设△ABC的内角A,B,C所对的边a,b,c成等比数列,则sinAcotC+cosAsinBcotC+cosB的范围是(  ) A.(0,+∞) B.(0,5+12) C.(5-12,5+12) D.(5-12,+∞)

问题描述:

设△ABC的内角A,B,C所对的边a,b,c成等比数列,则

sinAcotC+cosA
sinBcotC+cosB
的范围是(  )
A. (0,+∞)
B. (0,
5
+1
2
)

C. (
5
-1
2
5
+1
2
)

D. (
5
-1
2
,+∞)

设三边的公比是q,三边为a,aq,aq2,原式=sinAcosCsinC+cosAsinBcosCsinC+cosB=sinAcosC+cosAsinCsinBcosC+cosBsinC=sin(A+C)sin(B+C)=sinBsinA=ba=q∵aq+aq2>a,①a+aq>aq2②a+aq2>aq,③解三个不等式可得q>5-...