已知tan(a π 4)=1 2,且-π.2大于a小于0,则2sin2a-搜答案-灵鹊做题网

tan(9π/4+a)*tan(3π/4+a)=tan(2π+π/4+a)*tan(π-π/4+a)=tan(a+π/4)*tan(a-π/4)=(1+tana)/(1-tana)*(1-tana)/

sina+cosa=1/5sin^2a+cos^2a=1sin^2a+(1/5-sina)^2=12sin^2a-2sina/5+1/25=12sin^2a-2sina/5-24/25=050sin^

1.f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-π-a)sin(-π-a)=cosa*sina*tana/-cota*sina=-tan^2acosa2.co

(1)f(a)=-cosa(2)f(a)=1/5用口诀“奇变偶不变,符号看象限”!

⑴f(a)=sinacosa(-tana)/(-cota)*sina=sina.

sin(a+π)=-4/5sina=-sin(a+π)=4/5a是第二象限的角cosa

tan(A+B)=4(tanA+tanB)/(1-tanAtanB)=4tanAtanB=1/2再联立tanA+tanB=2又tanA

∵已知方程x2+4ax+3a+1=0（a＞1）的两根为tanα，tanβ，∴tanα+tanβ=-4a＜0，tanα•tanβ=3a+1＞4．∴tan（α+β）=tanα+ tanβ&nbs

集合A={α|tanα=3/4且sinαcotα0且sinαcotα

sina/cosa=4/3->sina=4/3cosa(1)a在第三象限->sina

（tan(α+π/4)=-1/2,(tanα+1)/(1-tanα)=-1/2,∴tanα=-3∴cosα=-√10/10(sin2α-2cos^2α)/sin(α-π/4)=(2sinαcosα-2

/>因为sina=3/5,a属于(π/2,π)所以cosa=-4/5(1)cos(a-π/4)=cosa*cosπ/4+sina*sinπ/4=-根号2/10；(2)sin(a/2)的平方+tan(a

tana=1∴a=π/4∴sina=根号2/2

(1+tana)=2(1-tana),tana=1/3,cosa=3sina,cos^2a=9sin^2a,cos^2a=9/10,sin^2a=1/10sin2acosa-sina/cos2a=2s

tan(a+π/4)=(tana+tan(π/4))/[1-tana*tan(π/4)]=(3+1)/(1-3*1)=-2tan(a-π/4)=(tana-tan(π/4))/[1+tana*tan(

f(a)=sinacosacotatana/(-sina)所以f(a)=-cosa

题目中的sin(a+x)应该是sin(a+π)sin(a+π)=4/5-sina=4/5sina=-4/5∵sinacosa＜0∴cosa=3/5[2sin(a-π)+3tan(3π-a)]/4cos

tan(π/4+a)=（1+tana）/（1-tana）=-1/2tana=-3sina=3根号10/10[sin2a-2(cosa)^2]/1+tana=[sin2a-2+2(sina)^2]/1+

tan(α+π/4)=(tanα+tan(π/4))/(1-tanα*tan(π/4))=(tanα+1)/(1-tanα)=1/2=>tanα=-1/3-π/2