sinx-2cosx=0.求sin2x-cos2x/1+sin^2x

可知sinx=2cosx
分子=sin2x-cos2x=2sinxcosx-（cos^2x-sin^2x）=4cos^2x-（cos^2x-4cos^2x）=7cos^2x
分母=sin^2x+cos^2x+sin2x=9cos2x
然后就是=7cos^2x/9cos2x=7/9

sinx-2cosx=0 可得：sinx=2cosx
(sin2x-cos2x)/(1+sin²x)
=(2sinxcosx-cos²x+sin²x)/(sin²x+cos²x+sin²x)
=(4cos²x-cos²x+4cos²x)/(4cos²x+cos²x+4cos²x)
=7cos²x/9cos²x
=7/9

将算式展开，然后利用条件，将所有的cos(x)用sin(x)来代替

sinx-2cosx=0
所以
tanx=2
所以
sin2x-cos2x/1+sin^2x
=（sin2x-cos2x）/[1+(1-cos2x)/2]
=2(sin2x-cos2x)/(3-cos2x)
=2(2tanx/(1+tan²x)-（1-tan²x)/(1+tan²x))/(3-(1-tan²x)/(1+tan²x))
=2(4/5+3/5)/(3+3/5)
=28/(18)
=14/9

sinx-2cosx=0
tanx = 2
sin2x-cos2x/1+sin^2x
=(2sinxcosx-cos²x+sin²x)/(cos²+2sin²x) [上下同除以cos²x]
=(2tanx-1+tan²x)/(1+2tan²x)
=(4-1+4)/(1+8)
=7/9