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)
