How do you evaluate #\tan ^ { - 1} \frac { x } { 2} = 2\tan ^ { - 1} x#?

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Answer 1 · 2018-03-20T03:09:58

The only solution is #x=0#.

We need the #tan# double angle formula:

#tan(2theta)=(2tantheta)/(1-(tantheta)^2)#

Here's the actual problem. To start, take the #tan# of both sides:

#tan^-1\frac{x}{2}=2tan^-1x#

#tan(tan^-1\frac{x}{2})=tan(2tan^-1x)#

#x/2=tan(2tan^-1x)#

#x/2=(2tan(tan^-1x))/(1-(tan(tan^-1x))^2)#

#x/2=(2x)/(1-x^2)#

#x=(4x)/(1-x^2)#

#x-x^3=4x#

#-3x-x^3=0#

#-x^3-3x=0#

#-x(x^2+3)=0#

The only real solution is #0# since #x^2+3#'s solutions are nonreal (#sqrt3i# and #-sqrt3i#).

We can verify this by looking at the graphs of #tan^-1(x/2)# and #2tan^-1(x)#, and seeing where they intersect; they only intersect at #(0,0)#:

(#tan^-1(x/2)# is in red and #2tan^-1(x)# is in blue.)

Hope this helped!