a + b = 45度
則 tan (a＋b) = tan 45度
∴ (tan a ＋ tan b)÷(1－tan a × tan b) = 1
tan a ＋ tan b ＋ tan a × tan b = 1
tan a ＋ tan b ＋ tan a × tan b ＋ 1 = 2
(tan a ＋ 1)(tan b ＋ 1) = 2
得：
(tan1° + 1) (tan2° + 1) (tan3° + 1) ...... (tan43° + 1) (tan44° + 1)
= [(tan1°+ 1)(tan44°+1)] [(tan2°+1)(tan43°+1)]...[(tan22°+1)(tan23°+1)]
= 2^22
(= 4194304)

怎知它們 = 2 ?