<s>[INST] Understanding expanding trig identities I already have the answer, but I am trying to figure out why because it's not making much sense. Or maybe my professor misspoke I am not sure.
I have $\sin^5x $ and I am trying to express it in terms containing only first powers of sine and cosine.
Expanding into:
$\sin x\cdot \sin^2x\cdot \sin^2x $
and further:
$\sin x(\frac{1-\cos2x}{2})(\frac{1+\cos2x}{2}) $
And this is the point where I get a little confused. I don't recall any identity saying that $\sin^2x+ \sin^2x $ = $\ (\frac{1-\cos2x}{2})(\frac{1+\cos2x}{2}) $
Anyways the answer is: $\ \frac{\sin x-\sin^2 x\cos 2x + \sin x\cos^2 (2x)}{4} $ But I'm not quite sure how to get there. I figured I would post the answer so people wouldn't think that this is homework, this is for my own understanding. 
 [/INST] Since $\sin^2 x = \frac{1 -\cos(2x)}{2}$, $$\sin^5 x = \sin x \cdot \left(\frac{1 - \cos(2x)}{2}\right)^2 = \sin x \cdot \frac{1 - 2\cos(2x) +\cos^2(2x)}{4} = \frac{\sin x - 2\sin x \cos(2x) + \sin x \cos^2(2x)}{4}.$$
Since $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$, 
\begin{align}\sin x - 2\sin x \cos(2x) + \sin x \cos^2(2x) &= \sin x - 2\sin x \cos(2x) + \frac{1}{2}\sin x + \frac{1}{2}\sin x \cos(4x)\\
&=\frac{3}{2}\sin x - 2\sin x \cos(2x) + \frac{1}{2}\sin x \cos(4x).
\end{align}
Therefore
$$\sin^5 x = \frac{3}{8}\sin x - \frac{1}{2}\sin x \cos(2x) + \frac{1}{8}\sin x \cos(4x).$$
Note: The answer you had written does not equal to $\sin^5 x$; at $x = \frac{\pi}{2}$, $\sin^5 x = 1$, but the expression $\frac{\sin x - \sin^2 x \cos(2x) + \sin x \cos^2(2x)}{4}$ equals $\frac{1 - 1(-1) + 1(-1)^2}{4} = \frac{3}{4}$.
</s>