1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?

题目

1.已知tan(a+b)=3,tan(a-b)=2,则sin4a=?
2.已知tana*tanb=13/7,tan[(a+b)/2]=（根号6）/2,求cos(a-b)
3.已知a,b∈（0,π）,且tan(a/2)=1/2,sin(a+b)=5/13,求cosb

答案

1、tan2a=[tan(a+b)+tan(a-b)]/[1-tan(a+b)·tan(a-b)]=(3+2)/(1-3*2)=-1
∴sin4a=2tan2a/[1+(tan2a)^2]=2(-1)/(1+1)=-1
2、cos(a+b)={1-tan[(a+b)/2]^2}/{1+tan[(a+b)/2]^2}=(1-6/4)/(1+6/4)=-1/5
又 cos(a+b)=cosa·cosb-sina·sinb=-1/5
13/7=tana·tanb=sina·sinb/(cosa·cosb)
∴-1/5=cosa·cosb-sina·sinb=cosa·cosb-13/7cosa·cosb=-6/7cosa·cosb
∴cosa·cosb=7/30
∴cos(a-b)=cosa·cosb+sina·sinb=cosa·cosb+13/7cosa·cosb=20/7cosa·cosb=2/3
3、sina=2tan(a/2)/{1+[tan(a/2)]^2}=2*1/2/(1+1/4)=4/5
cosa={1-[tan(a/2)]^2}/{1+[tan(a/2)]^2}=(1-1/4)/(1+1/4)=3/5
又 sin(a+b)=5/13
∴cos(a+b)=±12/13
∴cosb=cos(a+b-a)=cos(a+b)·cosa+sin(a+b)·sina=56/65 或 -16/65