Question 831646
For angles terminating in quadrant IV, sine is negative and cosine is positive.
For any angle {{{theta}}} , {{{sin^2(theta) + cos^2(theta)=1}}} .
So, If {{{cos(theta)=1/2}}} ,
{{{sin^2(theta) + (1/2)^2=1}}}
{{{sin^2(theta) + 1/4=1}}}
{{{sin^2(theta)=1-1/4}}}
{{{sin^2(theta)=3/4}}}
Since we know that {{{sin(theta)>0}}} ,
{{{sin(theta)=-sqrt(3/4)}}} is not a solution.
The solution is
{{{sin(theta)=sqrt(3/4)}}}
{{{highlight(sin(theta)=sqrt(3)/2)}}}
 
NOTE:
We also know that one of the possible values of {{{theta}}} is {{{-60^o}}} or {{{-pi/3}}},
because a 30-60-90 right triangle (one with angles measuring 30, 60, and 90 degrees) is half of an equilateral triangle>
{{{drawing(300,300,-0.1,1.1,-0.6,0.6,
triangle(0,0,0.866,0.5,0.866,-0.5),
green(line(0,0,0.866,0)),
green(rectangle(0.816,0,0.866,0.05)),
locate(0.88,0.3,1/2),locate(0.88,-0.2,1/2),
locate(0.44,0.25,1),locate(0,44,-0.27,1),
red(arc(0.866,0.5,0.3,0.3,90,150)),
locate(0.72,0.38,red(60^o))
)}}}
So for an equilateral triangle with side length of 1,
we get a right triangle where the leg opposite the {{{30^o}}} angle (adjacent to the {{{30^o}}} angle) measures {{{1/2}}} ,
meaning that {{{sin(30^o)=cos(60^o)=1/2}}} .