sinx+sin^2x=1求cos^2x+cosx的6次方+cosx的8次方的值!

(1)因为 sinx+sin^2x=1
又因为
sin^2x+cos^2x=1
两式相减得:
sinx-cos^2x=0
sinx=cos^2x
则:cos^2x+cos^6x+cos^8x
=cos^2x+(cos^2x)^3+(cos^2x)^4
=sinx+(sinx)^3+(sinx)^4
=sinx+[(sinx)^3+(sinx)^4]
=sinx+sin^2x[sinx+sin^2x]
=sinx+sin^2x[cos^2x+sin^2x]
=sinx+sin^2x
=cos^2x+sin^2x
=1
(2)3cos^2x+cos^8x
=3cos^2x+(cos^2x)^4
=3sinx+(sinx)^4
=3sinx+(sin^2x)^2
=3sinx+(1-sinx)^2
=3sinx+(1+sin^2x-2sinx)
=sin^2x+sinx+1
=sin^2x+cos^2x+1
=2