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.0

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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.0

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.0
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.
0

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.0
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

f(A)=2sinAcos^2θ/2+cosAsinθ-sinA =sinA(1+cosθ)+cosAsinθ-sinA =sin(A+θ) 即f(x)=sin(x+θ) =√3/2 A+θ=π/3或2π/3 正弦

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f(A)=sin(A+θ)=根号下3/2