tan2πα等于多少
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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
这需要用到万能公式tanα=tan[2*(α/2)]=2tan(α/2)/[1-tan(α/2)^2]=[2tan(α/2)]/[1-(tanα/2)^2]tan(a+π/4)=(tana+1)/(1
Tan2(x-∏/4)={2tan(x-∏/4)}/{1-tan2(x-∏/4)}下求tan(x-∏/4)的值.tan(x-∏/4)={tanx-tan∏/4}/{1+tanx*tan∏/4}=(3-
(1)tan1ºtan2ºtan3º...tan89º=(tan1º*tan89º)(tan2º*tan88º)(tan
2cos^2Θ/2-sinΘ-1/√2sin(Θ+π/4)=(cosΘ-sinΘ)/(cosΘ+sinΘ){在分子和分母上同时除以cosΘ}=(1-tanΘ)/(1+tanΘ)tan2Θ=-2√2=2
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
[[[1]]]由题设可知sina=3/5,cosa=-4/5∴tana=(sina)/(cosa)=-3/4.∴tan2a=(2tana)/[1-tan²a]=(-3/2)/[1-(9/16
sin(θ+7π/4)+cos(θ-3π/4)=根号5/2∴sin(θ-π/4+2π)+cos(θ+π/4-π)=根号5/2∴sin(θ-π/4)-cos(θ+π/4)=根号5/2∴sin(θ-π/4
利用三角万能公式可得1/cos2α+tan2α=(1+tan^2α)/(1-tan^2α)+2tanα/(1-tan^2α)=(1+tanα)^2/[(1-tanα)(1+tanα)]=(1+tanα
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)
由tanα=2得cos^2α=1/5,则cos2α=-3/5,sin2α=4/5原始=2sin2αcos2α*(sin2α+cos2α)/cos2α=2(sin^22α+sin2αcos2α)=-6/
tan(α+π/4)+tan(α+3π/4)=(tanα+tanπ/4)/(1-tanαtanπ/4)+(tanα+tan3π/4)/(1-tanα+tan3π/4)=(tanα+1)/(1-tanα
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)]/[
用word打出来了,这样好看些 参考资料是我传到自己的相册,有大图
tan(π/4+α)-cot(π/4+α)=tan(π/4+α)-1/tan(π/4+α)=[tan²(π/4+α)-1】/tan(π/4+α)=-2/tan(π/2+2α)=-2cot(π
知识:tan(a+b)=(tana+tanb)/(1-tanatanb)原式=tan2a[tan(π/6-a)+tan(π/3-a)]+tan(π/6-a)tan(π/3-a)=tan2a*tan(π
tanα-1/tanα=sinα/cosα-cosα/sinα=(sin²α-cos²α)/(sinαcosα)=-cos2α/(1/2sin2α)=-2cos2α/sin2α=-