- 当0<A<π时,求证2SIN2A<COTA/2
- 当0<A<π时,求证2SIN2A<COTA/2
2sin2A-cgt(A/2)
=2sin2A-[sinA/(1-cosA)]
=4sinAcosA-[sinA/(1-cosA)]
=sinA{4cosA-[1/(1-cosA)]}
=sinA*[-(2cosA-1)^2/(1-cosA)]
=-sinA(2cosA-1)^2/(1-cosA)
∵0<A<π
∴sinA>0;(1-cosA)>0且(2cosA-1)^2≥0
∴2sin2A-cgt(A/2)=-sinA(2cosA-1)^2/(1-cosA)≤0
∴2sin2A≤cgt(A/2)
怎么多了个等号?(在A=60°时相等)