sin1+cos1等于多少
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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