Simplify $Sin[2tan^-1(sqrt((1-x)/(1+x)))]$ ?

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Answer 1 · 2016-12-24T09:37:53

$sin[2tan^-1(sqrt(1-x)/sqrt(1+x))]=-sqrt(1-x^2)$

$sin[2tan^-1(sqrt(1-x)/sqrt(1+x))]$

As $sqrt(1-x)$ , is defined, we have $x<1$ and hence let $x=sin2A=2sinAcosA$

then $sqrt(1-x)/sqrt(1+x)=sqrt(sin^2A+cos^2A-2sinAcosA)/sqrt(sin^2A+cos^2A-2sinAcosA)$

= $(sinA-cosA)/(sinA+cosA)$

= $(sinA/cosA-1)/(sinA/cosA+1)$

= $(tanA-tan(pi/4))/(tanAtan(pi/4)+1)$

= $tan(A-pi/4)$

Hence $2tan^-1(sqrt(1-x)/sqrt(1+x))=2(A-pi/4)=2A-pi/2$ and

$sin[2tan^-1(sqrt(1-x)/sqrt(1+x))]$

= $sin(2A-pi/2)=-sin(pi/2-2A)=-cos2A=-sqrt(1-sin^2 2A)=-sqrt(1-x^2)$

Answer 2 · 2016-12-24T13:06:30

let $x = cos 2A$

So given expression becomes

$Sin[2tan^-1(sqrt((1-x)/(1+x)))]$

$=Sin[2tan^-1(sqrt((1-cos2A)/(1+cos2A)))]$

$=Sin[2tan^-1(sqrt((2sin^2A)/(2cos^2A)))]$

$=Sin[2tan^-1(tanA)]$

$=Sin(2A)$

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

$=sqrt(1-x^2)$

Please note
We have taken $x =cos2A$

And $sqrt((1-x)/(1+x))$ to be real the domain of this function will be $1>= x> -1$ $=>2A in [0,pi)=>A in [0,pi/2)$

So $sin2A$ or $Sin[2tan^-1(sqrt((1-x)/(1+x)))]$ should be $color(red)(POSITIVE)$

Simplify Sin[2tan^-1(sqrt((1-x)/(1+x)))]Sin[2tan^-1(sqrt((1-x)/(1+x)))]?