在ΔABC中,A,B,C为三个内角,f(B)=4cosB*sin(π/4+B/2)的平方+√3*cos2B-2cosB拜
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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
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