sin(330 x)=2sin(270 x)

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在mathematica里输入Plot[Sin[x] Sin[x + 2] - Sin[x + 1]Sin[x + 1]

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

matlab画y=sin(x)+sin(2*x)+...+sin(20*x)的图像

x=0:0.01:1;y=0;fori=1:20y=y+sin(i*x);endplot(y);

sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos

那个前半括号里面相加等于一

三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

sin(x+△x)-sinx=2sin△x/2.cos(x+△x/2)

由和差化积公式sinα-sinβ=2cos[(α+β)/2]sin[(α-β)/2],得:sin(x+△x)-sinx=2cos[(x+△x+x)/2]sin[(x+△x-x)/2]=2cos[(2x

化简!f(x)=sin(pai-x)cos(3/2pai+x)+sin(pai+x)sin(3/2pai-x)

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

一道三角恒等式证明题请证明sin(x+y)sin(x-y)=sin^2(x)-sin^2(y)

左边=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=sin²xcos²y-cos²xsin²y=sin²x(1-sin

泰勒公式的为什么㏑( 1 + sin X ) = sin X - ( sin X )²/2 +(sin X )

你好,第一:首先将㏑(1+X)用麦克劳林公式(泰勒公式的推广)分解开就是X-(X)²/2+(X)³/3-(X)∧4+o(∧4X),第二:将㏑(1+X)中的X换为sinX就ok了,很

已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x)

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

求导f(x) = cos(3x) * cos(2x) + sin(3x) * sin(2x).

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

y=sin[sin(x^2)] 则dy/dx=?

dy/dx相当于对x进行求导:dy/dx=y'=2x*cos[sin(x^2)]*cos(x^2)由于:sinx=cosx,sin(x^2)=2x*cos(x^2)

s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)

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)

为什么|sin(2x+pi)|=|sin(2x)|呢?

sin在(0,PI)是中心对称的.意思是SIN(X+PI)=-SIN(X)加绝对值就相等了.

证明sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((x+y)/2)sin((x+

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

5sin^2(X)+sin(2X)-cos^2(X)=1, 求解X

5(sinx)^2+sin2x--(cosx)^2=15(sinx)^2+2sinxcosx--(cosx)^2=(sinx)^2+(cosx)^24(sinx)^2+2sinxcosx--2(cos

已知函数fx=(1+1/tanx)sin^x-2sin(x+π/4)sin(x-π/4)

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

求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2

证明:∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co

已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

解方程 sin 2x + sin x = 0

sin2x+sinx=02sinxcosx+sinx=0sinx(2cosx+1)=0①党sinx=0x=kπ②党(2cosx+1)=0cosx=-1/2x=kπ/2+π/3