1.在三角形ABC中,A,B,C为三个内角,f(B)=4cosB*sin^2(π/4+B/2)+(根号3)cos2B－2

1.f(B)=4cosBsin^2(π/4+B/2)+更号3*(cos2B)-2cosB
= 4cosB * [ 1 - cos(π/2 + B)]/2 + √3 (cos2B) - 2cosB
= 2cosB * [ 1 - cos(π/2 + B)] + √3 (cos2B) - 2cosB
= 2cosB - 2cosBcos(π/2 + B)] + √3 (cos2B) - 2cosB
= - 2cosBcos[π -(π/2 - B)] + √3 (cos2B)
= 2cosBcos(π/2 - B) + √3 (cos2B)
= 2cosBsinB + √3 (cos2B)
= sin(2B) + √3 cos(2B)
= 2 * [(1/2) * sin(2B) + (√3 /2) cos(2B)]
= 2 * [cos(π/3)*sin(2B) + sin(π/3)cos(2B)]
= 2 sin(2B + π/3)
(1)
f(B) = 2
2 sin(2B + π/3) = 2
sin(2B + π/3) = 1
B ∈(0,π)
2B + π/3 ∈ ( π/3,7π/3)
2B + π/3 = π/2
B = π/12
(2)若f(B)-m>2恒成立,求实数m的取值范围.
2 sin(2B + π/3) - m > 2
2sin(2B + π/3) > m+2
2B + π/3 ∈ ( π/3,7π/3)
sin(2B + π/3) ∈ [-1,1]
f(B) ≥ -2
f(B) > m + 2 恒成立,即 即使对最小值 f(B) = -2 也成立
-2 > m + 2
m < -4
2.(1)sin[(B+C)/2]=sin[90-(B+C)/2]=sin[(180-B-C)/2]=sin(A/2)
所以sin[(B+C)/2]平方=sin(A/2)平方=(1-cosA)/2=1/3
cos2A=2(cosA)平方-1=-7/9
所以 sin＾2【/2】+cos2A=1/3-7/9=-4/9
(2) cosA=1/3 所以 sinA=2倍根号2/3
正弦定理 a/sinA=b/sinB=c/sinC
所以由等比定理得 a/sinA=(b+c)/(sinB+sinC)=根号(27/8)=M
所以 b+c=M(sinB+sinC)
因为 bc≤[(b+c)平方]/2 此时b=c
所以 sinB=sinC
cosA=1/3 所以cos(B+C)=cos(2B)=cosA=-1/3
cosB=根号3/3
所以 sinB=根号6/3 sinC=根号6/3
所以 b=c=M*sinB=3/2
所以 bc最大=9/4

1.f(B)=2cosB*(1-cos(π/2+B))+√3cos2B-2cosB=sin2B+√3cos2B=2sin(2B+π/3)=2,B=π/12
2sin(2B+π/3)-m＞2恒成立可转化为2sin(2B+π/3)＞2+m恒成立，求出2sin(2B+π/3)的值域可知m≤-√3/2-2

2.求sin^2［(B+C)/2］+cos2A=sin^2（π/2-A/...