第1个回答 2013-04-22
原题应为(cosA)^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1
其中A,B,C∈[0,π/2],且A+B+C=π
对(cosA)^2+(cosB)^2=[cos2A+1]/2 + [cos2B +1]/2
=[cos2A +cos2B]/2 +1
=cos(A+B)cos(A-B) +1
=(cosAcosB-sinAsinB)(cosAcosB+sinAsinB)+1
=(cosAcosB)^2-(sinAsinB)^2+1
原式化为
(cosAcosB)^2-(sinAsinB)^2 +1 +(cosC)^2+2cosAcosBcosC=1
即(cosAcosB+cosC)^2-(sinAsinB)^2=0
其中A,B,C∈[0,π/2] 故cosAcosB+cosC>0 ,sinAsinB>0
所以 cosAcosB+cosC=sinAsinB
即cos(A+B)+cosC=0
∵A+B+C=π
∴A+B=π-C
显然 cos(A+B)+cosC=0 成立