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三个 arctan 的和
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大家拿出计算机, 算算
arctan(1/2) + arctan(1/5) + arctan(1/8)
哈哈, 认得它吗? |
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发表于 14-5-2004 12:21 PM
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楼主 |
发表于 14-5-2004 12:45 PM
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pipi 于 14-5-2004 12:21 PM 说 :
不就是 pi/4 吗?
对了. 或许把问题改改会清楚点
4 * (arctan (1/2) + arctan(1/5) + arctan(1/8) ) |
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发表于 14-5-2004 12:54 PM
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发表于 14-5-2004 01:24 PM
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楼主 |
发表于 14-5-2004 01:58 PM
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发表于 14-5-2004 04:23 PM
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可以呀!
let A=arctan(1/2) --> tan A = 1/2
B=arctan(1/5) --> tan B = 1/5
C=arctan(1/8) --> tan C = 1/8
arctan(1/2) + arctan(1/5) + arctan(1/8) = D
=> A + B + C = D
=> A + (B + C) = D
=> tan( A + (B + C) ) = tan D
=> [tan A + tan (B+C)]/[1 - (tan A)(tan (B+C)] = tan D
=> [ 1/2 + E ]/[ 1 - (1/2)*E ] = tan D {let E = tan (B+C)}
===> E = tan (B+C)
= [tan B + tan C]/[1 - (tan B(tan C)]
= [1/5 + 1/8] / [1 - (1/5)(1/8)]
= [13/40] / [39/40]
= 1/3
=> [ 1/2 + 1/3 ]/[ 1 - (1/2)*(1/3) ] = tan D
=> [ 5/6 ] / [ 1 - (1/6) ] = tan D
=> 1 = tan D
=> D = arctan(1)
=> D = (pi/4)
=> arctan(1/2) + arctan(1/5) + arctan(1/8) = pi/4 |
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楼主 |
发表于 14-5-2004 05:28 PM
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很好!
应用当试题不错嘛 ... |
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发表于 15-5-2004 03:52 AM
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这反正切公式用在电脑上来算 pai 的近似值的。
因为它的展开式允许计算到任何位数。 |
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