y=2cos(2x+π/3)+1,x∈(0,π/2)
当x=0时,有最大值=2×1/2+1=2
当x=π/3时,有最小值=2×(-1)+1=-1y=2cos(2x+π/3)+1 求化简过程y=2cos(2x+π/3)+1 =2cos2xcosπ/3-2sin2xsinπ/3+1=cos2x-√3sin2x+1求下列函数的最大值和最小值:y=(sinx-1)/(sinx-2) 膜拜y=3cos2x-4sinx+1 =3(1-2sin²x)-4sinx+1=3-6sin²x-4sinx+1=-6sin²x-4sinx+4=-6(sin²x+2/3sinx)+4=-6(sinx+1/3)²+4+2/3=-6(sinx+1/3)²+13/3当sinx=-1/3；原式有最大值=13/3当sinx=1；原式有最小值=-6×16/9+13/3=-19/3