Question 1026041
<pre>
{{{3pi/2 <= u < 2pi}}} tells us that angle u is in
the fourth quadrant. 

Since {{{cot(u)=adjacent/opposite}}} and we are given that
{{{cot(u)=-5}}}, we consider the {{{-5}}} as {{{("" + 5)/(-1)}}},
the we can make the numerator be the adjacent side of +5 and 
the denominator be -1.   

So we draw a right triangle in the fourth quadrant whose adjacent
side is +5 and whose opposite side is -1.  [We knew to consider
the -5 as {{{("" + 5)/(-1)}}} and not {{{(-5)/("" + 1)}}} because
+5 goes to the right and -1 goes down.] 

The angle u is represented by the red arc drawn counterclockwise
from the right side of the x-axis to the terminal side which is
the hypotenuse of the right triangle.

The hypotenuse is calculated from the Pythagorean theorem:
{{{c^2=a^2+b^2}}}
{{{c^2=5^2+(-1)^2}}}
{{{c^2=25+1}}}
{{{c^2=26}}}
{{{c=sqrt(26)}}} <-- the hypotenuse is always positive!

{{{drawing(500,250,-2,8,-3,2,
red(locate(-.6,.6,u)),
graph(500,250,-2,8,-3,2),
locate(1.8,.5,adjacent="" + 5),locate(5.1,-.4,opposite=-1),
line(0,0,5,0), line(5,0,5,-1), line(0,0,5,-1),
locate(.1,-.55,hypotenuse=sqrt(26)),
red(arc(0,0,1.2,-1.2,0,349))  )}}}

{{{sin(u) = opposite/hypotenuse = (-1)/sqrt(26)}}}
{{{cos(u) = adjacent/hypotenuse = ("" + 5)/sqrt(26)}}}

Now we use some double angle identities:

{{{sin(2u)=2sin(u)cos(u) = 2((-1)/sqrt(26))(("" + 5)/sqrt(26))=-10/26=-5/13}}}

{{{cos(2u)=cos^2(u)-sin^2(u) = (("" + 5)/sqrt(26))^2-((-1)/sqrt(26))^2}}}{{{""=""}}}

{{{25/26-1/26=24/26=12/13}}}

We could use the formula for tan(2u), but it's easier
to use 

{{{tan(2u)=sin(2u)/cos(2u)=matrix(1,9,-5/13,"÷",12/13,""="",-5/13,""*"",13/12,""="",-5/12)) }}}

Edwin</pre>