cosθ+cos2θ+cos3θ
答:cos2的值等于-0.5。cos2是一个三角函数,表示的是一个角的余弦值。cos2的值等于-0.5,这是因为cos2实际上是cos(2π/3),而cos(2π/3)的值可以用三角函数的周期性质简化为cos(2π/3+2π),也就是cos(4π/3)。
答:因cos(A+B)=cosAcosB-sinAsinB,cos(A-B)=cosAcosB+sinAsinB,故原式=【cos(3x+2x)/2*cos(3x-2x)/2-sin(3x+2x)/2*sin(3x-2x)/2】-【cos(3x+2x)/2*cos(3x-2x)/2+sin(3x+2x)/2*sin(3x-2x)/2】=-2sin(3x+2x)/2*sin(3x-2x)/2=-2sin(5x/2)sin(x...
答:cosα=-cos(180°-α)cos1°+cos2°+cos3°+...+cos180°=0
答:=cos1°²+cos2°²+cos3°²+cos4°²... +cos44°²+cos45°²+cos46°²+…+cos89°²=cos1°²+cos2°²+cos3°²+cos4°²... +cos44°²+cos45°²+cos46°²+…+cos89°²=cos...
答:cos3a=cos(2a+a)=cos2a*cosa-sin2a*sina =cos2a*cosa-2sina*cosa*sina =cos2a*cosa-2sin^2a*cosa =(2cos^2(a)-1)*cosa-2(1-cos^2(a))*cosa =2cos^3(a)-cosa-2cosa+2cos^3(a)=4cos^3(a)-3cosa.
答:这个是和差化积公式的直接套用啊。本题中,α为3α,β为α cos(3α)+cosα=2cos[(3α+α)/2]cos[(3α-α)/2]=2cos(2α)cosα
答:设AB=c、BC=a、AC=b AB=c=2 1.2sin(2C)cosC-sin(3C)=√3(1-cosC)2sin(2C)cosC-sin(2C)cosC-cos(2C)sinC=√3(1-cosC)sin(2C)cosC-cos(2C)sinC)=√3(1-cosC)sinC=√3(1-cosC)sinC+√3cosC=1 (1/2)sinC+(√3/2)cosC=1/2 sin(C+π/3)=1/2 C为三角形内角,0...
答:由于余弦的周期是2π,cos(2π+α)相当于cosα,而α视作锐角(0<α<π/2),故根据余弦函数图像可以知道α角在此范围内cosα的值大于0,为正;根据奇变偶不变,符号看象限,cos(3π+α)相当于cos(π+α),角度π+α的值大于π小于3π/2,在第三象限,而余弦在二三象限为负,故cos(π...
答:f(A)=[2cos³A+sin²A+cosA-3]/[2+2cos²A+cosA]sin(π/3)=√3/2 cos(π/3)=1/2 ∴f(A)=-0.5
答:原式=cos(2x+x)+cosx=cos2x*cosx-sin2x*sinx+cosx =cos2x*cosx-(2sinxcosx*sinx-cosx)=cos2x*cosx+cosx-2sinx^2*cosx =cos2x*cosx+cosx(1-2sinx^2)=cos2x*cos+cosx*cos2x=2cos2x*cosx 就是半角公式
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仉疤15182636443:
利用欧拉公式证明cosθ+cos2θ+cos3θ+···+cosnθ= - 1/2+sin(n+1/2)θ/sin(θ/2) -
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:[答案] cosθ=[e^(iθ)+e^(-iθ)]/2 cos2θ=[e^(2iθ)+e^(-2iθ)]/2 cos3θ=[e^(3iθ)+e^(-3iθ)]/2 ... cosnθ=[e^(inθ)+e^(-inθ)]/2
仉疤15182636443:
cosθ+cos2θ+······+cosnθ=? -
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:[答案] (cosx+cos2x+cos3x+...+cosnx)cosx=(cosxcosx+cosxcos2x+...+cosxcosnx)=(1/2)(cos2x+1)+(1/2)(cos3x+cosx)+(1/2)(cos4x+cos3x)+...+(1/2)(cosn+1)x+cosnx)=(1/2)(1+cos(n+1)x) +(1/2)(cosx+cos2x+...+cosnx)(cosx+c...
仉疤15182636443:
化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) -
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: 原式=cosθ+cosθcos2π/3-sinθsin2π/3+cosθcos4π/3-sinθsin4π/3=cosθ+cosθ*(-1/2)-sinθ*√3/2+cosθ*(-1/2)-sinθ*(-√3/2)=0
仉疤15182636443:
cosθ+cos2θ+cos3θ=0,求θo<θ<90° -
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:[答案] 原式=cos(x+2x)=cosxcos2x-sinxsin2x=cosx((cosx)^2-(sinx)^2)-2(sinx)^2cosx =(cosx)^3-3cosx(sinx)^2 =cosx(1-4(sinx)^2) cosx=0 或者(sinx)^2=1/4 x=(k+1/2)pi 或者x=(k/2+1/4)pi k属于Z
仉疤15182636443:
已知cosθ 求 cos2θ -
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: ^^有公式cos2θ= 2cosθ^-1 =2*(4/5)^-1 =2*(16/25)-1 =32/25-1 =7/25 ok?爽了分拿来吧~~~~~~~~
仉疤15182636443:
职一数学!求解 设θ=10°,试求cosθ+cos2θ+cos3θ+cos4θ+cos5θ.... -
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: 利用三角函数和差化积公式,将cosθ和cos17θ配对,cos2θ和cosθcos16θ配对...以此类推 注意到θ=10° cosθ+cos17θ=2cos(17θ+θ)/2cos(17θ-θ)/2=2cos9θcos8θ=0 同理,cos2θ+cosθcos16θ=2cos(16θ+2θ)/2cos(16θ-2θ)/2=2cos9θcos7θ=0......所以最后的结果为cos18θ=cos180°=-1
仉疤15182636443:
请数学高手解答 cosθ+cos2θ+······+cosnθ=? -
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: (cosx+cos2x+cos3x+...+cosnx)cosx=(cosxcosx+cosxcos2x+...+cosxcosnx)=(1/2)(cos2x+1)+(1/2)(cos3x+cosx)+(1/2)(cos4x+cos3x)+...+(1/2)(cosn+1)x+cosnx)=(1/2)(1+cos(n+1)x) +(1/2)(cosx+cos2x+...+cosnx)(cosx+cos2x+...+cosnx)(cosx-1/2)=(1/2)(1+cos(n+1)x cosx+cos2x+...+cosnx=(1+cos(n+1)x)/(2cosx-1)
仉疤15182636443:
关于诱导公式的一道题cosθ+cos^2θ=1求sin^2θ+sin^6θ+sin^8θ=? -
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:[答案] cosθ+cos^2θ=1 sin^2θ=1-cos^2θ=cosθ 所以 sin^2θ+sin^6θ+sin^8θ =cosθ+cos^3θ+cos^4θ =cosθ+cos^2θ*(cosθ+cos^2θ) =cosθ+cos^2θ*1 =1
仉疤15182636443:
1+cosθ+cos2θ+cos3θ+···+cosnθ=1/2+sin(n+1/2)θ/2sin(θ/2) -
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: 令z=e^jθ,代入恒等式,取实部就好
仉疤15182636443:
数学2cosθcos2θ - cosθ=? -
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: 倒着推 cos3θ=cos(θ+2θ)=cosθcos2θ-sinθsin2θ=cosθcos2θ-sinθ(2sinθcosθ)=cosθcos2θ-(2sinθsinθ)cosθ=cosθcos2θ-(1-cos2θ)cosθ=cosθcos2θ-cosθ+cosθcos2θ=2cosθcos2θ-cosθ