求证3+cos4α-4cos2α=(8sinα)^4
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/20 06:04:11
求证3+cos4α-4cos2α=(8sinα)^4
证(1-cos2α)/(1+cos2α)=tan²α
(1+sin2α)/(cosα+sinα)=cosα+sinα
证(1-cos2α)/(1+cos2α)=tan²α
(1+sin2α)/(cosα+sinα)=cosα+sinα
![求证3+cos4α-4cos2α=(8sinα)^4](/uploads/image/z/20082285-45-5.jpg?t=%E6%B1%82%E8%AF%813%2Bcos4%CE%B1-4cos2%CE%B1%3D%288sin%CE%B1%29%5E4)
证明:1)3+cos4α-4cos2α=2+(1+cos4α)-4cos2α
=2+2(cos2α)^2-4cos2α
=2[1+(cos2α)^2-2cos2α]
=2(1-cos2α)^2
=2*{1-[(cosα)^2-(sinα)^2}^2
=2*{1-(cosα)^2+(sinα)^2}^2
=2*{2(sinα)^2}^2
=8*(sinα)^4
2)(1-cos2α)/(1+cos2α)={1-[(cosα)^2-(sinα)^2]}/{1+[(cosα)^2-(sinα)^2]}
={1-(cosα)^2+(sinα)^2}/{1-(sinα)^2+(cosα)^2}
=2(sinα)^2/{2(cosα)^2}
=tan²α
3)(1+sin2α)/(cosα+sinα)=(sin²α+cos²α+2sinαcosα)/(cosα+sinα)
=(cosα+sinα)^2/(cosα+sinα)
=cosα+sinα
=2+2(cos2α)^2-4cos2α
=2[1+(cos2α)^2-2cos2α]
=2(1-cos2α)^2
=2*{1-[(cosα)^2-(sinα)^2}^2
=2*{1-(cosα)^2+(sinα)^2}^2
=2*{2(sinα)^2}^2
=8*(sinα)^4
2)(1-cos2α)/(1+cos2α)={1-[(cosα)^2-(sinα)^2]}/{1+[(cosα)^2-(sinα)^2]}
={1-(cosα)^2+(sinα)^2}/{1-(sinα)^2+(cosα)^2}
=2(sinα)^2/{2(cosα)^2}
=tan²α
3)(1+sin2α)/(cosα+sinα)=(sin²α+cos²α+2sinαcosα)/(cosα+sinα)
=(cosα+sinα)^2/(cosα+sinα)
=cosα+sinα
求证2sin(α-2β)sin(α+2β)=cos4β-cos2α
sin4α-cos4α=sin2α-cos2α求证
求证:sin4α+cos4α=1-2sin2αcos2α
求证cos^8α-sin^8α 1/4sin2αsin4α=cos2α
cos2α=4分之5,则cos4次方α+sin4次方α=
已知sinθ+cosθ=2sinα,sinθcosθ=(sinβ)^2,求证4(cos2α)^2=(cos2β)^2
已知α∈(π/2,π),且sinα=3/5,则sin^2α/2+(sin4αcos2α)/(1+cos4α)的值为多
数学4道化简题1.(1+cos4α)/cos2α2.(1-cos4α)/2sin2α3.(1+cos6A)/cos3A4
证明sin4α-cos4α=sin2α-cos2α
已知cos2α=1/4,则cos4α+sin4α+sin2αcos2α的值为?
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3
[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]=?