Question

在△ABC中,a,b,c分别是角A,B,C的对边,bcosC,acosA,ccosB成等比数列.求角A的弧度数

Answer 1

(acosA)^2=(bcosC)(ccosB)=(bcosB)(ccosC)

(sinAcosA)^2=(sinBcosB)(sinCcosC)

(sin2A)^2=sin2Bsin2C

(1-cos2A)/2=(cos(2B-2C)-cos(2B+2C))/2=(cos(2B-2C)-cos2A)/2

1=cos(2B-2C)

因为|2B-2C|