若tan(A B)=2tanA,求证3sinB=sin(2A B)

再答：

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

sinb=(根号10/10）,ab是锐角,则有cosb=√(1-sin^2b)=3√10/10,tanb=sinb/cosb=1/3.tan2b=2tanb/(1-tan^2b)=3/4.tan（a+

∵tana-cota=2,∴(tana-cota)^2=4tan²a-2tanacota+cot²a=4tan²a-2+cot²a=4∴tan²a+c

tan(A+B)=2tanAsin(A+B)*cosA=2sinA*cos(A+B)①sin(A+B)*cosA-sinA*cos(A+B)=sinA*cos(A+B)sinB=sinA*cos(A+

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

∵tana-cota=2,∴(tana-cota)^2=4,tan^2a+cot^2a-2=4∴tan^2a+cot^2a=6再问：2从哪里来？再答：(tana-cota)^2=4，tan^2a+co

tan(5π/2-a)=tan(5π/2-a-2π)=tan(π/2-a)=cota=1/tana=1/2,顺便帮忙采纳一下.

1.∵tan(a/2)=2∴tana=[2tan(a/2)]/{1-[tan(a/2)]^2}=(2×2)/(1-2^2)=-4/3∴tan(a+π/4)=[tana+tan(π/4)]/[1-tan

sin2a=2sinacosa=2sinacosa/(sin²a+cos²a)=2tana/(1+tan²a)(上下同时除以cos²a)cos2a=cos

tana+cota=a两边平方tan^2a+2*tana*cota+cot^2a=a^2因为tana*cota=1所以tan^2a+2+cot^2a=a^2所以tan^2a+cot^2a=a^2-2

结果:tana*tanb=1/2.过程也不复杂,把tana移项,然后展开tan(a+b),再全部通分,两边合并同类项.

tana+1/tana=3可化成sina/cosa+cosa/sina=3化简得sin^2a+cos^2a/sinacosa=3可得出sinacosa=1/3由此可得出sina+cosa=根号15/3

tanA=2tan(A/2)/[1-(tan(A/2))^2]-2/tanA=-2*[1-(tan(A/2))^2]/[2tan(A/2)]=[(tan(A/2))^2-1]/(tanA/2)tan(

tan(a/2)-1/(tana/2)=sin(a/2)/cos(a/2)-cos(a/2)/sin(a/2)通分=[sin²(a)-cos²(a/2)]/[sin(a/2)cos

tan(90度-a)=cota=1/tana=1/2再问：你和我的答案一样哦，只是我不确定，谢谢啊

tan^2A+2tanA=3（tanA-1)(tanA+3）=0tanA=1或者-3tanA=1时,A=kπ+π/4tanA=-3时,A=kπ-arctan3即是A=kπ+π/4或者kπ-arctan

α/2为锐角,tan(α/2)>0tan（α）=2tan(α/2)/{[1-tan(α/2)]^2}设tan(α/2)=xx^2+x-1=0x=(根号5)/2-1/2=tan(α/2)

∵tanA-1/tanA=2∴平方,tan²A-2+1/tan²A=4∴tan²A+1/tan²A=6

tan(45+A)=(tan45+tanA)/(1-tan45tanA)tan45=1tan(45+A)=(1+tanA)/(1-tanA)=1/(2+√3)=2-√3