1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 04:43:25
1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)

1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)

1)tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
原式=-tanα/sinα(-cosα)-sinα(-sinα)/(-cosα)cosα
=(sinα/cosα)/sinαcosα-sin²α/cos²α
=(1-sin²α)/cos²α
=1

tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
=tan(-α)/(-sinαcosα)-sin(-α)sinα/(-cosα)cosα
=tanα/(sinαcosα)-sin^2α/cos^α
=1/cos^2α-sin^2α/cos^2α
=cos^2α/cos^2...

全部展开

tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+α)cos(2π+α)
=tan(-α)/(-sinαcosα)-sin(-α)sinα/(-cosα)cosα
=tanα/(sinαcosα)-sin^2α/cos^α
=1/cos^2α-sin^2α/cos^2α
=cos^2α/cos^2α
=1
很高兴为您解答,祝你学习进步!【学习宝典】团队为您答题。
有不明白的可以追问!如果您认可我的回答。
请点击下面的【选为满意回答】按钮,谢谢!

收起

若a属于(0,π/2)试比较tanα、tan(tanα)、tan(sinα) 已知α∈(0,π|2),2tanα+3sinβ=7,且tanα-6sinβ=1,求sinα的值 已知α,β∈(0,π/2),且2tanα+3sinβ=7,tanα-6sinβ=1,则sinα= 1.求证tanαsinα/tanα-sinα=tanα+sinα/tanαsinα2.已知sinα+cosα=-1/5(π/2 已知α,β属于(0,π/2)且sinβ=cos (α+β)sinα,(1)求证tanβ=sinαcosα/1+sin^2 α(2)将tanβ表示成关于tanα的函数(3)求tanβ的最大值,并求当tanβ取最大值时,tan(α+β)的值 sin(2π-α)tan(α+3π)tan(-α-π)/tan(α-3π)cos(π-α) 化简[sin(π-α)cos(3π-α)tan(-α-π)]/[tan(4π-α)sin(5π+α)] (1+tanα)/(1-tanα)=3+2√2,求cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α-π)的值 (1+tanα)/(1-tanα)=3+2√2,求cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α-π)的值 证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tanα) 已知tanα/2=2,求(1)tan(α-4/π)的值,(2)(3sinα+cosα)/(3cosα-sinα) 已知α,β属于(0,π/2),且SINβ/SINα=COS(α+β)(1)求证tan(α+β)=2tanα(2)求证tanβ=sinαcosα/(1+sin²α)(3)将tanβ表示成tanα的函数关系式 若sinα=3sin(π/2-α),则tanα+1/tanα-1= β∈(0,π/4),4tan(π/2)=1-tan^2(π/2),3sinβ=sin(2α+β)…… 高一必修4三角函数sin420*cos750+sin(-330)*cos(-660)tan675+tan765-tan(-330)+tan(-690)sin(25π/6)+cos(25π/3)+tan(-25π/4)2.已知sin(π+α)=-1/2 计算(1)sin(5π-α) (2)sin(π/2+α) (3)cos(α-3π/2) (4)tan(π/2-α) ①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明:sin(-θ)=-sinθ②证明tanθsinθ∕tanθ-sinθ=1+cosθ∕sinθ③已知sinα-2cosα+1=0,α≠kπ+π∕2,k∈z求:tan(3π-α)和1∕sin2α-sinαcosα+1的值‍ 已知tan(α+π/4)=1/3已求出tanα=-1/2求2sin²α-sin(π-α)sin[(π/2)-α]+sin²[(3π/2)+α]的值. 化简:cos(2π-α)sin(-π-α)/sin(π/2+α)tan(3π-α)