Question

已知向量M=（2，-1），N=（sinA\2,cos(B+C)）,A,B,C为△ABC的内角，其对应的边分别为a,b,c

1，当M×N去读最大值时，求角A的大小
2，在（1）成立的条件下，当a=根号3时，求b²+c²的取值范围

Answer 1

(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 <sinBsinC<= 3/4

by sine rule
b/sinB = a/sinA
b= asinB/sinA =2sinB
c=2sinC
bc = 4sinBsinC

a=√3
by cosine -rule
a^2=b^2+c^2-2bccosA
3=b^2+c^2-bc
b^2+c^2 = 3+bc
3+4(0)< b^2+c^2<= 3+ 4(3/4)
3 < b^2+c^2<= 6