Question

函数f(x)=sinx(sinx-cosx)的单调递减区间?

Answer 1

f(x)=(sinx)^2-sinxcosx=(1-cos2x)/2-(sin2x)/2=-(sin2x+cos2x)/2+1/2=-(根号2)sin(2x+π/4)/2+1/2
2x+π/4在[-π/2+2kπ,π/2+2kπ]递减（注意，上一行的前面有个负号）
所以递减区间为[-3π/8+kπ,π/8+2kπ]

Answer 2

f(x)=sinx(sinx-cosx) =负的（根号2/2）sin(四分之π+2X)+0.5所以他的单调递减区间是（-3π/8+kπ，π/8+kπ）打得好蛋疼！望楼主采纳!

Answer 3

就是求f'(x)=cosx(cosx+sinx)<0
变形得cosx*cosx + cosx*sinx <0 ===> ( cos2x + 1)/2 + sin2x/ 2 < 0
===>cos2x + sin2x < -1==>√2sin(2x + π/4) < -1

Answer 4

此题考察的是函数的导数求增减区间类的问题。求f(x)导数 f'(x)=cosx(sinx-cosx)+sinx(cosx+sinx)=2cosxsinx-(cosx)^2+(sinx)^2=sin2x-cos2x 令 f'(x)=0 求出x=π/8 则x=π/8是稳定点，当x<π/8时f'(x)<0 单调减

Answer 5

不对，应该是f(x)=sin^2x-sinxcosx=sin^2x-1/2sin2x
然后把sin^2x化成含有sin2x的式子是那个倍角公式然后再求导就行其他的没事

Answer 6

f(x)=sinx(sinx-cosx)
=sin^2x-sinxcosx
=1/2-1/2cos2x-1/2sin2x
=1/2-
√2/2[sin(2x+π/4)]
其递减区间就是[sin(2x+π/4)]
的递增
2kπ-π/2<2x+π/4<2kπ+π/2
[kπ-3/8π,kπ+1/8π]为所求区间