1.化简(1+cosa-sina)/(1-cosa-sina)+(1-cosa-sina)/(1+cosa-sina)2.求tan20+4sin20的值3.求证:tanx+1/cosx=tan(x/2+∏/4)4.已知sina=4/5,cos(a+b)=-3/5,a、b∈(0,∏/2)则sinb=
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![1.化简(1+cosa-sina)/(1-cosa-sina)+(1-cosa-sina)/(1+cosa-sina)2.求tan20+4sin20的值3.求证:tanx+1/cosx=tan(x/2+∏/4)4.已知sina=4/5,cos(a+b)=-3/5,a、b∈(0,∏/2)则sinb=](/uploads/image/z/7209093-21-3.jpg?t=1.%E5%8C%96%E7%AE%80%EF%BC%881%2Bcosa-sina%EF%BC%89%2F%EF%BC%881-cosa-sina%EF%BC%89%2B%EF%BC%881-cosa-sina%EF%BC%89%2F%EF%BC%881%2Bcosa-sina%EF%BC%892.%E6%B1%82tan20%2B4sin20%E7%9A%84%E5%80%BC3.%E6%B1%82%E8%AF%81%EF%BC%9Atanx%2B1%2Fcosx%3Dtan%EF%BC%88x%2F2%2B%E2%88%8F%2F4%EF%BC%894.%E5%B7%B2%E7%9F%A5sina%3D4%2F5%2Ccos%EF%BC%88a%2Bb%EF%BC%89%3D-3%2F5%2Ca%E3%80%81b%E2%88%88%EF%BC%880%2C%E2%88%8F%2F2%EF%BC%89%E5%88%99sinb%3D)
1.化简(1+cosa-sina)/(1-cosa-sina)+(1-cosa-sina)/(1+cosa-sina)2.求tan20+4sin20的值3.求证:tanx+1/cosx=tan(x/2+∏/4)4.已知sina=4/5,cos(a+b)=-3/5,a、b∈(0,∏/2)则sinb=
1.化简(1+cosa-sina)/(1-cosa-sina)+(1-cosa-sina)/(1+cosa-sina)
2.求tan20+4sin20的值
3.求证:tanx+1/cosx=tan(x/2+∏/4)
4.已知sina=4/5,cos(a+b)=-3/5,a、b∈(0,∏/2)则sinb=
1.化简(1+cosa-sina)/(1-cosa-sina)+(1-cosa-sina)/(1+cosa-sina)2.求tan20+4sin20的值3.求证:tanx+1/cosx=tan(x/2+∏/4)4.已知sina=4/5,cos(a+b)=-3/5,a、b∈(0,∏/2)则sinb=
(1+cosa-sina)/(1-cosa-sina)+(1-cosa-sina)/(1+cosa-sina)=2sina/(cos^2a-1) 这部分没问题是对的 但后面错了 最后=2sina/(1-sin^2a-1)=-2/sina
tan20'+4sin20'=(sin20+2sin40)/cos20=
(sin20+sin40+sin40)/cos20=
(2sin30cos10+sin40)/cos20=
(cos10+sin40)/cos20=
(sin80+sin40)/cos20=2sin60cos20/cos20=2sin60=根号3
secx-tanx=1/cosx-sinx/cosx=(1-sinx)/cosx
=(coxx/2-sinx/2)^2/(cosx/2)^2-(sinx/2)^2
=(coxx/2-sinx/2)/(coxx/2-sinx/2)
tan(π/4 - x/2)=(1-tanx/2)/(1+tanx/2)
=(coxx/2-sinx/2)/(coxx/2-sinx/2)
∴secx - tanx = tan(π/4 - x/2)
sinA+sinB)(sinA+sinB)=1+2sinAsinB=9/25
sinAsinB=-8/25
(cosA+cosB)(cosA+cosB)=1+2cosAcosB=16/25
cosAcosB=-9/50
cos(A-B)=cosAcosB+sinAsinB=-1/2
SIN(A+B)+COS(A+B)=或者=根号2倍的cos(A+B-45度)=根号2倍的[cos(A+B)cos45度+sin(A+B)sin45]
=1/2[cos(A+B)+sin(A+B)]
=2分之根号2倍的cos0
=2分之根号2