tan4分之π tan(-6分之17π)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 05:30:26
用和角来算因为tan(a+β)=2/5,tan(β-π/4)=1/4a+β-(β-π/4)=a+π/4所以tan(a+π/4)=[tan(a+β)-tan(β-π/4)]/[1+tan(a+β)tan
问题补充:是求tan(α-2β)的值这是公式sin(α-β)=sinαtan(π-β)=1/2β=90sin(α/2)-cos(α/2)=5分之根号10公式忘
原式=sin(π/6)+cos(π/3)-tan(π/4)=1/2+1/2-1=0
题目不全已知tan(x+y)=2/5,tan(y-π/4)=1/4,求tan(x+π/4)的值解令a=x+y,b=x+π/4tan(x+π/4)=tan[(x+y)-(y-π/4)]=tan(a-b)
cos(19π/6)tan(-16π/3)=cos(12π/6+7π/6)[-tan(16π/3)]=-cos(7π/6)tan(12π/3+4π/3)=-cos(π+π/6)tan(π+π/3)=-
sin(-11/6π)+cos12/5π×tan4π=sin(-11/6π+2π)+cos12/5π×tan(4π-4π)=sin(1/6π)+cos12/5π×tan0=1/2+cos12/5π×0
原式=[-sin(a+5π)]×[sina]-[-tan(3π/2-a]×[-tana]=[sina]×[sina]-[1/tana]×[tana]=sin²a-1=-cos²a
(3)√3sin(x/2)+cos(x/2)=2{[√3sin(x/2)]/2+[cos(x/2)]/2}=2[sin(x/2)*cos(π/6)+cos(x/2)*sin(π/6)]=2sin(x/
判断角所在的象限,一、三象限的角的正切值是正的,如果你题中的角单位是读的话符号是负如果是弧度的话符号是正你就自己判断一下吧
sin2分之π+cos2分之3π-tanπ=1+0-0=1对不起,刚才算错了
cos(-60°)=cos60°=1/2sin2π/3=sin(π-2π/3)=sinπ/3=√3/2tan5π/4=tan(5π/4-π)=tanπ/4=1cos11π/6=cos(2π-11π/6
=cos(1/4π+2π)-tan(11/6π)=cos(1/4π)-tan(2π-1/6π)=2^0.5/2+tan(1/6π)=2^0.5/2+3^0.5/3二分之根号2+三分之根号三
tan^4α+6tanα=49/16或tan^4α+6tanα=28sinα*cosα=(sinα*cosα)/(sin^2α+cos^2α)=(tanα)/(tan^2α+1)=2/52tan^2α
sinθ=3/5,cosθ=-4/5,tanθ=-3/4,tanψ=1/2tan(θ+ψ)=(tanθ+tanψ)/(1-tanθtanψ)=(-3/4+1/2)/[1-(-3/4)x1/2]=(-1
令kπ-π/2
根据所学公式展开后得(tanA+1)/(1-tanA)-(1-tanA)/(tanA+1).接着分别同分后同时相减就可得结果:4tanA/tan^2A-1
tan(-17π/6)=tan(4π-17π/6)=tan(7π/6)=tan(π+π/6)=tanπ/6=√3/3
sin(25pai/6)=sin(1pai/6)=0.5cos(25pai/3)=cos(1pai/3)=0.5tan(-25pai/4)=-tan(1pai/4)=-1第一题0.5+0.5-1=0第
这个看看你的课本中关于诱导公式这一节你就会解了嘿嘿符号太难打了