Question

在三角形ABC中,BC=根5,AC=3,4cos2A减cos2C=3.求AB的值

Answer 1

4cos2A-cos2C=3,4[1-2(sinA)^2]-[1-2(sinC)^2]=3,

(sinC-2sinA)*(sinC+2sinA)=0,)sinC≠-2sinA,

sinC-2sinA=0,sinC/sinA=2,根据正弦定理,sinC/sinA=c/a=2,c=2a,

a=BC=√5,AB=c=2a=2√5.