sin(2x 45) 2cos(3x-30)的周期
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 07:06:53
再问:再问:在你答题的时候我蛋疼做了一遍,结果好像不一样……再问:不过还是辛苦施主了
2sin^2α-cos^α+sinα*cosα-6sinα+3cosα=0(2sinα-cosα)(sinα+cosα)-3(2sinα-cosα)=0(2sinα-cosα)(sinα+cosα-3
(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa
5/3上下同除tan
f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx
(cosα-sinα)/(cosα+sinα)+(cosα+sinα)/(cosα-sinα)=[(cosα-sinα)^2+(cosa+sinα)^2]/[(cosα)^2-(sinα)^2]=2[
3sina+cosa=03sina=-cosatana=sina/cosa=-1/31、(3cosa+5sina)/(sina-cosa)【分子分母同除以cosa】=[3+5tana]/[tana-1
sina=-2cosatana=-2sin²a-3sinacosa+1=(sin²a-3sinacosa+sin²a+cos²a)/(sin²a+co
sin^6α+cos^6α+3sin^2α*cos^2α=(sin^2α+cos^2α)(sin^4α-sin^2α*cos^2α+cos^4α)+3sin^2α*cos^2α=sin^4α+2sin
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
2sinθ+3cosθ=2两边平方有:4(sinθ)^2+12sinθcosθ+5(cosθ)^2=4(12sinθ+5cosθ)cosθ=0所以有:cosθ=0,代入原式,得sinθ=1或者12si
已知tan=2(sinα+cosα)/(2sinα-cosα)=(sinα/cosα+cosα/cosα)/(2sinα/cosα-cosα/cosα)=(tanα+1)/(2tanα-1)=(2+1
3/2cosx-3/2(sinx)^2
已知两边同除以余弦得到Tanα=1/3sin²α-2sinαcosα+3cos²α+1=(sin²α-2sinαcosα+3cos²α+sin²α+c
sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin
因为cosα=1.5sinα,所以原式=0.5/2.5+2.5/0.5=1/5+5=5.2
Cos(&-3派/2)=Cos(3派/2-&)余弦在第三象限,为负值所以Cos(&-3派/2)=Cos(3派/2-&)=-sin&
(sinα^32cosα)他们是相乘还是相减啊?
sinα+3cosα=2平方:(sinα+3cosα)^2=4sinα^2+9cosα^2+6sinαcosα=4sinα^2+4cosα^2(-3sinα^2+5cosα^2+6sinαcosα)/
y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2