已知tanα=1/2,求tan2α,tan(2α+π/3)的值

已知tanα=1/2,求tan2α,tan(2α+π/3)的值
已知tanα=1/2,求tan2α,tan(2α+π/3)的值

已知tanα=1/2,求tan2α,tan(2α+π/3)的值
因tanα=1/2,故tan2α=2tanα/(1-tan^2 α)=4/3
tan(2α+π/3)=(tan2α+tanπ/3)/(1-tan2αtanπ/3)
=[(4/3)+根号3]/[1-(4/3)(根号3)]
=- (48+25√3)/39

4/3 (48+25√3)/39

α＜π/2 , 所以sinα = √(1-cos α) = √[1-(1/7) ] = 4 √ 3 / 7 所以tanα = sinα / cosα = 4 √ 3 (2) 根据两倍角的

tan2a=2tana/(1-tana^2)
=2乘以1/2除以(1-1/2乘以1/2)
=4/3