Answer
$\frac{2~(sin~x-sin^3~x)}{cos~x} = sin~2x$
Work Step by Step
$\frac{2~(sin~x-sin^3~x)}{cos~x}$
When we graph this function, it looks like the graph of $~~sin~2x$
We can verify this algebraically:
$\frac{2~(sin~x-sin^3~x)}{cos~x}$
$= \frac{2~sin~x(1-sin^2~x)}{cos~x}$
$= \frac{2~sin~x(cos^2~x)}{cos~x}$
$= 2~sin~x~cos~x$
$=sin~2x$
