作业帮sin(3 x) sin(2 x)极限
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 11:10:13
1和3等价无穷小替代,sinxx,答案为2/5和w,第二题用洛必达法则,答案cosa,过程应该会写吧,我用手机回答的,输入不方便,请谅解
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
原式=1/2sinx+根号3/2cosx+sinx-根号3cosx+根号3/2cosx-3/2sinx（先全部展开）=0（最后合并同类项得0）
用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos
第一步,二倍角公式cos2@=1-2sin²@2-（1/2）（cos6x+cos10x+cos14x+cos18x）=2得cos6x+cos18x+cos10x+cos14x=0第二步,和差
f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s
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)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx
x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)
sin(2x+π/3)
π/6+2kπ
sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3（1/2-cos^2y）+sin^2y=1.5-3cos^2y）+sin^2y又有sin^2y+c
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
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
3/2cosx-3/2(sinx)^2
-cosx
sin^2(x)+cos^2(x+30)+sin(x)cos(x+30）=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30
y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x