Question 1112371
{{{alpha}}} is in III quadrant

use Pythagoras theorem

{{{sqrt(8^2+15^2)}}} = hypotenuse =17

cos{{{alpha}}} = 15/17

{{{2cos^2((alpha)/2)}}}= {{{1+cos(alpha)}}}

{{{2cos^2((alpha)/2)}}}= {{{1+(15/17)}}}


{{{2cos^2((alpha)/2)}}}= {{{32/17)}}}


{{{cos^2((alpha)/2)}}}= {{{16/17)}}}

cos {{{alpha}}} = {{{4/sqrt(17)}}}
{{{alpha}}} is in III quadrant so ....