求1 cos^4
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(1+X4)COSx先求导数再乘以dx就行了
sina=±根15/4,tana=根15
f(x)=(6cos^4x-5cos^2x+1)/(2cos^2x-1)=3cos^2x-1,2cos^2x-1≠0cos2x≠02x≠kπ+π/2,k∈Z解出x得定义域f(x)=3cos^2x-1=
cos(α+β)=1/3,cos(α-β)=1/4,求cosα-cosβcos(α+β)=cosαcosβ-sinαsinβ=1/3.(1)cos(α-β)=cosαcosβ+sinαsinβ=1/4
一、cot(α/2)-tan(α/2)=cos(α/2)/sin(α/2)-sin(α/2)/cos(α/2)=[cos^(α/2)-sin^(α/2)]/[sin(α/2)cos(α/2)]=2co
(2sinα-cosα)/(2cosα+sinα)=12sinα-cosα=2cosα+sinαtana=1/3tan2a=2tana/(1-tanatana)=3/4cos2a=(1-tanatan
(2)2/3sin²x+1/8*2cos²x=1/2(1-cos2x)+1/8(1+cos2x)=1/2+1/8-1/2cos2x+1/8cos2x=5/8-3/8cos2xcos
sin⁴θ+cos⁴θ=(sin²θ+cos²θ)²=sin⁴θ+cos⁴θ+2sin²θcos²θ所
(sinx)^4+(cosx)^4=1即(sinx)^4+(cosx)^4+2(sinx)^2(cosx)^2-2(sinx)^2(cosx)^2=1即[(sinx)^2+(cosx)^2]^2-2(
sin(θ+kπ)=-2cos(θ+kπ),可得tanQ=-24sinθ-2cosθ/5cosθ+3sinθ(分子分母同时除以cosQ)=10⑵(1/4)sin平方θ+(2/5)cos平方θ(分子分母
因为cos(x+π/4)=1/3cos(x-π/4)=cos(x+π/4-π/2)=sin(x+π/4)所以cos(2x)/cos(x-π/4)=sin(2x+π/2)/sin(x+π/4)=sin[
sinθ-cosθ=-1/5两边平方得1-2sinθcosθ=1/25sinθcosθ=24/50=12/25sin²θ+cos²θ=1两边平方得sinθ^4+cosθ^4=1-2
(1)sin^3θ+cos^3θ=(sinθ+cosθ)(sin^2θ-sinθcosθ+cos^2θ)=(sinθ+cosθ)[-(sinθ+cosθ)^2/2+3/2]//令sinθ+cosθ=x
tan(θ+π/4)=-2即(tanθ+1)/(1-tanαθ)=-2解得tanθ=3cos²θ+sinθcosθ-1=(cos²θ+sinθcosθ)/(sin²θ+c
sinx*cosx=1/2sin2x
sin(2π+a)cos(-π+a)/cos(-a)tana=sin(a)cos(π-a)/cos(a)(sina/cosa)=-sinacosa/sina=-cosa=-1/4
因为cos(A+B)cos(A-B)=(1/2)(cos2A+cos2B)=(1/2)[2(cosA)^2-1+2(cosB)^2-1]=(cosA)^2+(cosB)^2-1=1/4所以cosA^2
sina=1-sin²a=cos²acos²a+cos^4a=cos²a+sin²a=1