Question #8d7d4

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Answer 1 · 2018-01-01T18:42:23

Let
#tan^-1(tan(x/2) tan(y/2))=theta#

#=>tantheta=(tan(x/2) tan(y/2))#

Now

#cos2theta=(1-tan^2theta)/(1+tan^2theta)#

#=(1-tan^2(x/2)tan^2(y/2))/(1+tan^2(x/2)tan^2(y/2))#

#=(cos^2(x/2)cos^2(y/2)-sin^2(x/2)sin^2(y/2))/(cos^2(x/2)cos^2(y/2)+sin^2(x/2)sin^2(y/2)#

#=(cos(x/2+y/2)cos(x/2-y/2))/(1/4(1+cosx)(1+cosy)+1/4(1-cosx)(1-cosy))#

#=(cos(x/2+y/2)cos(x/2-y/2))/(1/4(2+2cosxcosy))#

#=(2cos(x/2+y/2)cos(x/2-y/2))/(1+cosxcosy)#

#=(cosx+cosy)/(1+cosxcosy)#

So

#2theta=cos^-1((cosx+cosy)/(1+cosxcosy))#

#=>2tan^-1(tan(x/2) tan(y/2))=cos^-1((cosx+cosy)/(1+cosx cosy))#