几道三角函数的问题…1:函数y=3sin(x+20)+5sin(x+80)的最大值是多少?2.已知 sina+sinb+

1个回答

  • 1.

    y=3sin(x+20°)+5sin(x+80°)

    =3sin(x+20°)+5sin(x+20°+60°)

    =3sin(x+20°)+5*[sin(x+20°)*cos60°+cos(x+20°)*sin60°]

    =3sin(x+20°)+2.5sin(x+20°)+2.5√3*cos(x+20°)

    =5.5sin(x+20°)+2.5√3*cos(x+20°)

    =5.5*[sin(x+20°)+(5√3/11)*cos(x+20°)]

    =5.5*[sin(x+20°)+tana*cos(x+20°)]

    =5.5*[sin(x+20°)*cosa+cos(x+20°)*sina]/cosa

    =7sin(x+a+20°)

    由于sin(x+a+20°)属于[-1,1]

    则:当sin(x+a+20°)=1时,

    y=3sin(x+20)+5sin(x+80)

    取最大值为7

    2.由sina+sinb+sinc=0

    得:sina+sinb=-sinc

    则:(sina+sinb)^2=(sinc)^2 ---①

    由cosa+cosb+cosc=0

    得:cosa+cosb=-cosc

    则:(cosa+cosb)^2=(cosc)^2 ---②

    ①+②,得:

    (sina)^2+(cosa)^2+(sinb)^2+(cosb)^2+2sinsinb+2cosacosb

    = (sinc)^2+(cosc)^2

    1+1+2cos(a-b)=1

    则:cos(a-b)=-1/2

    3.

    由和差化积公式,得:

    sinx+siny

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

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

    由于0