化简cos^4 α+sin^4 α-1/4(1+cos4α)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 10:43:28
化简cos^4 α+sin^4 α-1/4(1+cos4α)
化简cos^4 α+sin^4 α-1/4(1+cos4α)
化简cos^4 α+sin^4 α-1/4(1+cos4α)
cos^4 α+sin^4 α-1/4(1+cos4α)
=1-2(sinαcosα)^2-(1/4)[1+1-2(sin2α)^2]
=1-(1/2)(sin2α)^2-1/2+(1/2)(sin2α)^2
=1/2
(sinx)^4+(cosx)^4
=[1-(cosx)^2](sinx)^2+[1-(sinx)^2](cosx)^2
=1-2(sinxcosx)^2
=1-1/2*(sin2x)^2
=3/4+1/4*cos4x
化简sin^4α+cos^4α+2sin^2α*cos^2α-1
化简:(1)sin^2·α+cos^4·α+sin^2·αcos^2·α(2)(1-cos^4·α-sin^4·α)/(1-cos^6·α-sin^4·α)
化简:[1+2sin(α+2π)*cos(α-2π)]/[sin(α+4π)+cos(α+8π)]
化简sin^4α+cos^2α+sin^2α*cos^2α
化简4sin²α-sinαcosα-3cos²α=0
1-cos^6(α)-sin^6(α)/1-cos^4(α)-sin^4(α)
求证:sinα^4+cosα=1-2sinα^2cosα^2
若sin^2α+sinα=1 则cos^4α+cos^2α=
已知cosα+cos²α=1则sin²α+sin^4α=
已知tanα=3,计算:(1)4sinα-2cosα/5cosα+3sinα (2)sinαcosα (3)(sinα+cosα)^2(急~)(1)4sinα-2cosα/5cosα+3sinα(2)sinαcosα(3)(sinα+cosα)^2
[1-(sinα)^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=?
已知tan=2,那么1/2sinαcosα+cosα²α和4sin²α+2sinαcosα-cos²α?
化解下列三角函数(1),(4-5cosα)/(3-5sinα)+(3+5sinα)/(4+5cosα)(2),(2+sin²α-2cos²α+sinα)/(3sinαcosα+cosα)
化简2Cos平方α-4sinαcosα+1
2sinα-3cosα/4sinα-9cosα=-1,则9sin^2α-3sinαcosα-5=
化简:2cos²α-1/【2sin(π/4-α)sin²(π/4+α)】
化简:2cos²α-1/【2sin(π/4-α)sin²(π/4+α)】
化简sin^4α+cos^4α+1/2sin²2α=?