已知x属于（0,π/2）,tanx=1/2,求tan2x和sin(2x+π/3）的值

tanx=1/2
tan2x=2tanx/[1-(tanx)^2]=1/(1-1/4)=4/3
x属于（0，π/2），2x属于（0，π）
由tan2x>0知2x则cos2x=1/√[1+(tan2x)^2]=1/√(1+16/9)=3/5
sin2x=√[1-(cos2x)^2]=4/5
sin(2x+π/3)=sin2xcos(π/3)+cos2xsin(π/3)
=(4/5)*(1/2)+(3/5)*(√3/2)
=(4+3√3)/10

tanx=1/2,所以x0
tan2x=2 tanx/(1+tan^2 x)=4/5
tanx=1/2,所以x0
sin(2x+π/3)=sin2x /2 + cos2x *√3 /2=(4+√3)/10

tan2x=2tanx/(1-tan²x)=4/3
sin(2x+π/3）=1/2sin2x+√3/2cos2x
=sinxcosx+√3/2(cos²x-sin²x)
=[sinxcosx+√3/2(cos²x-sin²x)]/(sin²x+cos²x)
=[tanx+√3/2（1-tan²x）]/(tan²x+1)
=8√3/15+2/5

tan2x=2tanx/(1-tan^2x)=4/3
sin(2x+π/3）=sin2xcosπ/3+cos2xsinπ/3=1/2sin2x+√3/2cos2x
=1/2*2tanx/(1+tan^2x)+√3/2(1-tan^2x)/(1+tan^2x)
=2/5+3√3/10

tan2x=2tanx/(1-tanx^2)=4/3tan2x=4/3=sin2x/cos2xsin2x^2+cos2x^2=1 x属于（0,π/2) 2x属于（0,π) tan2x>0sin2x=4/5 cos2x=3/5sin(2x+π/3）=sin2x cosπ/3+cos2x sinπ/3=（4+3√3）/10