证明cos(4β)-4cos(4α)=3 具体题目见内容已知2sinα=sinθ+cosθ(sinβ)^2=sinθcosθ证明cos(4β)-4cos(4α)=3
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证明cos(4β)-4cos(4α)=3 具体题目见内容已知2sinα=sinθ+cosθ(sinβ)^2=sinθcosθ证明cos(4β)-4cos(4α)=3
证明cos(4β)-4cos(4α)=3 具体题目见内容
已知
2sinα=sinθ+cosθ
(sinβ)^2=sinθcosθ
证明
cos(4β)-4cos(4α)=3
证明cos(4β)-4cos(4α)=3 具体题目见内容已知2sinα=sinθ+cosθ(sinβ)^2=sinθcosθ证明cos(4β)-4cos(4α)=3
2sinα=sinθ+cosθ 得到
(2sinα)^2=(sinθ+cosθ)^2=(sinθ)^2+(cosθ)^2+2sinθcosθ
而(sinβ)^2=sinθcosθ
所以(2sinα)^2-2(sinβ)^2=(sinθ)^2+(cosθ)^2=1
根据半角公式展开,化简得到
2cos(2α)=cos2β
两边平方,得到
[2cos(2α)]^2=(cos2β)^2
再用一次半角公式展开化简,得到cos(4β)-4cos(4α)=3
注：
半角公式:
(sinβ)^2=0.5[1-cos2β]
(cosβ)^2=0.5[1+cos2β]
(sinθ+cosθ)²-2sinθcosθ=sin²θ+cos²θ=1
4sin²α-2sin²β=1
cos(4β)-4cos(4α)
=2cos²(2β)-1-4(2cos²(2α)-1)
=3+2(1-2sin²β)²-8(1-2sin²α)²
全部展开
(sinθ+cosθ)²-2sinθcosθ=sin²θ+cos²θ=1
4sin²α-2sin²β=1
cos(4β)-4cos(4α)
=2cos²(2β)-1-4(2cos²(2α)-1)
=3+2(1-2sin²β)²-8(1-2sin²α)²
=3+2*((1-2sin²β)²-(2(1-2sin²α))²)
=3+2*(1-2sin²β+2(1-2sin²α))(1-2sin²β-2(1-2sin²α))
=3+2*(-4sin²α-2sin²β+3)(4sin²α-2sin²β-1)
=3+2*(-4sin²α-2sin²β+3)*0
=3
得证
收起
4cos(4α)=4[1-2sin^2(2α)]
=4[1-8(sinαcosα)^2]
=4-32sin^2α(1-sin^2α)]
=4-32sin^2α+32sin^4α
=4-8(sinθ+cosθ)^2+2(sinθ+cosθ)^4
=4-8(1+2sinθcosθ)+2(1+2sinθcosθ)^2
=4-8(1+2sin^2β)+2(1...
全部展开
4cos(4α)=4[1-2sin^2(2α)]
=4[1-8(sinαcosα)^2]
=4-32sin^2α(1-sin^2α)]
=4-32sin^2α+32sin^4α
=4-8(sinθ+cosθ)^2+2(sinθ+cosθ)^4
=4-8(1+2sinθcosθ)+2(1+2sinθcosθ)^2
=4-8(1+2sin^2β)+2(1+2sin^2β)^2
=-4-16sin^2β+2(1+4sin^2β+4sin^4β)
=-2-8sin^2β+8sin^4β
cos(4β)=1-2sin^2(2β)
=1-2(2sinβcosβ)^2=1-8sin^2β+8sin^4β
cos(4β)-4cos(4α)=3
收起