Question 1131784
Find {{{sin (x/2)}}}, {{{cos( x/2)}}}, and{{{ tan (x/2)}}} from the given information:


 {{{sec (x )= 9/8}}}, {{{270}}}° < {{{x}}} < {{{360}}}° 

{{{sec (x )= 9/8=hypotenuse/adjacent_side}}}
{{{hypotenuse=9}}}
{{{adjacent_ side=8}}}
{{{opposite_side=sqrt(9^2-8^2)=sqrt(81-64)=sqrt(17)}}}


{{{sin (x)=-sqrt(17)/9}}} (in quadrant IV where {{{sin(x) < 0}}})

{{{cos( x)=8/9}}}


Identities you need to use:

{{{sin (x/2)}}}= ± {{{sqrt((1-cos (x))/2)}}}


choose positive root because {{{(x/2)}}} is in quadrant II where{{{ sin(x) > 0}}}

{{{sin (x/2)}}}= ± {{{sqrt((1-cos x)/2)=sqrt((1-8/9)/2)=sqrt(1/18)}}}


{{{cos (x/2)}}}= ± {{{sqrt((1+cos( x))/2)}}}


choose negative root because {{{(x/2)}}} is in quadrant II where {{{cos(x)< 0}}}

{{{cos (x/2)=-sqrt((1+cos (x))/2)=-sqrt((1+8/9)/2)=-sqrt(17/18)}}}

{{{tan( x/2)=sin(x)/(1+cos (x))=-sqrt(17)/9/(1+8/9)=-sqrt(17)/17}}}


How to check answers with calculator:

{{{sec (x)=9/8}}}
{{{cos (x)=8/9}}}

{{{cos^-1(8/9)}}}&#8776; {{{0.47588225}}} radians &#8776;{{{27.27}}}° (reference angle in specified quadrant IV)


standard position of angle={{{360-27.27°=332.73}}}°
{{{x/2=332.73/2=166.37}}}º


reference angle={{{180-166.37=13.63}}}º


{{{sin (x/2)=sin (13.63)}}} &#8776;{{{0.2357}}}.......(in quadrant II where {{{sin(x) > 0}}})

and above we have
{{{sin (x/2)=sqrt(1/18)=0.2357}}}.....which confirms our solution


you can check {{{cos (x/2)}}} and tan {{{(x/2)}}} in the same way