首页 >>  正文

sin^2cos^2

来源:baiyundou.net   日期:2024-07-03

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。

","gnid":"9e20d6b99a498e2ea","img_data":[{"flag":2,"img":[{"desc":"","height":"600","title":"","url":"https://p0.ssl.img.360kuai.com/t01064161b657855156.jpg","width":"800"},{"desc":"","height":"533","title":"","url":"https://p0.ssl.img.360kuai.com/t010cd13f31d94a5cfd.jpg","width":"800"},{"desc":"","height":"600","title":"","url":"https://p0.ssl.img.360kuai.com/t01c9eaa186ea6c5fed.jpg","width":"800"}]}],"original":0,"pat":"art_src_0,sexf,sex4,sexc,fts0,sts0","powerby":"hbase","pub_time":1686838681000,"pure":"","rawurl":"http://zm.news.so.com/53d3ef3d5c980ba390514ed8e79c4628","redirect":0,"rptid":"75c2aae867576aa2","rss_ext":[],"s":"t","src":"仁新数学","tag":[],"title":"函数y=1/sin(5x+6)的性质及其图像

章侍司4976sin^2(x/2)]*[cos^2(x/2)]等于多少啊 -
长点屈13643788524 ______ [sin^2(x/2)]*[cos^2(x/2)] =[sin(x/2)cos(x/2)]^2 =[(1/2)*2*sin(x/2)cos(x/2)]^2 =[(1/2)sinx]^2 =(1/4)(sinx)^2

章侍司4976函数最小值Y=1/(SINX)^2+2/(COSX)^2的最小值
长点屈13643788524 ______ 因为sin^2+cos^2=1 则sin^2=1-cos^2 然后用换元法,设cosx=t y=1/1-t^2+1/t 后面就用解一元二次方程做了

章侍司4976(sinx)^2 (cosx)^2 (tanx)^2 周期分别是什么~ -
长点屈13643788524 ______ 前两个利用2倍角公式 第3个先化成余弦,然后利用二倍角公式,或者化成余弦利再用前两个的结论也行 答案是:π π π 这都不会 怎么学的啊

章侍司4976化简/sin^2 70° 已知 - sinx=3/2cosx.求2cos^2x - sin2x的值 -
长点屈13643788524 ______[答案] 式子太长不好打,我就分下来解了 sin50°=cos40° cos40°(1+√3tan10°)=cos40°*2*(1/2+√3/2tan10°) =2cos40°*(sin30°cos10°+cos30°sin10°)/cos10° =2cos40°sin40°/cos10° =sin80°/cos10°=1 sin^2 70°=(cos20°)^2=(cos40°+1)/2 (倍角公式) ...

章侍司4976y=sin^2x和y=cosx^2求二阶导数等于多少? -
长点屈13643788524 ______[答案] y=sin^2x的导数:y'=2sinxcosx 二阶导数为: y''=2cos^2-2sin^x=2cos2xy=cosx^2的导数:y'=-2xsinx^2 二阶导数为: y''=-2(sinx^2+2x^2cosx^2)=-2sinx^2-4x^2cosx^2

章侍司4976sin^x/2= cos^x/2= tan^x/2= cot^x/2= -
长点屈13643788524 ______ sin(x/2)^2=(1-cosx)/2 cos(x/2)^2=(1+cosx)/2 tan(x/2)^2=(1-cosx)/(1+cosx)=(1-cosx)^2/sinx^2=sinx^2/(1+cosx)^2 cot(x/2)^2=(1+cosx)/(1-cosx)=(1+cosx)^2/sinx^2=sinx^2/(1-cosx)^2

章侍司4976sinx^2cosx^2的化简推导过程如何推导的结果为 1 - cos4x -
长点屈13643788524 ______[答案] sin²xcos²x =(1/4)[2sinxcosx]² =(1/4)sin²2x =(1/8)[2sin²x] =(1/8)[1-cos4x] =(1/8)-(1/8)cos4x

章侍司4976证明sin^2x+cosx^2=1
长点屈13643788524 ______ sin^2x+cosx^2=(y/r)^2+(x/r)^2=(y^2+x^2)/r^2=r^2/r^2=1

章侍司4976y=sinx^2乘以cosx^2,求导题目是y=sin平方x乘以cos平方x,进行求导,正确答案是sin2xcosx^2 - 2xsin平方xsinx^2, -
长点屈13643788524 ______[答案] 方法一: y=(sinx)^2·(cosx)^2, y'=[(sinx)^2]'·(cosx)^2+(sinx)^2·[(cosx)^2]' =2sinx·(sinx)'·(cosx)^2+(sinx)^2·2cox·(cosx)' =2sinx(cosx)^3-2(sinx)^3cosx 备注:化简看看 =2sinxcosx[(cox)^2-(sinx)^2] =sin2x·cos2x =1/2·2sin2xcos2x =1/2 ·sin4x ...

章侍司4976若sin^2 x - sinxcosx - 2cos^2 x=0,求x? 请给出详细的步骤! 谢谢 ~~ -
长点屈13643788524 ______[答案] (sinx)^2-2(cosx)^2=sinxcosx [1-3(cosx)^2]^2=[1-(cosx)^2](cosx)^2 (cosx)^2=t (1-3t)^2=t-tt 9tt-6t+1=t-tt 10tt-7t+1=0 (2t-1)(5t-1)=0 t=1/2或t=1/5 然后你自己做吧

(编辑:自媒体)
关于我们 | 客户服务 | 服务条款 | 联系我们 | 免责声明 | 网站地图 @ 白云都 2024