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cos1和sin1等于多少

来源:baiyundou.net   日期:2024-08-22

函数y=1/sin(5x+6)的性质及其图像


主要内容:

本文主要介绍函数y=1/sin(5x+6)的定义域、单调性、凸凹性等性质,并解析函数的单调区间和凸凹区间。

※.函数定义域

根据函数特征,函数自变量x在分母,则有sin(5x+6)≠0,此时有:

5x+6≠kπ,k∈Z,即x≠(kπ-6)/5,

所以函数的定义域为:{x|x≠(kπ-6)/5 ,k∈Z。}


※.函数单调性

根据正弦函数的单调性,可知其取倒数的函数y=1/sin(5x+6)单调性。

对于函数y=sin(5x+6)的单调性及单调区间为:

(1)单调增区间

2kπ-π/2≤5x+6≤2kπ+π/2,

2kπ-π/2-6≤5x≤2kπ+π/2-6

(4k-1)π/10-6/5≤x≤(4k+1)π/10-6/5,

(2)单调减区间

2kπ+π/2≤5x+6≤2kπ+3π/2,

2kπ+π/2-6≤5x≤2kπ+3π/2-6

(4k+1)π/10-6/5≤x≤(4k+3)π/10-6/5,

由此可知,函数y=1/sin(5x+6)的单调性如下:

(1)函数的减区间为:(4k-1)π/10-6/5≤x≤(4k+1)π/10-6/5,

(2)函数的增区间为:(4k+1)π/10-6/5≤x≤(4k+3)π/10-6/5。

※.函数的凸凹性

用导数知识来解析函数的凸凹性

∵y=1/sin(5x+6),

∴dy/dx=-5cos(5x+6)/sin^2(5x+6),继续求导有:

d^2y/dx^2=-5\n[-5sin(5x+6)sin^2(5x+6)-5cos(5x+6)*2sin(5x+6)cos(5x+6)]/sin^4(5x+6)],

=5^2[sin(5x+6)sin^2(5x+6)+cos(5x+6)*2sin(5x+6)cos(5x+6)]/sin^4(5x+6)],

=5^2[sin^2(5x+6)+cos(5x+6)*2cos(5x+6)]/sin^3(5x+6)],

=5^2*[1+cos^2(5x+6)]/sin^3(5x+6),

此时函数的凸凹性如下:

(1)当sin(5x+6)>0时,d^2y/dx^2>0,此时函数为凹函数,即:

2kπ<5x+6<2kπ+π,

2kπ-6<5x<2kπ+π-6

2kπ/5-6/5<x<(2k+1)π/5-6/5,

(2)当sin(5x+6)<0时,d^2y/dx^2<0,此时函数为凸函数,即:

2kπ+π<5x+6<2kπ+2π,

2kπ+π-6<5x<2kπ+2π-6

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n(2k+1)π/5-6/5<x<(2k+2)π/5-6/5。

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