Question 1155044
<pre>
We know that  270° < θ < 360° means that θ is in quadrant IV.
We also know that the tangent is the opposite/adjacent or y/x.
So we can draw a right triangle in the 4th quadrant so that
the opposite side is the numerator +5 of the tangent (-5/12) 
taken as (+5)/(-12), and the denominator is -12 of the tangent 
taken as (+5)/(-12).  So we have the drawing: 

{{{drawing(500,500,-14,14,-14,14,

graph(500,500,-14,14,-14,14), locate(2.2,1.1,x=""+5),
locate(5,-12,"(5,-12)"),locate(5.1,-6,y=-12),locate(.9,-6,"r=?"),
triangle(5,0,5,-12,0,0)  )}}}

Next we use the Pythagorean theorem to calculate r:

{{{matrix(6,1,
matrix(1,3,r^2,""="",x^2+y^2),
matrix(1,3,r^2,""="",(""+5)^2+(-12)^2),
matrix(1,3,r^2,""="",25+144),
matrix(1,3,r^2,""="",169),
matrix(1,3,r,""="",sqrt(169)),
matrix(1,3,r,""="",13) )}}}

Fill in the value of r=13:

{{{drawing(500,500,-14,14,-14,14,

graph(500,500,-14,14,-14,14), locate(2.2,1.1,x=""+5),
locate(5,-12,"(5,-12)"),locate(5.1,-6,y=-12),locate(.9,-6,"r=13"),
triangle(5,0,5,-12,0,0)  )}}}

Now we use the formula for sin(2θ):

{{{sin(2theta)=2sin(theta)cos(theta)}}}

and the fact that the sine is opposite/hypotenuse or y/r = (-12)/(13) = -12/13
and the fact that the cosine is adjacent/hypotenuse or x/r = (+5)/(13) = 5/13

{{{sin(2theta)=2(-12/13)(5/13)}}}

{{{sin(2theta)=2(-12/13)(5/13)=-120/169}}}

We use the formula for cos(2θ):

{{{cos(2theta)=cos^2(theta)-sin^2(theta)}}}


{{{cos(2theta)=(-5/13)2-(-12/13)^2}}}

{{{cos(2theta)=25/169-144/169}}}
 
{{{cos(2theta)=-119/169}}}

Edwin</pre>