在△ABC中,角A,B,C所对边分别为a,b,c.若b2=ac,求y=1+sin2B/sinB+cosB的取值范围.
在△ABC中,角A,B,C所对边分别为a,b,c.若b2=ac,求y=
的取值范围.
1+sin2B |
sinB+cosB |
数学人气:977 ℃时间:2019-12-14 03:12:09
优质解答
∵b2=ac,∴cosB=a2+c2−b22ac=a2+c2−ac2ac=12(ac+ca)-12≥12.∴0<B≤π3,y=1+sin2BsinB+cosB=(sinB+cosB)2sinB+cosB=sinB+cosB=2sin(B+π4).∵π4<B+π4≤7π12,∴22<sin(B+π4)≤1.故1<y≤2....
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