已知sin(π-α)-cos(π+α)=根号2/3 求下列各式的值 (1)sinα-cosα (

1个回答

  • (1) sin(π-α)-cos(π+α) = sinα + cosα =√2sin(α+π/4) = (√2)/3 所以:sin(α+π/4) = 1/3

    而sinα - cosα = -(√2)cos(α+π/4)

    所以当 α+π/4在第一象限时,有:sinα - cosα = -(√2)√(1-1/9) = -4/3

    所以当 α+π/4在第二象限时,有:sinα - cosα = (√2)√(1-1/9) = 4/3

    (2) sin^3(2π-α)-cos^3(2π-α) = -sin^3(α) - cos^3(α) = -(sinα + cosα )(sin^2(α) + sinαcosα + cos^2(α))

    = - (sinα + cosα )(1 + sinαcosα)

    而 sinαcosα =[ (sinα + cosα)^2 -1 ] /2 =[ 2/9 -1 ]/2 =-7/18

    以根据(1)有:sin^3(2π-α)-cos^3(2π-α) = -(√2)/3( 1 - 7/18) = -(11√2)/54