已知sin的平方2a+sin2acosa-cos2a=1,a属于(0,pai/2),(1)求tana的值;(2)求使函数

3个回答

  • (1) sin的平方2a+sin2acosa-cos2a=1

    sin2acosa-cos2a=1-(sin2a)^2=(cos2a)^2

    sin2acosa=cos2a(1+cos2a)

    2sina(cosa)^2=cos2a*2(cosa)^2

    sina=cos2a=1-2(sina)^2

    2(sina)^2+sina-1

    sina=-1,sina=1/2

    ∵a∈(0,π/2)

    ∴sina=1/2,a=π/6

    tana=tan(π/6)=√3/3

    (2)f(x)=sin的平方(x+a)+sin(a+x)cos(a+x)

    =1/2[sin(2x+2a)-cos(2x+2a)+1]

    =√2/2[sin(2x+2*π/6)cos(π/4)-cos(2x+2*π/6)sin(π/4)]+1/2

    =√2/2sin(2x+2*π/6-π/4)+1/2

    =√2/2sin(2x+π/12)+1/2

    sin(2x+π/12)=1时,f(x)取得最大值

    2x+π/12=2kπ+π/2

    x=kπ+5π/24

    因此,{x|x=kπ+5π/24}