F.4 附加數學 - 二倍角公式

一條不容易的附加數學問題
(a)tan3A

=tan(2A+A)

=(tan2A+tanA)/(1-tanAtan2A)

=[2tanA/(1-tan^2A)+tanA ]/[(1-[2tan^2A/(1-tan^2A)]]

=(2tanA+tanA-tan^3A)/(1-3tan^2A]

= 3 tanA - tan^3 A / 1 - 3 tan² A 。

(b)

代A=18
tan54 = 3 tan18 - tan^3 18 / 1 - 3 tan² 18
1/tan36= tan18 - tan^3 18 / 1 - 3 tan² 18
(1-tan^2 18)/2tan18=(3tan18 - tan^3 18) / 1 - 3 tan² 18
(1-tan^2 18)(1 - 3 tan² 18)=2tan18(3tan18 - tan^3 18)
1-4tan^2 18 + 3 tan^4 18=6tan^2 18 - 2tan^4 18
1-10tan^2 18 + 5 tan^4 18=0
tan^2 18
=(10-√80)/10
=(10-4√5)/10
tan^2 36
=4tan^218/(1-tan^2 18)^2
=(5-2√5)
tan^2 36
=1/tan^2 36
=1/(5-2√5)
=(5+2√5)/5



2008-01-31 00:47:55 補充：
倒數4行tan^2 36應是tan^2 54