已知函数f(x)=sin(x+4分之7派)+(x+4分之3派),x属于R.求f(x)的最小正周期和最小值

1个回答

  • f(x)=sin(x+2π-π/4)+cos(x-3π/4)

    =sin(x-π/4)+cos(x-3π/4)

    =sin(x-π/4)+cos(x-π/4-π/2) \公式 cos(x-π/2)=sinx\

    =sin(x-π/4)+sin(x-π/4)

    =2sin(β-π/4)

    所以最小正周期为2π,函数最小值为 -2

    (2)f(β)=2sin(β-π/4)

    ∴【f(β)】方=4sin?(β-π/4) \二倍角公式\

    =2cos(2β-π/2) +2

    =2 sin(2β)+2 \公式 cos(x-π/2)=sinx\

    下面求sin(2β):

    根据条件:cos(β-α)=4/5,cos(β+α)= - 4/5

    sin(2β)=sin [(β-α)+(β+α) ]

    =sin(β-α)cos(β+α) + cos(β-α)sin(β+α)

    =3/5×4/5 -4/5×3/5=0 \这里这种方法很常见,也是很喜欢考的技巧之一,就是已知

    (β-α)和(β+α)的正弦或者余弦,求2β或者2α的

    正弦或者余弦\

    【其中,∵0<α<β≤2分之π,∴0<β-α<π/2 ∴sin(β-α)>0 , ∴ sin(β-α)=3/5

    ∵ cos(β+α)=-5分之4