已知A,B,C为△ABC的三个内角,且cos^2A+cos^2B+cos^2C=t

不妨设C为钝角
2cos²A=1+cos2A,2cos²B=1+cos2B
则t=1+(cos2A+cos2B+2cos²C)/2
=1+[2cos(A+B)cos(A-B)+2cos²C]/2
=1+cosC[cosC-cos(A-B)]
因cosC<0且|A-B|<90°,
则cosC-cos(A-B)<0
则t=1+cosC[cosC-cos(A-B)]>1
显然cos(A-B)≥1,又cosC<0
则cosC[cosC-cos(A-B)]≤cosC(cosC-1)
则t≤1+cosC(cosC-1)=(cosC-1/2)²+3/4
t<(-1-1/2)²+3/4=3(因0>cosC>-1)
故1<t<3