Question

设函数f(x)=2sinxcos^2θ/2+cosxsinθ-sinx

在△ABC中，a，b，c分别是角A,B,C的对边，已知a=1，b=根号2，f（A）=根号3/2，求角C. 过程请仔细 ；谢谢、
0<θ<π，在x=π处取最小值

Answer 1

f(x)=2sinxcos^θ/2+cosxsinθ-sinx
=sinx(cosθ+1)+cosxsinθ-sinx
=sinxcosθ+cosxsinθ+sinx-sinx
=sin(x+θ)
f(A)=sin(A+θ)
=√3/2 所以A+θ=π/3或2π/3
由于0<θ<π，在x=π处取最小值，所以有f（π)=sin(π+θ) =-sinθ
故θ =π/2 所以A=π/6
a/sinA=b/sinB B=π/4 所以C=7π/12

Answer 2

f(x)=2sinxcos²θ/2+cosxsinθ-sinx
=sinx(cosθ+1)+cosxsinθ-sinx
=sinxcosθ+cosxsinθ+sinx-sinx
=sin(x+θ)
θ没告诉 没法求出<A
f(A)=sin(A+θ)=根号下3/2

Answer 3

θ是不是B
f(A)=sin(A+B)=根号3/2
A+B=π/3或2π/3
C=π/3或2π/3