2secA=sec(A+B)+sec(A-B),0＜A＜(π/2)＜B＜π,求证:cosA=(根号2)×cos(B/2)2secA_数学解答

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)

人气：117 ℃　时间：2020-06-30 23:05:34

解答

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)