新浪何处无玻璃
幼苗
共回答了21个问题采纳率:95.2% 举报
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年前
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