已知向量a=(cosα ,sinα ),向量b=(根号2-sinα ,cos α),α属于(-π/2,π/2)

1个回答

  • a=(cosα,sinα),b=(√2-sinα,cosα)

    a·b=(cosα,sinα)·(√2-sinα,cosα)

    =√2cosα-sinαcosα+sinαcosα

    =√2cosα

    1

    |a+b|^2=|a|^2+|b|^2+2a·b=1+2+sinα^2-2√2sinα+cosα^2+2√2cosα

    =4-2√2sinα+2√2cosα

    =4-2√2(sinα-cosα)

    =4-4sin(α-π/4)=4+2√3

    即:sin(α-π/4)=-√3/2

    α∈(-π/2,π/2),则:α-π/4∈(-3π/4,π/4)

    故:α-π/4=-π/3或-2π/3

    即:α=-π/12或-5π/12

    2

    c=(√2,sinα),则:a-c=(cosα-√2,0)

    则:(a-c)·b=(cosα-√2,0)·(√2-sinα,cosα)

    =√2cosα-sinαcosα-2+√2sinα

    =-sin(2α)/2+2sin(α+π/4)-2

    =cos(2α+π/2)/2+2sin(α+π/4)-2

    =(1-2sin(α+π/4)^2)/2+2sin(α+π/4)-2

    =-sin(α+π/4)^2+2sin(α+π/4)-3/2

    =-(sin(α+π/4)-1)^2-1/2

    α∈(-π/2,π/2),则:α+π/4∈(-π/4,3π/4)

    即:sin(α+π/4)∈(-√2/2,1]

    故当sin(α+π/4)=1时,(a-c)·b取得最大值:-1/2