tan(π/4+α)=k,则2α=cos2α=sin(π/2+2α)=sin[2(π+α)]那后来呢一步步来本人数学不怎么好.
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若tan(-a)=k^2(k≠0),|cos(π+α)|=-cosα,则1/cos(π+α)的值为
[sinπ(/2+α)-cos(3π/2-α)]/[tan(2kπ -α)+cot(-kπ+α)]=[sin(4kπ-α)sin(π/2 -α)]/[cos(5π+α)-cos(π/2+α)
tan(kπ+α)=?cos(kπ+α)=?
已知cos(α+β)=sin(α-β),且α≠π/4+kπ,k属于 Z,β≠π/2+kπ,k属于Z,则tanβ
知道tanα=k,求sinα,cosα
求证sin(π/2 +α)-cos(3π/2 -α)/tan(2kπ -α)+cot(-kπ+α)=sin(4kπ -α)sin(π/2 -α)/cos(5π+α)sin(π/2 +α)-cos(3π/2 -α)/tan(2kπ -α)+cot(-kπ+α)=sin(4kπ -α)sin(π/2 -α)/cos(5π+α)-cos(π/2 + α)谁帮我做做
函数y=sinα+cosα-4sinαcosα+1,且2sin^2α+sin2α/1+tanα=k,π/4
cos(α+k·2π)= sin(α+k·2π)= tan(α+k·2π)= sin(-α)= cos(-α)= tan(-α)=要和书上的一样.
求证(1-cosα)/sinα=sinα/(1+cosα)=tan(α/2)(α≠kπ,k∈z) 快回答!
书上说tan(2kπ+α)=tanα,那可不可以tan(2kπ-α)=-tanα
若α∈(-π/2+2kπ,2kπ)(k∈Z),则sinα,cosα,tanα的大小关系是A.tanα>sinα>cosαB.tanα>cosα>sinαC.tanα<sinα<cosαD.tanα<cosα<sinα
已知sinα=4sin(α+β),α+β≠kπ+π/2(k∈Z).求证tan(α+β)=sinβ/(cosβ-4)
若2sinα=1+cosα,α≠kπ(k属于Z),则tan(α/2)=
已知tanα=2,则cos(α+π)等于
若cos(α+β)/cos(α-β)=2/3则tanαtanβ=
cosα=-4/5 则tanα/2
已知tan(π/4+α)=-1/2,求4cosα(sinα+cosα)/(1-tanα)
求证:tanα+1/tan(π/4+α/2)=1/cosα帮个忙!