求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值

1：分析cos^4x+sin^4x=（cos^2x+sin^2x）^2 -2cos^2x sin^2x
=1-2cos^2x sin^2x
分子=1-1+2cos^2x sin^2x=2cos^2x sin^2x
cos^6x+sin^6x=(cos^2x+sin^2x)(cos^4x-cos^2x sin^2x+sin^4x)
=1×[(cos^2x+sin^2x)^2-3cos^2x sin^2x]
=1-3cos^2x sin^2x
分母=1-1+3cos^2x sin^2x=3cos^2x sin^2x
所以原式=2cos^2x sin^2x/3cos^2x sin^2x =2/3
2:解tan(π/4+a)=(1+tana)/(1-tana)=2,
1+tana=2-2tana,tana=1/3.
1/(2sinacosa+cos^2a)=(sin^2a+cos^2a)/(2sinacosa+cos^2a)
=(tan^2a+1)/(2tana+1)=(1/9+1)/(2/3+1)=2/3.
3：1/2=sin(π/6),√3/2=cos(π/6),因此可对表达式化简：
y=(1/2)(cosx)^2+(√3/2)sinxcosx+1
=cosx[sin(π/6)cosx+cos(π/6)sinx]+1
=sin(x+π/6)cosx+1 ………………………………………………………（1）
sin(2x+π/6)=sin(x+π/6+x)=sin(x+π/6)cosx+cos(x+π/6)sinx ………（2）
1/2=sin(π/6)=sin(x+π/6-x)=sin(x+π/6)cosx-cos(x+π/6)sinx ………（3）
（2）+（3）可得：sin(x+π/6)cosx=[sin(2x+π/6)]/2+1/4 ……………（4）
把（4）代入（1）继续化简：
sin(x+π/6)cosx+1=[sin(2x+π/6)]/2+1/4+1=[sin(2x+π/6)]/2+5/4
因此：y=[sin(2x+π/6)]/2+5/4
（1）y取最大值时,sin(2x+π/6)=1,即2x+π/6=2kπ+π/2,求得x=kπ+π/6（k∈Z）,
因此所求x的集合为：{x|x=kπ+π/6（k∈Z）}
（2）由函数表达式y=[sin(2x+π/6)]/2+5/4可知变换顺序：
sinx → sin(x+π/12) → sin[2(x+π/12)]=sin(2x+π/6) → [sin(2x+π/6)]/2 → [sin(2x+π/6)]/2+5/4
即将函数y=sinx的图像先整体左移π/12个单位,然后横向压缩一倍（即左右压缩）,之后纵向压缩一倍（即上下压缩）,最后整体上移5/4个单位,就可得到题设函数的图像.或者：sinx → sin2x → sin(2x+π/6) → [sin(2x+π/6)]/2 → [sin(2x+π/6)]/2+5/4即将函数y=sinx的图像先横向压缩一倍,然后整体左移π/12个单位,之后纵向压缩一倍,最后整体上移5/4个单位,也可得到题设函数的图像.

1-cos^4x-sin^4x
=1-(cos^4x+sin^4x)
=1-[(cos^2x+sin^2x)^2-2cos^2xsin^2x]
=1-(1-2cos^2xsin^2x)
=2cos^2xsin^2x
1-cos^6x-sin^6x
=1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)
=1-[(cos^2x+sin^2x)^2-3cos^2xsin^2x]
=1-[1-3cos^2xsin^2x]
=3cos^2xsin^2x
原式=2cos^2xsin^2x/3cos^2xsin^2x=2/3

不断的用sin^2x=1-cos^2x

原式＝[1-(sin^2x+con^2x)^2-2sin^2x*con^2x]/[1-(sin^2x+con^2x)*(sin^4x+con^4x-sin^2x*con^2x)]=(2sin^2x*con^2x)/{1-[(sin^2x+con^2x)^2-2sin^2x*con^2x-sin^2x*con^2x]}=(2sin^2x*con^2x)/(3sin^2x*con^2x)=2/3

已知sin^2x+cos^2x=1 分子配方 1-sin^4x-cos^4x=1+2sin^2xcos^2x-(sin^2x+cos^2x)^2=1+2sin^2xcos^2x-1=2sin^2xcos^2x 分母先用立方和公式,再配方 1-sin^6-cos^6x=1-(sin^2x+cos^2x)*(sin^4x-sin^2xcos^2x+cos^4x)=1-(sin^4x-sin^2xcos^2x+cos^4x)=1+3sin^2xcos^2x-(sin^2x+cos^2x)^2=3sin^2xcos^2x 因此原式=2/3