化简(1+cos2a)/(tana/2)-(cota/2)
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化简(1+cos2a)/(tana/2)-(cota/2)
(1+cos2a)/(tana/2)-(cota/2)
=(1+cos2a)*(cota/2)-(cota/2)
=cos2acota/2
是不是打错了
(1+cos2A)/[tan(A/2)-cot(A/2)]
=2cosA^2/[sin(A/2)/cos(A/2)-cos(A/2)/sin(A/2)]
=2cosA^2*sin(A/2)*cos(A/2)/[sin(A/2)^2-cos(A/2)^2]
=cosA^2*sinA/(-cosA)
=-cosA*sinA
=-sin(2A)/2
=(1+cos2a)*(cota/2)-(cota/2)
=cos2acota/2
是不是打错了
(1+cos2A)/[tan(A/2)-cot(A/2)]
=2cosA^2/[sin(A/2)/cos(A/2)-cos(A/2)/sin(A/2)]
=2cosA^2*sin(A/2)*cos(A/2)/[sin(A/2)^2-cos(A/2)^2]
=cosA^2*sinA/(-cosA)
=-cosA*sinA
=-sin(2A)/2
已知tana+cota=5/2,求cos2a和sin(2a+π/4)
化简根号sin2a(1+cota)+cos2a(1+tana)
化简 cos2a/(1-2tana*cos^2a)
1.化简[2sin2a/(1+cos2a)]*1/tana
证明 tanA-cotA=(1-2cos^2A)/(sinAcosA)
sina+cosa=1/2,tana+cota=
证明下列各式:(1+tana+cota)/(1+tan^2 a+tana)-cota/(1+tan^2 a)=sinac
1若cota+csca=5,则sina= ,tana(1-cota^2)+cota(1-tana^2)= ,若tanx=
证明tanA+cotA=2/sin2A
tana+cota=2,sina×cosa=
tanA=1/ tanA=sinA/ cotA=cosA/ sin^2A+cos^2A=
已知(1+tana)/(1-tana)=3,则[(sina+cosa)^2-1]/(cota-sina*cosa)=?