(1-sinx)/cosx 化简成tanx/2的式子
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(1-sinx)/cosx 化简成tanx/2的式子
(1-sinx)/cosx 化简成tanx/2的式子
(1-sinx)/cosx 化简成tanx/2的式子
(1-sinx)/cosx
=[sin²(x/2)+cos²(x/2)-2sin(x/2)cos(x/2)]/[cos²(x/2)-sin²(x/2)]
=[cos(x/2)-sin(x/2)]²/{[cos(x/2)+sin(x/2)]*[cos(x/2)-sin(x/2)]}
=[cos(x/2)-sin(x/2)]/[cos(x/2)+sin(x/2)]
分子分母同时除以cos(x/2)
=[1-tan(x/2)]/[1+tan(x/2)]
(1-cosx)/sinx=tan什么
求证:tan(x/2)=(1-cosx+sinx)/(1+cosx+sinx)
求证:(1+sinx-cosx)/(1+sinx+cosx)=tan(x/2)
已知(1-cosx+sinx)/(1+cosx+sinx)=-2,求tan的值
(sin^x/sinx-cosx)-sinx+cosx/tan^2x-1
证明:tan(x/2)=sinx/1+cosx
求证sin^2x/sinx-cosx-(sinx+cosx)/(tan^2-1)=sinx+cosx谢谢GGJJ
求证 sin^2x/(sinx-cosx)-(sinx+cosx)/tan^2 x-1=sinx+cosx
求证:(sinX+cosX+1)/(1+cosX)=1+tan(X/2)
化简[sin^2(x)]/(sinx-cosx)-(sinx+cosx)/[tan^2(x)-1]-sinx
已知tan(a/2)=5,求(1+sinx-cosx)/(1+sinx=cosx)已知tan(x/2)=5,求(1+sinx-cosx)/(1+sinx+cosx)
化简:sinx/(sinx-cosx) -(sinx+cosx)/(tanx-1)
tanx=—3/4,tan(sinx)>tan(cosx),则sinx=?
(sinx+cosx)(tan^2x+1/tanx)=1/cosx+1/sinx
求证:tan(x/2)= sinx/(1+cosx)=(1-cosx)/sinx
请问tan(x/2)=sinx/(1+cosx)=(1-cosx)/sinx这是怎得到的?Thanks
化简(sin^2x/sinx-cosx) -(sinx +cosx/tan^2x-1)
化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1