Question #27fc5

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Answer 1 · 2016-12-24T12:41:59

let $x = cos 2A$

So given expression becomes

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

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

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

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

$=Sin(2A)$

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

$=sqrt(1-x^2)$

Alternative

Let $tan^-1(sqrt((1-x)/(1+x)))=theta$

$=>(sqrt((1-x)/(1+x)))=tantheta$
So
$Sin[2tan^-1(sqrt((1-x)/(1+x)))]$

$=Sin[2theta]$

$=(2tantheta)/(1+tan^2theta)$

$=(2sqrt((1-x)/(1+x)))/(1+(sqrt((1-x)/(1+x)))^2)$

$=(2sqrt((1-x)/(1+x)))/(1+(1-x)/(1+x))$

$=(2sqrt((1-x)/(1+x)))/((1+x+1-x)/(1+x))$

$=2sqrt((1-x)/(1+x))xx(1+x)/2$

$=sqrt(1-x)xxsqrt(1+x)$

$=sqrt(1-x^2$