锐角△ABC的内角A,B,C对边分别为a,b,c,a=2bsinA.
(1)角B的大小
(2)cosA+sinC的取值范围?
锐角△ABC的内角A,B,C对边分别为a,b,c,a=2bsinA.
(1)角B的大小
(2)cosA+sinC的取值范围?
(1)a/sinA=b/sinB
2bsinA/sinA=b/sinB
sinB=1/2
B=60度
(1)a=2bsinA,所以sinA/2sinB=sinA,2sinB=1,B=π/6或5π/6.
因为ΔABC是锐角三角形,所以B只能为π/6
(2)因为B=π/6,所以A+C=5π/6.cosA+sinC=cosA+sin(5π/6-A)
展开得,cosA+sinC=cosA+sin(5π/6-A)=cosA+sin(5π/6)cosA-cos(5π/6)sinA
=cosA+(1/2)cosA-(√3/2)sinA=3/2cosA-(√3/2)sinA=(-√3/2)(sinA-√3cosA)=√3/4sin(π/6-A)
很明显,cosA+sinC的取值范围为[-√3/4,√3/4]
30度