已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急

来源：学生作业帮助网编辑：六六作业网时间：2024/11/25 01:36:08

已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急已知

已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急
已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值
急

已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急
由已知可得 tanα=tan[(α+π/4)-π/4]=[tan(α+π/4)-tan(π/4)] / [1+tan(α+π/4)tan(π/4)]
=(1/5-1)/(1+1/5)= -2/3 ,
所以 [sin(2α)-(sinα)^2] / [1-cos(2α)]
=sinα(2cosα-sinα) / [2(sinα)^2]
=(2cosα-sinα) / (2sinα)
=1/tanα-1/2
= -3/2-1/2
= -2 .

tan(π/4+α)=1/5
(1+tanα)/(1-tanα)=1/5
tanα=-2/3
(sin2α-sin²α)/(1-cos2α)
=(2sinαcosα-sin²α)/2sin²α
=(2tanα-tan²α)/2tan²α
=

tan(a+π/4)=[tana+tan(π/4)]/[1-tanatan(π/4)]=(tana+1)/(1-tana)=1/5
1-tana=5tana+5
6tana=-4
tana=-2/3
[sin(2a)-sin²a]/[1-cos(2a)]
=(2sinacosa-sin²a)/(2sin²a)
=(2cosa-sina)/(2sina)
=(2-tana)/(2tana)
=[2-(-2/3)]/[2(-2/3)]
=(8/3)/(-4/3)
=-2

已知tan(α+β)=2/5,tan(β-π/4)=1/4,求1+tanα/1-tanα的值大神们帮帮忙已知tan(A+B)=2/5,tan(B-π/4)=1/4,求tan(A+π/4已知tan(A+B)=2/5,tan(B-π/4)=1/4,求tan(A+π/4) 已知tanα/2=2,求tanα与tan（α+π/4）已知tan(α＋β)=2/5,tan(β-π/4)=1/4,求tan(α+π/4)的值已知tan（α+β）=2/5,tan（β-π/4)=1/4,求tan（α+π/4）的值快已知tan（α+ β）=2/5,tan（β -π/4）=1/4,求tan（α+π/4）已知tan(α+π/4)=2,求tanα和tan(α-π/4)的值已知tanα=1/4,tan(α-β)=1/3,求tanβ的值已知tan(π/4+α)=-1/2,求4cosα(sinα+cosα)/(1-tanα) 已知 α 为锐角,sin α =4/5,tan( α -β)=1/3,求cos2 α ,tanβ 已知 α 为锐角,sin α =4/5,tan( α -β)=1/3,求cos α ,tanβ 已知tan（π/4+α）=2,tanβ=1/2求tan2α的已知tanα=1/2,求tan2α,tan(2α+π/3)的值已知tanα=2,求tan(α-π/4)的值已知tanα=3,求tan（α-π/4）已知tanα＝-1/2,求tan（α＋π/4）已知3π/4＜α＜π,tanα+1/tanα=-10/3.求tanα的值已知tan（α-β）=1/2,tan（ π/4 -β）=1/3求 tan（α-π/4）如果可以麻烦都算一下- --