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问题：已知向量m=(根号3,1),向量n=(cos（x/3）,-sin（x/3）),记f(x)=2*向量m*向量n*sin（x/3）（1）若x∈[π/4,π],求函数f(x)的值域；（2）在△ABC中,角A,B,C所对的边分别为a,b,c,且a,b,c成等比数列,若f(c)=1,求sinA的
答案：↓↓↓ 董杰的回答： 网友采纳 向量m.向量n=√3cos(x/3)-sin(x/3). f(x)=2m.n*sin(x/3). =2[√3cos(x/3)-sin(x/3)]sin(x/3). =2[√3sin(x/3)cos(x/3-sin^2(x/3)] =2{(√3/2)2sin(x/3)cos(x/3)-[(1-cos2x/3)/2]}. =2[(√3/2)sin(2x/3)+(1/2)cos(2x/3)]-1. ∴f(x)=2sin(2x/3+π/6)-1. (1)∵x∈[π/4,π],∴2x/3+π/6∈[0,π/2],sin(2x/3+π/6)在x∈[π/4,π/2]单调递增,在x∈[π/2,π],减. ∴sin(2x/3+π/6)在x=π/2处取得最大值1,即f(x)max=2*1-1=1.sin(2x/3+π/6)在x=π处取得最小值1/2.、 f(x)min=2*1/2-1=0. ∴f(x)在x∈[π/4,π],f(x)∈[0,1].---所求函数的值域. （2）∵f(C)=2sin(2C/3+π/6)=1.∴2C/3+π/6=π/2.2C/3=π/2-π/6=π/3,∴C=π/2. ∵三角形的三边a,b,c成等比数列,∴b^2=ac,. 由正弦定理,得；(sinB)^2=sinAsinC.∵sinC=sin90°=1.∴(sinB)^2=sinA.(1). (sinB)^2=[sin(A+C)]^2 =sinAcosC+cosAsinC. =sinA*cos90°+cosAsin90°. ∴(sinB)^2=cosA.(2). ∵(1)=(2),∴sinA=cosA. sinA/cosA=1.(cosA≠0) tanA=1, ∴∠A=π/4. | 
