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tAn十二分之五派

根据周期性:tan(5π/4) = tan[(5π/4) - π] = tan(π/4) = 1 ∴原式 = {tan(π/4) + tan(5π/12)}/{1 - [tan(π/4)][tan(5π/12)]} = tan[(π/4) + (5π/12)] = tan(2π/3) = -√3

tan5π/12=tan(3π/4-π/3)={tan3π/4-tanπ/3}/{1+tan3π/4tanπ/3}={-1-3}/{1+(-1)3}=(3+1)/(3-1)=(3+1)/2=2+3

你好:tan(12分之5π)=2+√3(tan 12分之5π减tan 3分之π)乘以6=(2+√3-√3)*6=12

(1-tan5π/12tanπ/4)/tan5π/12+tanπ/4=1/[tan5π/12+tanπ/4/(1-tan5π/12tanπ/4)]=1/tan(tan5π/12+π/4)=1/tan2π/3=-1/根号3 扬州慢的 作者姜夔(jiāng kuí )

tan22.62°=5/12

其实tan4分之5派可以化成1,也就是等于tan4分之1派,然后可以化成(tan4分之1派加上tan12分之5派)除以(1-tan12分之5派乘以tan4分之1派)后面乘以的tan4分之1派是常数1不影响结果就可以化成tan(4分之1派+12分之5派)等于tan3分之2派也就是负的根号3

因为 tan(a+π/3)=tan(a+π/12+π/4)=[tan(a+π/12 )+tanπ/4]/[1-tan( a+π/12 )*tanπ/4 ]=(5+1)/(1-5*1)=6/(-4)=-3/2所以tan(a+π/3)=-3/2希望能帮到你,祝学习进步

sin(5π/3)=sin(2π/3+π)=-sin(2π/3)=-根号3除以2cos(5π/3)=cos(2π/3+π)=-cos(2π/3)=-1/2tan(5π/3)=tan(2π/3+π)=tan(2π/3)=-根号3

(tan5π/4+tan5π/12)/(1-tan5π/12)∵tan5π/4=tanπ/4=1∴原式=(tan5π/4+tan5π/12)/(1-tan5π/4*tan5π/12) =tan(5π/4+5π/12) =tan5π/3 =tan2π/3=-√3

Tan(5*π/12)=tan(π/6+π/4) =[sin(π/6+π/4)]/cos(π/6+π/4) =(sin30cos45+cos30sin45)/(cos30sin45-sin30sin45)

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