已知sinα-cosα=m,求sin∧4α+cos∧4α的值

sin∧4α+cos∧4α=(sin^2a+cos^2a)-2sin^2acos^2a
=1-2sin^2acos^2a
sinα-cosα=m,
(sinα-cosα)^2=m^2
2sinacosa=1-m^2
sin^2acos^2a=(1-m^2)/4
sin∧4α+cos∧4α=1-2sin^2acos^2a
=1-(1-m^2)/2
=(1+m^2)/2