Let #arcsin(1/3)=theta#
#=>tan(2arcsin(1/3))# becomes #color(brown)(tan(2theta))#
and #tan(2theta)=(2tan(theta))/(1-tan^2(theta)) color(red)(larr"this is what we are looking for"#
- So we need #color(brown)tan(theta)#
Here's how we do:
#color(green)"Recall:" arcsin# is the inverse trig function of #sin#
So since, #arcsin(1/3)=theta#
#=>sin(theta)=1/3#
#=>1/sin(theta)=3=csc(theta)#
#=>csc(theta)=3#
#=>(csc(theta))^2=(3)^2#
#=>csc^2(theta)=9#
Use the trig identity : #cos^2(theta)+sin^2(theta)=1#
Divide all through by #sin^2(theta)#
#=>cot^2(theta)+1=csc^2(theta)#
#=>1/tan^2(theta)+1=csc^2(theta)#
#=>tan^2(theta)=1/(csc^2(theta)-1)#
#=>tan(theta)=sqrt(1/(csc^2(theta)-1))#
#=>tan(theta)=sqrt(1/(9-1))=sqrt(1/8)=sqrt(8)/8=(2sqrt(2))/8=color(brown)(sqrt2/4)#
Finally,
#=>tan(2theta)=(2tan(theta))/(1-tan^2(theta))=(2(sqrt2/4))/(1-(sqrt2/4)^2)=(8sqrt2)/(16-2)=color(blue)((4sqrt2)/7)#
