sin2αtan2α=tan2α-sin2α 恒等式怎么做 2是平方-搜答案-拍题作业网

题目有误,α∈（-π/2,0）cosα=1/5α∈（-π/2,0）sinα

∵sin2α=2sinαcosα=-sinα，α∈（π2，π），∴cosα=-12，sinα=1−cos2α=32，∴tanα=-3，则tan2α=2tanα1−tan2α=3．故选：A．

cosα=-4/5,且α为第三象限角可得Sinα=－3／5tanα=3／4sin2α=2sinαcosα=24／25cos2α=cos²α－sin²α=7／25tan2α=24／7

tan2α=1/32tana/(1-tana^2)=1/36tana=1-tana^2tana^2+6tana-1=0tana=[-6+/-根号(40)]/2=-3+/-根号10

sinα/cosα=1/2cosα=2sinα代入sin²α+cos²α=1所以sin²α=1/5cos²α=4/5sin2α=2sinαcosα=2sinα(

tan2α-sin2α=(sin2a+cos2a)(tan2α-sin2α)//导入sin2a+cos2a=1=sin2atan2α-(sin2α)^2+cos2atan2a-sin2acos2a=s

sin2a=2sinacosa=-sina,cosa=-1/2,sina=√3/2,cos2a=2(cosa)^2-1=-1/2tan2a=(-√3/2)/(-1/2)=√3

答案如图片再问：根号2乘上sin(π/4+α)为什么等于sinα+cosα？再答：用公式展开

Sin2A=2SinA•CosACos2A=CosA^2-SinA^2=1-2SinA^2=2CosA^2-1tan2A=2tanA/（1-tanA^2）

则cosa=-3/5sin2a=2sinacosa=-24/25cos2a=cosa方-sina方=-7/25tan2a=sin2a/cos2a=24/7

（1+tan2α）/（1-tanα）=2010=>{1+2tanα/[(1-tanα)^2]}/(1-tanα)=1-(tanα)^2+2tanα=2010(1+tanα)=>2009+(tanα)^

tanα=3,α属于(π/4,π/2)sinα=3/√10cosα=1/√10sin2α=2sinαcosα=2(3/√10)(1/√10)=3/5cos2α=(cosα)^2-(sinα)^2=1/

(1)cosα=1/7,因为0＜α＜π/2,所以sinα=√(1-cosα)=√[1-(1/7)]=4√3/7所以tanα=sinα/cosα

sin2a=2sinacosacos2a=cosa平方-sina平方tan2a=2tana比上1-tana平方

tan2α-sin2α=sin2α/cos2α-sin2α=sin2α-sin2αcos2α/cos2α=sin2α(1-cos2α)/cos2α=tan2αsin2α

sin2α=2√2/3cos2α=1/3

由sinα=5/13,α∈(π/2,π)则cosα=-12/13(由sinα^2+cosα^2=1,又cosα

α∈(π/2,π)cosα=-√(1-1/9)=-2√2/3sin2α=2sinα*cosα=1/3*-2√2/3=-2√2/9cos2α=2cos²α-1=2*8/9-1=7/9tan2α

∵coα∈(π/2,π)∴cos