tan2α拆开等于
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tanα=三分之一倍角公式:tan2α=2tanα/[1-(tanα)^2]=2*1/3/[(1-(1/3)^2]=3/4
(1)tan1ºtan2ºtan3º...tan89º=(tan1º*tan89º)(tan2º*tan88º)(tan
tan2α=1/32tana/(1-tana^2)=1/36tana=1-tana^2tana^2+6tana-1=0tana=[-6+/-根号(40)]/2=-3+/-根号10
tan2α-sin2α=(sin2a+cos2a)(tan2α-sin2α)//导入sin2a+cos2a=1=sin2atan2α-(sin2α)^2+cos2atan2a-sin2acos2a=s
tan2分之2α=tanα.sin4a=2sin2αcos2α=4sinαcosα[1-2(sinα)^2].
tan@*tan2@/(tan2@-tan@)=1/[(tan2@-tan@)/(tan2@*tan@)]=1/[1/tan@-1/tan2@]=1/[1/tan@-(1-tan^2@)/(2*tan
35-5=30.
答案如图片再问:根号2乘上sin(π/4+α)为什么等于sinα+cosα?再答:用公式展开
tan2α=(2tanα)/(1-(tanα)^2)=1/3可以解得tanα=-3+根(10)或者tanα=-3-根(10)
(1+tan2α)/(1-tanα)=2010=>{1+2tanα/[(1-tanα)^2]}/(1-tanα)=1-(tanα)^2+2tanα=2010(1+tanα)=>2009+(tanα)^
证:2sinβ/(cosα+cosβ)=[(sinα+sinβ)-(sinα-sinβ)]/(cosα+cosβ)=(sinα+sinβ)/(cosα+cosβ)-(sinα-sinβ)/(cosα+
(1)cosα=1/7,因为0<α<π/2,所以sinα=√(1-cosα)=√[1-(1/7)]=4√3/7所以tanα=sinα/cosα
已知tan(α+β)=1/3,tan2β=-2;求tan2α的值tan[2(α+β)]=2tan(α+β)/[1-tan²(α+β)]=(2/3)/(1-1/9)=3/4tan2α=tan[
tan(α+2α)=tan(3α)=tan(3·π/9)=tan(π/3)=√3又tan(α+2α)=[tanα+tan(2α)]/[1-tanα·tan(2α)]因此[tanα+tan(2α)]/[
tan2α=tan(α+β+α-β)=[tan(α+β)+tan(α-β)]/[1-tan(α+β)tan(α-β)]=8/(1-15)=-4/7tan2β=tan[(α+β)-(α-β)]=)=[t
tan2α-sin2α=sin2α/cos2α-sin2α=sin2α-sin2αcos2α/cos2α=sin2α(1-cos2α)/cos2α=tan2αsin2α
a平方加b平方加2a
用正弦和余弦的二倍角公式tan2a=sin2a/cos2a=2sinacosa/(cosa^2-sina^2)=2tana/(1-(tana)^2)(上下同时除以(cosa)^2)
tanα-1/tanα=sinα/cosα-cosα/sinα=(sin²α-cos²α)/(sinαcosα)=-cos2α/(1/2sin2α)=-2cos2α/sin2α=-