已知向量m=(2,-1),n=(sin二分之A,cos(B+c)),A,B,C为三角ABC的内角,其所对的边分别为a,b
已知向量m=(2,-1),n=(sin二分之A,cos(B+c)),A,B,C为三角ABC的内角,其所对的边分别为a,b,c.(1) 当m,n取得最大值时,求角A的大小 (2)在(1)成立的条件下,当a=根号3时,求b的二次方+c的二次方的取值范围 一定要做对啊.
分还是给这位仁兄了
分还是给这位仁兄了
赞 温迪37 幼苗
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(1)
M.N
=(2,-1).(sin(A/2),cos(B+C))
=2sin(A/2)-cos(B+C)
=2sin(A/2)+cosA
=2sin(A/2) + (1-2[sin(A/2)]^2)
= -2[sin(A/2)]^2+2sin(A/2)+1
=-2(sin(A/2)-1/2)^2+1/2
max M.N at sin(A/2) = 1/2
sin(A/2) = 1/2
A/2 = π/6
A = π/3
(2)
sin(B) = sin(2π/3-C)
=(√3/2)cosC +(1/2)sinC
sinBsinC
= [(√3/2)cosC +(1/2)sinC]sinC
=(√3/4)(sin2C)+ (1/4)(1-cos2C)
=(1/2)[(√3/2)sin2C-(1/2)cos2C] + (1/4)
= (1/2)sin(2C-π/6) + (1/4)
0
M.N
=(2,-1).(sin(A/2),cos(B+C))
=2sin(A/2)-cos(B+C)
=2sin(A/2)+cosA
=2sin(A/2) + (1-2[sin(A/2)]^2)
= -2[sin(A/2)]^2+2sin(A/2)+1
=-2(sin(A/2)-1/2)^2+1/2
max M.N at sin(A/2) = 1/2
sin(A/2) = 1/2
A/2 = π/6
A = π/3
(2)
sin(B) = sin(2π/3-C)
=(√3/2)cosC +(1/2)sinC
sinBsinC
= [(√3/2)cosC +(1/2)sinC]sinC
=(√3/4)(sin2C)+ (1/4)(1-cos2C)
=(1/2)[(√3/2)sin2C-(1/2)cos2C] + (1/4)
= (1/2)sin(2C-π/6) + (1/4)
0
1年前
10
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