已知sin(π-α)-cos(π+α)=根号2⼀3 求下列各式的值

sin(π-α)-cos(π+α)=√2/3
sinα+cosα=√2/3
(sinα+cosα)^2=2/9
(sinα)^2+2sinαcosα+(cosα)^2=2/9
1+2sinαcosα=2/9
2sinαcosα=-7/9
sinαcosα=-7/18

1-2sinαcosα=16/9
(sinα-cosα)^2=16/9
sinα-cosα=±4/3

sin^3(π/2-α)-cos^3(π/2+α)
=cos^3α+sin^3α
=(cosα+sinα)(sin^2α-sinαcosα+cos^2α)
=√2/3(1-sinαcosα)
=√2/3(1+7/18)
=√2/3*25/18
=25√2/54

1)
sin(π-a)-cos(π+a)
=sina+cosa=√2/3
∴同平方得sin2a=-7/9
∴(sina-cosa)²=(sina+cosa)²-4sinacosa
=16/9
∵sina>0,cosa<0
∴sina-cosa=4/3
2)
由（1）知 sina+cosa=√2/3
sina-cosa=4/3
∴sin³(π/2-α)-cos³(π/2+α)
=cos³a+sin³a
=cos²acosa+sin²asina
=(1+cos2a)/2*cosa+(1-cos2a)/2*sina
=[sina+cosa+cos2a(cosa-sina)]/2
=(√2/3+4/3cos2a)/2
∵cos2a=cos²a-sin²a=(cosa+sina)(cosa-sina)=-4√2/9
∴(√2/3+4/3cos2a)/2
=-7√2/54
额… 这样吧