已知函数f(x)=2-sin(2x+π/6)-2sin^2x x∈R记三角型ABC内角A,B,C的对边长分别为a,b,c若F(B/2)=1 b=1 c=√3求a的值

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已知函数f(x)=2-sin(2x+π/6)-2sin^2x x∈R记三角型ABC内角A,B,C的对边长分别为a,b,c若F(B/2)=1 b=1 c=√3求a的值

已知函数f(x)=2-sin(2x+π/6)-2sin^2x x∈R记三角型ABC内角A,B,C的对边长分别为a,b,c若F(B/2)=1 b=1 c=√3求a的值
已知函数f(x)=2-sin(2x+π/6)-2sin^2x x∈R记三角型ABC内角A,B,C的对边长分别为a,b,c若F(B/2)=1 b=1 c=√3
求a的值

已知函数f(x)=2-sin(2x+π/6)-2sin^2x x∈R记三角型ABC内角A,B,C的对边长分别为a,b,c若F(B/2)=1 b=1 c=√3求a的值
f(x)=2-sin(2x+π/6)-2sin^2x
=2-sin2xcos(π/6)-cos2xsin(π/6)-(1-cos2x)
=1-sin2xcos(π/6)+cos2xsin(π/6)
=1-sin(2x-π/6)
F(B/2)=1-sin(B-π/6)=1
B=π/6
b²=a²+c²-2accosB
1=a²+3-2a*√3*(√3/2)
a²-3a+2=0
(a-1)(a-2)=0
a=1或a=2

F(B/2)=2-sin(2x+π/6)-2sin^2x
=1-sin(B+π/6)-cosB=1
所以sin(B+π/6)+cosB=0
化简√3*sin(B+π/3)=0
因为B∈(0,π)
所以B=2π/3
根据余弦定理
3=1+a^2-2*a*cos2π/3
(a+2)(a-1)=0
所以a=1