Question #39d7c

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Answer 1 · 2017-11-24T14:34:26

#[1+(1+x)^(1/2)]tanx =[ 1+(1-x)^(1/2)]#

#=>tanx =[ 1+(1-x)^(1/2)]/[ 1+(1+x)^(1/2)]#

To make simplification easy let us put #x=cos2theta# in RHS

So we have

#=>tanx =[ 1+(1-cos2theta)^(1/2)]/[ 1+(1+cos2theta)^(1/2)]#

#=>tanx =[ 1+(2sin^2theta)^(1/2)]/[ 1+(2cos^2theta)^(1/2)]#

#=>tanx =[ 1+sqrt2sintheta]/[ 1+sqrt2costheta]#

#=>tanx =[ 1/sqrt2+sintheta]/[ 1/sqrt2+costheta]#

#=>tanx =[ sin(pi/4)+sintheta]/[ cos(pi/4)+costheta]#

#=>tanx =[ 2sin(pi/8+theta/2)cos(pi/8-theta/2)]/[ 2cos(pi/8+theta/2)cos(pi/8-theta/2)]#

#=>tanx =tan(pi/8+theta/2)#

#=>x=npi+pi/8+theta/2" where " n in ZZ#

#=>4x=4npi+pi/2+2theta#

#=>sin(4x)=sin(4npi+pi/2+2theta)#

#=>sin(4x)=cos(2theta)#

#=>sin(4x)=x#