已知sinx＋cosx＝m 绝对值m≤根号2 且绝对值m不等于1,且x是第二象限角...

1
sinx+cosx=m 平方一下得
sin^2x+2sinxcosx+cos^2x=m^2
sinxcosx=(m^2-1)/2
sin^3x+cos^3x
=(sinx+cosx)(sin^2-sinxcosx+cos^2x)
=m(1-(m^2-1)/2)
=m(3-m^2)/2
2
sin^2x+cos^2x=1
sin^2x-2sinxcosx+cos^2x=1-2sinxcosx=1-2*(m^2-1)/2=2-m^2
(sinx-cosx)^2=2-m^2
因为x是第二象限角
所以sinx>0 cosx0
所以sinx-cosx=√(2-m^2)
sin^4x-cos^4x
=(sin^2x+cos^2x)(sinx+cosx)(sinx-cosx)
=1*m*√(2-m^2)
=m√(2-m^2)