已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x+a-1(a为常数),若f(x)的最大值

1个回答

  • f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2*x+a-1

    =sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+2cos^2x-1+a

    =2sin2xcosπ/3+2cos^2x-1+a

    =sin2x+cos2x+a

    =√2sin(2x+π/4)+a

    当sin(2x+π/4)=1时,取最大值,即√2+a=√2+1 a=1

    (2)f(x)的对称中心就是 =√2sin(2x+π/4)与x轴的交点(因为图像向上平移了一个单位,所以对称中心的纵坐标都为1)

    即√2sin(2x+π/4)=0解得x=kπ/2-π/8.

    所以对称中心的坐标为(kπ/2-π/8,1)

    (3)g(x)=f(x+3/8π)+2=√2sin(2(x+3/8π)+π/4)+1+2=√2sin(2x+π)+3

    =-√2sin2x+3

    求-√2sin2x的减区间就行了,也就是求√2sin2x的增区间

    2kπ-π/2《2x《2kπ+π/2 ,解出来 kπ-π/4《x《kπ+π/4 所以g(x)的减区间为【kπ-π/4,kπ+π/4】