Question

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

Answer 1

(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)/54,7,已知sin(π-α)-cos(π+α)=根号2/3 求下列各式的值 (1)sinα-cosα (
2)sin^3(2π-α)-cos^3(2π-α)