Why are the two graph #f(x)=sin(arctan2x)# and #g(x)=(2x)/(sqrt(1+4x^2))# equal?

1 Answer
Mar 5, 2017

Please see below

Explanation:

#arctan2x# stands for an angle say #A#, whose tangent ratio is #2x#. In other words, #arctan2x=A# means #tanA=2x#.

As #tanA=2x#, #sin(arctan2x)=sinA#

= #sqrt(1-cos^2A)#

= #sqrt(1-1/sec^2A)#

= #sqrt(1-1/(1+tan^2A))#, but #tanA=2x#, hence

#sin(arctan2x)=sqrt(1-1/(1+(2x)^2))#

or #sin(arctan2x)=sqrt(1-1/(1+4x^2))#

or #sin(arctan2x)=sqrt((4x^2)/(1+4x^2))#

or #sin(arctan2x)=(2x)/sqrt(1+4x^2)#

Hence the graph of #f(x)=sin(arctan2x)# and #g(x)=(2x)/sqrt(1+4x^2)#

are same and appears as follows.
graph{(2x)/sqrt(1+4x^2) [-10, 10, -5, 5]}