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 $(-5-sqrt26)/sqrt(52+10sqrt26)$