sin2x除以1 sin^2(x)的积分
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 22:48:33
不好意思上传上去自动旋转了
f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,
(sinx)^2表示sinx的平方(sinx)^2+2(sinx)^2cosx+(cosx)^2-(sinx)^2=12(sinx)^2cosx+(cosx)^2=(sinx)^2+(cosx)^22
注:题中f(x)=(√3)sin2x-2sin²x-1 解.(1)f(x)=(√3)sin2x-2sin²x-1 =(√3)sin2x+cos2x-2 =2sin(2x+π/
y=1/2sin2x+1/2(1-cos2x)=1/2(sin2x-cos2x+1)=1/2[√2sin(2x-45º)+1]值域[1/2(1-√2),1/2(1+√2)]
∵y=[1+(sinx)^2]/sin2x=[1+(1-cos2x)/2]/sin2x=(3-cos2x)/(2sin2x)=3(csc2x)/2-(cot2x)/2∴y'=[-2*3(csc2x)(
/>Y=1/2sin2x+sin^2x=1/2sin2x+(1-cos2x)/2=1/2+√2/2(√2/2*sin2x-√2/2*cosx)=1/2+√2/2sin(2x-π/4),因为sin(2x
再问:分母是sin2x不是2x再答:x趋向0时sin2x~2x再问:哦哦哦对哦!!谢谢
xsin(1/x^2)-x/sin2x1/x^2→∞,所以sin(1/x^2)在-1到1之间震荡而x→0,所以xsin(1/x^2)极限是0x/sin2x=(1/2)*(2x)/sin2xx→0则2x
Y=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)=((sin^2x+cos^2x)^2-sin^2xcos^2x)/(2-sin2x)=(1-sinxcosx)(1+s
f(x)=0.5sin^2x+cos^2x+根号3除以4乘以sin2xf(x)=0.5(sin^2x+cos^2x)+0.5cos^2x+(√3/4)sin2xf(x)=0.5+0.5cos^2x+(
y=1/2sin2x+(1-cos2x)/2=1/2(sin2x-cos2x)+1/2=√2/2*(√2/2*sin2x+√2/2zos2x)+1/2=√2/2*(sin2xzosπ/4+cos2xs
f(x)=sin2x+cos2x-1=√2sin(2x+π/4)-1.1、最小正周期是π,最大值时2x+π/4=2kπ+π/2,即x=kπ+π/4,k是整数.再问:已知函数f(x)=2sin(∏-X)
若有不懂请追问,如果解决问题请点下面的“选为满意答案”.
设t=sin2x,由x属于(0,45°)得0
原式=[((sinx)^2+(cosx)^2)^2-(sinxcosx)^2]/2(1-sinxcosx)=[1-(sinxcosx)^2]/2(1-sinxcosx)=(1-sinxcosx)(1+
将函数f(X)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π/4)化简得:=(1+cosx/sinx)*2sinxcosx+m(sinxcosπ/4+cosxsinπ/4)(
cos(π/4-x)=-4/5得(√2/2)(cosx+sinx)=-4/5所以cosx+sinx=-4√2/5-----------(1)cos(π/4+x)=sin(π/2-π/4-x)=sin(
(1)、f(x)的定义域为sinx≠0,即x≠kπ;如果定义x=kπ时f(x)等于-1,可将定义域扩大至整个实数域;f(x)=[2√3sinxcosx-2cos²x]+1=√3sin2x-c
f(x)=(1+cos2x+8sin^2x)/sin2xf(x)=(1+2cosx²-1+8sin²x)/2sinxcosxf(x)=(cosx²+4sin²x