Why are the two graph f(x)=sin(arctan2x)f(x)=sin(arctan2x) and g(x)=(2x)/(sqrt(1+4x^2))g(x)=2x1+4x2 equal?

1 Answer
Mar 5, 2017

Please see below

Explanation:

arctan2xarctan2x stands for an angle say AA, whose tangent ratio is 2x2x. In other words, arctan2x=Aarctan2x=A means tanA=2xtanA=2x.

As tanA=2xtanA=2x, sin(arctan2x)=sinAsin(arctan2x)=sinA

= sqrt(1-cos^2A)1cos2A

= sqrt(1-1/sec^2A)11sec2A

= sqrt(1-1/(1+tan^2A))111+tan2A, but tanA=2xtanA=2x, hence

sin(arctan2x)=sqrt(1-1/(1+(2x)^2))sin(arctan2x)=111+(2x)2

or sin(arctan2x)=sqrt(1-1/(1+4x^2))sin(arctan2x)=111+4x2

or sin(arctan2x)=sqrt((4x^2)/(1+4x^2))sin(arctan2x)=4x21+4x2

or sin(arctan2x)=(2x)/sqrt(1+4x^2)sin(arctan2x)=2x1+4x2

Hence the graph of f(x)=sin(arctan2x)f(x)=sin(arctan2x) and g(x)=(2x)/sqrt(1+4x^2)g(x)=2x1+4x2

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