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发表于 2014-4-24 17:57
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本帖最后由 大车司机 于 2014-4-24 19:42 编辑
5 p( C! T: w3 v& x) g+ x
! c1 _% L1 i" P% }6 l. cTan22.5°=厂2-1=0.4142
3 n" `- T+ f* s9 I& ^. @2 zTan15°=2-厂3=0.2682 X& S" W, R' n5 D9 d8 j k1 h$ n
假如直角三角形的两个直角边的长度为X和1,边长1对应的角为θ,那么
! l% R/ O' h# E* VTanθ/2=厂(X^2+1)-X (夹θ角的两边长之差)
! l5 \6 k; n' z# e* uIf X=1, we have three lengths 1,1 and 厂2,So θ=45°,Tan22.5°=厂2-1
5 M2 n( _+ R2 b% ]- M- `( [8 @If X=厂3=1.732,X^2+1=4, so we have three lengths 1, 厂3,2, θ=30° Tan15°=2-1.732.....# _& x) c* x7 }: Q# `7 o; @
, P/ I# V6 N0 u! G# EI hope this will help you memorize ... 相邻的两个整数的平方根之差!.....30°的直角三角形1,1.732,2 --->Tan15°=2-1.732.......22.5°...1,1,1.4142---->Tan22.5°=1.4142-1 (夹45°角的两边长之差)。; [- h4 B4 s" N6 W0 l5 i
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