trigonometry - Show that $sin^2 theta cdot cos^2theta = (1/8)[1 - cos(4 theta)]$.

I have these two problems I'm working on!

First of the Double Angle Formula! This formula I attempted to do a lot but couldn't get to the identity!
$$\sin^2 \theta \cdot \cos^2\theta = \tfrac18[1 - \cos(4 \theta)]$$

For Above question you can only use the following:

\begin{align}
\sin^2\theta &= \tfrac12 (1-\cos(2 \theta)) \\
\cos^2\theta &= \tfrac12(1 + \cos( 2 \theta ))
\end{align}

And lastly this Sum And Difference Formula! I tried this one so much, I'm leaning toward it being impossible (it's obviously not.... because it's a question):
$$\cos(a-b) \cdot \cos(a + b) = (\cos^2a - \sin^2b).$$