化简sina^8-cosa^8-sin^6a+cos^6a+sin^4a

令 x=(sin a)^2,
则 (cos a)^2 =1-x.
所以 原式 =x^4 -(1-x)^4 -x^3 +(1-x)^3 +x^2
=[ x^2 +(1-x)^2 ] [ x^2 -(1-x)^2 ] -x^3 +(1 -3x +3x^2 -x^3) +x^2
=(2x^2 -2x +1) (2x -1) -2x^3 +4x^2 -3x +1
=2x^3 -2x^2 +x
=2 (sin a)^6 -2 (sin a)^4 +(sin a)^2
= = = = = = = = =
换元法,注意次数.
可惜最后更麻烦,再检查一下题目.