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sin∧8x

sin^8x+cos^8x= =(sin^4x-cos^4x)^2+2sin^4xcos^4x =[(sin^2x-cos^2x)(sin^2x+cos^2x)]^2+2sin^4Xcos^4x =(sin^2x-cos^2x)^2+2sin^4Xcos^4x =sin^4x-2sin^2Xcos^2x+cos^4x+2sin^4Xcos^4x=sin^4x+cos^4x+2sin^4Xcos^4x-2sin^2Xcos^2x=(sin^4x+

解:∫(sinx)^8dx=∫((sinx)^2)^4dx =∫((1-cos(2x))/2)^4dx (应用倍角公式) =∫(1-4cos(2x)+6(cos(2x))^2-4(cos(2x))^3+(cos(2x))^4)dx =∫(1-4cos(2x)+6(cos(2x))^2+(cos(2x))^4)dx-4∫(cos(2x))^3dx =∫(1-4cos(2x)+3(1+cos(4x))+((1+cos(4x))/2)^2)dx-4∫(cos(2x)

解令t=3x+8m=sinty=m^3.

∫(sinx)^8dx=-∫ (sinx)^7 dcosx= -cosx (sinx)^7 + ∫ 7(cosx)^2(sinx)^6 dx=-cosx (sinx)^7 +7∫ (1- (sinx)^2)(sinx)^6 dx8∫(sinx)^8dx= -cosx (sinx)^7 + 7∫(sinx)^6dx∫(sinx)^8dx = (1/8) [-cosx (sinx)^7 + 7∫(sinx)^6dx]= (1/8) {-cosx (sinx)^7 + (7/6)[-cosx (sinx)^5

1)已知:f、g的周期都是:Tf(x+T)=f(t)g(x+T)=g(t)h(t)=f(t)+g(t)h(t+T)=f(t+T)+g(t+T)=f(t)+g(t)=h(t)那么f+g的周期还是:T.2) sinx 和 cosx 的周期均为:2π; sin^8x 和 cos^8x 的周期都是:π; 那么由1):sin^8x+cos^8x 的周期还是:π3)化简: sin^8x+

x是趋于0的不?那么在x趋于0时,sinx就等价于x,所以原极限=lim(x趋于0) x^5/ x^8= lim(x趋于0) 1/x^3于是极限值不存在

我来回答sin[θ-(3π/2)]=cos(2π-θ)=cos(-θ)=cosθ告诉你一个正弦和余弦转化的秘诀就是sin(a)=cos(π/2-a),两个角度之和是π/2{sin(θ-5π)cos[(-π/2)-θ]cos(8π-θ)}/sin[θ-(3π/2)]sin(-θ-4π)=-sin(θ)* -sin(θ)* cos(θ) /cos(θ) * -sin(θ)=-sinθ

∫(sint)^2dt=(1/2)∫(1-cos2t)dt=(1/2)[t-(sin2t)/2]|=(1/2)[x-(sin2x)/2].d/dx(∫sin^2tdt)=(sinx)^2.

cos^8x-sin^8x-cos2x =(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x)-cos2x =(cos^4x+sin^4x)*1*cos2x-cos2x =[(cos^2x+sin^2x)^2-2(sinxcosx)^2]cos2x-cos2x =[1-2(sinxcosx)^2]cos2x-cos2x =[1-2(sinxcosx)^2-1]cos2x =-(2sinxcosx)^2/2*cos2x=-sin^

原式=(2sin^2(2x)cos2x)/4+(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x) =2sin^2xcos^2xcos2x+(cos^4x+sin^4x)cos2x =cos2x(cos^4x+2sin^2xcos^2x+sin^4x) =cos2x(cos^2x+sin^2x)^2 =cos2x

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