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.

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!