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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 # ]9 L5 R7 ]1 p
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Tan22.5°=厂2-1=0.4142
' v5 Y9 a3 H& WTan15°=2-厂3=0.2680 ^4 f: s& U+ z9 u% w, b
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
W: S; j6 [4 v kTanθ/2=厂(X^2+1)-X (夹θ角的两边长之差)
( e% [1 S5 [" [# T! i. zIf X=1, we have three lengths 1,1 and 厂2,So θ=45°,Tan22.5°=厂2-1
+ k' a/ G) b& U1 q5 d8 nIf X=厂3=1.732,X^2+1=4, so we have three lengths 1, 厂3,2, θ=30° Tan15°=2-1.732.....
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( {0 Q# }5 H" V( o0 d' H BI 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)。
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