谁能帮我解决几道高中数学必修4-三角恒等变换的几道题?

2个回答

  • 1.

    -1=tan(α+β)=tanα+tanβ/(1-tanαtanβ)

    tanαtanβ-1=tanα+tanβ)

    tanαtanβ-tanα-tanβ=1

    tanαtanβ-tanα-tanβ+=2

    (1-tanα)(1-tanβ)=2

    2.

    sin^4θ+cos^4θ=(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ

    =1-(1/2)*(2sinθcosθ)^2

    =1-(sin2θ)^2/2=5/9

    sin2θ=2√2/3 sin2θ=-2√2/3

    3.

    sin(α/2)-cos(α/2)=1/5,两边平方得

    1-sinα=1/25

    sinα=24/25

    f(X)=sin(X+π/6)+sin(X-π/6)+cosX+a

    =2sinxcosπ/6+cosx+a

    =√3sinx+cosx+a

    =2sin(x+π/6)+a

    最大值为1

    a=-1

    f(X)≥0

    sin(x+π/6)≥1/2

    2kπ+π/6≤x+π/6≤2kπ+5π/6

    2kπ≤x≤2kπ+2π/3