Given Cotx=5, how do you find sin (x/2), cos (x/2), tan (x/2)?

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Answer 1 · 2017-02-13T06:28:13

#sin(x/2)=(-5+sqrt26)/sqrt(52-10sqrt26)# #cos(x/2)=1/sqrt(52-10sqrt26)# and #tan(x/2)=-5+sqrt26#

or #sin(x/2)=(-5-sqrt26)/sqrt(52+10sqrt26)#, #cos(x/2)=1/sqrt(52+10sqrt26)# and #tan(x/2)=-5-sqrt26#

As #cotx=5#, #tanx=1/5#

but as #tanx=(2tan(x/2))/(1-tan^2(x/2))#

#(2tan(x/2))/(1-tan^2(x/2))=1/5#

or #10tan(x/2)=1-tan^2(x/2)#

or #tan^2(x/2)+10tan(x/2)-1=0#

and using quadratic formula

#tan(x/2)=(-10+-sqrt(10^2-4xx1xx(-1)))/2#

= #(-10+-sqrt104)/2=-5+-sqrt26#

i.e. #tan(x/2)=-5+sqrt26# or #-5-sqrt26#

Hence #sec(x/2)=sqrt(1+(-5+sqrt26)^2)#

= #sqrt(1+25+26-10sqrt26)=sqrt(52-10sqrt26)# or

#sec(x/2)=sqrt(1+(-5-sqrt26)^2)#

= #sqrt(1+25+26+10sqrt26)=sqrt(52+10sqrt26)#

and #cos(x/2)=1/sqrt(52-10sqrt26)# or #1/sqrt(52+10sqrt26)#

and #sin(x/2)=(-5+sqrt26)/sqrt(52-10sqrt26)# or #(-5-sqrt26)/sqrt(52+10sqrt26)#