sin(x-2π)
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 15:35:32
楼上都错了,图像没问题这个表达式实际是个常数,你可以运行TrigReduce[Sin[x]Sin[x+2]-Sin[x+1]^2]看看,结果为1/2(-1+Cos[2])只不过Plot的自动选择坐标系
2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[
(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间:π/2+2
y=sin(x+π3)sin(x+π2)=(sinxcosπ3+cosxsinπ3)cosx=12sinxcosx+32cos2x=14sin2x+32•1+cos2x2=34+12sin(2x+π3
令a=π-x则a趋于0sin3x=sin(3π-3a)=sin3asin2x=sin(2π-2a)=-sin2a所以原式=-lim(a→0)sin3a/sin2asin3a和sin2a的等价无穷小是3
f(x)=2sin(π-x)sin(π/2-x)=2sinxcosx=sin2x1)最小正周期=2π/2=π2)在区间[-派/6,派/2]上x=π/4时,有最大值=sinπ/2=1x=-π/6时,有最
f(x)=2sinx*sin(π/2+x)-2sin^2x+1=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)因为f(x0/2)=根2/3所以sin(x0+π/4)
f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si
sin(π/4+x/2)sin(π/4-x/2)=sin(π/4+x/2)sin[π/2-(π/4+x/2)]∵π/4=π/2-π/4∴sin(π/4-x/2)=sin(π/2-π/4-x)=sin[
sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π
f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T
fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3(cos*2x-sin*2x)=sin2x+根号3cos2x=2sin(2x+派
sin(2x+π/3)
(Ⅰ)f(x)=sinx•cosx+12cos2x+12=12sin2x+12cos2x+12=22sin(2x+π4)+12∴函数f(x)的最小正周期T=2π2=π(Ⅱ)当x∈[−π8,3π8]时,
原式=sin(x+π/3)+√3cos(x+π/3)+2sin(x-π/3)=2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3)=2sin(x+π/3+π/6)+2
-cosx
原式=[sinx*(-cosx)*sinx]/[(-cosx)*sinx*(-sinx)]=-1
f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s
(1)当x∈(0,π/2)时:y=|sin(x+π/2)|-|sinx|=sin(x+π/2)-sinxy′=cos(x+π/2)-cosx(2)当x∈(π/2,π)时:y=-sin(x+π/2)-s
sin2x+sin(π-x)/2cos2x+2sin^2x+cos(-x)=sin2x+sin(x)/2cos2x+2sin^2x+cos(x)=(2cosxsinx+sinx)/2cos²