■49663 / inTopicNo.3) |
Re[2]: オイラーの公式
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□投稿者/ らすかる 一般人(27回)-(2019/07/15(Mon) 04:49:48)
| l[n]=2nrsin(π/n)なので (l[2n])^2={4nrsin(π/(2n))}^2 =(4nr)^2・{sin(π/(2n))}^2 =(4nr)^2・{1-cos(π/n)}/2 (∵半角の公式) =2・(2nr)^2・{1-cos(π/n)} =2・(2nr)^2・{1-√{1-(sin(π/n))^2}} (∵cos(π/n)>0,(sinx)^2+(cosx)^2=1) =2・2nr・{2nr-2nr√{1-(sin(π/n))^2}} =4nr{2nr-√{(2nr)^2-(2nrsin(π/n))^2}} =4nr{2nr-√{4n^2r^2-(l[n])^2}}
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