原式=sinA/COSA+sinB/cosB
=(sinAcosB+sinBcosB)/cosAcosB
=sin(A+B)/cosAcosB
=sin(180-120)/cosAcosB
=sin60/cosAcosB
=√3/2cosAcosB
=√3/2cosAcos(60-A)
=√3/[1-(sinA)^2+√3sinA ] 令sinA=x,且sinA在(0.√3/2)
F(x)=√3/[1-x^2+√3x]
下面的 就自己解决了撒
原式=sinA/COSA+sinB/cosB
=(sinAcosB+sinBcosB)/cosAcosB
=sin(A+B)/cosAcosB
=sin(180-120)/cosAcosB
=sin60/cosAcosB
=√3/2cosAcosB
=√3/2cosAcos(60-A)
=√3/[1-(sinA)^2+√3sinA ] 令sinA=x,且sinA在(0.√3/2)
F(x)=√3/[1-x^2+√3x]
下面的 就自己解决了撒