2secA=sec(A+B)+sec(A-B),0<A<(π/2)<B<π,求证:cosA=(根号2)×cos(B/2)
问题描述:
2secA=sec(A+B)+sec(A-B),0<A<(π/2)<B<π,求证:cosA=(根号2)×cos(B/2)
2secA=sec(A+B)+sec(A-B),
0<A<(π/2)<B<π,
求证:cosA=(根号2)×cos(B/2)
答
2/cosA = 1/cos(A+B)+1/cos(A-B) 通分
2/cosA = 2cosAcosB/cos(A+B)cos(A-B) 化简
cosAcosAcosB=cosAcosAcosBcosB-sinAsinAsinBsinB
cosAcosA(1-cosB)=sinBsinB
cosAcosA=sinBsinB/(1-cosB)
cosAcosA=1+cosB 因为[1+cosB=cos(B/2)cos(B/2)]
cosA=(根号2)×cos(B/2)