Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.
I know that $\sin^2x=(1/2)(1-\cos2x)$,
so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$
However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$
I am not sure if that is correct. 
 A: Rather than messing about with trig identities, it may be simpler to use complex exponentials.  If $z = e^{ix} = \cos(x) + i \sin(x)$, so $\sin(x) = (z - z^{-1})/(2i)$ and $\cos(x) = (z + z^{-1})/2$,
$$\eqalign{\sin^4 x - \sin^6 x &= \frac{(z - z^{-1})^4}{16} + \frac{(z - z^{-1})^6}{64}\cr
&= \frac{z^4 - 4 z^2 + 6 - 4 z^{-2} + z^{-4}}{16} \\
&\qquad+\frac{z^6 - 6 z^4 + 15 z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6}}{64}\cr
&= \frac{z^6 + z^{-6}}{64} - \frac{z^4 + z^{-4}}{32} - \frac{z^2 + z^{-2}}{64} + \frac{1}{16}\cr
&= \frac{\cos(6x)}{32} - \frac{\cos(4x)}{16}  - \frac{\cos(2x)}{32} + \frac{1}{16}\cr}$$
A: I have noticed the other answers either don't answer your question (and get many upvotes none-the-less) or are much more complicated than necessary (even more complicated than you understand based on the comment by Raymond).  So, here is the answer to your actual question.
You dropped the 1/2.
$$\sin^6 x = (\sin^2 x)^3 = \left(\frac{1 - \cos{2x}}{2}\right)^3 = \frac{1}{8} (1 - \cos{2x})^3$$
so you end up missing $1/8$ in the end.  Also, you messed up the sign on one term.  In general
$$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$
so here we have
\begin{align*}
  (1 - \cos(2x))^3 &= \bigg(1 + (-\cos(2x))\bigg)^3 \\
  &= 1 + 3(-\cos(2x)) + 3(-\cos(2x))^2 + (-\cos(2x))^3 \\
  &= 1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x).
\end{align*}
Therefore, your final answer is
\begin{align*}
  \sin^4x - \sin^6x &= \frac{1}{4}\bigg(1 - 2\cos(2x) + \cos^2(2x)\bigg) \\
  &- \frac{1}{8}\bigg(1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x)\bigg) \\
  &= \frac{1}{8}\bigg(1  - \cos(2x) - \cos^2(2x) + \cos^3(2x)\bigg)
\end{align*}
A: $\sin^4x-\sin^6x=\sin^4x(1-\sin^2x)=\sin^4x \cos^2x=\sin^2x(\sin^2x \cos^2x)$ $$=\frac{\sin^2 x\,\sin^2 2x}{4}=\left(\frac{\sin x \sin 2x}{2}\right)^2=\left(\frac{\cos x-\cos3x}{4}\right)^2$$
A: There are various ways to transform $\sin^4 x-\sin^6 x$ to some equivalent form. For example, $\sin^4 x-\sin^6 x=\sin^4 x(1-\sin^2 x)=\sin^4 x\,\cos^2 x$.
If you are interested in expressing the difference as a sum of trigonometric functions of multiples of $x$, this will provide an efficient start. For then you can use $\sin^2 x=\frac{1-\cos 2x}{2}$ and $(\sin x\cos x)^2=\frac{(\sin 2x)^2}{4}$. 
