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#