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)#