Question 965646
Given cos&#952;={{{1/4}}} and 270° < &#952; < 360° degrees, find sin2&#952; exactly 
<pre>
Since {{{cos(theta)=x/r}}}, we draw x = numerator of {{{1/4}}}, which is 1,
and r = the denominator of {{{1/4}}}, which is 4. Since 270° < &#952; < 360°, &#952; is in
the 4th quadrant, we draw angle &#952; in the 4th quadrant like this, with the
red arc indicating angle &#952;:

{{{drawing(400,400,-5,5,-5,5,line(-7,0,7,0),line(0,7,0,-7),
line(0,0,1,-sqrt(15)),line(1,-sqrt(15),1,0),
red(arc(0,0,3,-3,0,285)), locate(1.1,-1.3,"y=?"), locate(.1,.5,x=1),
locate(-.3,-1.8,r=4),red(locate(-1.3,1.3,theta))  )}}}{{{matrix(19,1,

Calculate,y,by,the,Pythagorean, theorem,x^2+y^2=r^2,1^2+y^2=4^2,1+y^2=16,
y^2=15,y=-sqrt(15),negative,because,y,goes,down,from,the,x-axis)

}}}{{{drawing(400,400,-5,5,-5,5,line(-7,0,7,0),line(0,7,0,-7),
line(0,0,1,-sqrt(15)),line(1,-sqrt(15),1,0),
red(arc(0,0,3,-3,0,285)), locate(1.1,-1.3,y=-sqrt(15)), locate(.1,.5,x=1),
locate(-.3,-1.8,r=4),red(locate(-1.3,1.3,theta))  )}}}

{{{sin(2theta)=2sin(theta)cos(theta)}}}
{{{sin(2theta)=2(y/r)(x/r)}}}
{{{sin(2theta)=2(-sqrt(15)/4)(1/4)}}}
{{{sin(2theta)=-2sqrt(15)/16}}}
{{{sin(2theta)=-sqrt(15)/8}}}

Edwin</pre>