作业帮已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/14 05:20:18
已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为
答案好像是（3-4√3）/10
答案好像是（3-4√3）/10
[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2
=cos(2π/3+2α)
(sinα-cosα)/(sinα+cosα)=1/3
(1-tanα)/(1+tanα)=1/3
(1-tanα)/(1+tanα)=2/(1+tanα)-1=1/3
1+tanα=3/2 tanα=1/2 (cosα)^2=1/[1+(tanα)^2]=4/5 cos2α=2(cosα)^2-1=3/5
sin2α=2sinαcosα=2*(1*2)/5=4/5
cos(2π/3+2α)=cos(2π/3)cos(2α)-sin(2π/3)sin(2α)=(-3/10)-(4√3/10)
再问： 额，答案好像是（3-4√3）/10
再答： 疏忽了一下，改正如下 (sinα-cosα)/(sinα+cosα)=1/3 (tanα-1)/(tanα+1)=1/3, 1-2/(tanα+1)=1/3, tanα+1=3 tanα=2 (cosα)^2=1/5, cos2α=-3/5 tan2α=2tanα/[1-(tanα)^2]=-4/3 sin2α=4/5 cos2π/3=-1/2所以cos(2π/3+2α)=)=(3/10)-(4√3/10)
=cos(2π/3+2α)
(sinα-cosα)/(sinα+cosα)=1/3
(1-tanα)/(1+tanα)=1/3
(1-tanα)/(1+tanα)=2/(1+tanα)-1=1/3
1+tanα=3/2 tanα=1/2 (cosα)^2=1/[1+(tanα)^2]=4/5 cos2α=2(cosα)^2-1=3/5
sin2α=2sinαcosα=2*(1*2)/5=4/5
cos(2π/3+2α)=cos(2π/3)cos(2α)-sin(2π/3)sin(2α)=(-3/10)-(4√3/10)
再问： 额，答案好像是（3-4√3）/10
再答： 疏忽了一下，改正如下 (sinα-cosα)/(sinα+cosα)=1/3 (tanα-1)/(tanα+1)=1/3, 1-2/(tanα+1)=1/3, tanα+1=3 tanα=2 (cosα)^2=1/5, cos2α=-3/5 tan2α=2tanα/[1-(tanα)^2]=-4/3 sin2α=4/5 cos2π/3=-1/2所以cos(2π/3+2α)=)=(3/10)-(4√3/10)
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