Answer
$-\frac{sin~2x+sin~x}{cos~2x-cos~x} = cot~\frac{x}{2}$
Work Step by Step
$-\frac{sin~2x+sin~x}{cos~2x-cos~x}$
When we graph this function, it looks like the graph of $~~cot~\frac{x}{2}$
We can verify this algebraically:
$-\frac{sin~2x+sin~x}{cos~2x-cos~x}$
$=-\frac{2~sin~x~cos~x+sin~x}{2~cos^2~x-1-cos~x}$
$=-\frac{sin~x~(2~cos~x+1)}{(2~cos~x+1)(cos~x-1)}$
$=-\frac{sin~x}{cos~x-1}$
$=\frac{sin~x}{1-cos~x}$
$=\frac{1}{tan~\frac{x}{2}}$
$=cot~\frac{x}{2}$