Question

在△ABC中,A,B,C为三个内角,f(B)=4cosBsin^2(π/4++B/2)+根号3cos2B-2cosB

（1）若f（B）=2，求角B
（2）若f（B）-m>2恒成立，求实数m的取值范围

Answer 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