<s>[INST] Fewest steps to show $\sin^2(s) = \frac{1}{2} - \frac{1}{2} \cos(2s)$? Is there a method for getting from this statement
$
\sin^2(s)
$
to this statement
$
\frac{1}{2}\ -\ \frac{1}{2}\cos(2s)
$
in fewer steps than the following?
$\begin{align}
\cos(2s) &= \cos^2(s) - \sin^2(s) \\
\frac{1}{2} \cos(2s) &= \frac{1}{2} \cos^2(s) - \frac{1}{2} \sin^2(s) \\
-\frac{1}{2} \cos(2s) &= -\frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} - \frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( 1 - \cos^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( \sin^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \sin^2(s) \\
\end{align}$
 [/INST] You can certainly eliminate several of the steps. I might write it as
$$\cos 2s = \cos^2 s - \sin^2 s\tag{dbl angle formula}$$
$$\cos 2s = 1-2\sin^2 s\tag{$\cos^2$ in terms of $\sin^2$}$$
$$1-\cos 2s = 2 \sin^2 s\tag{subtr both sides from $1$}$$
$$\tfrac12 - \tfrac12\cos 2s = \sin^2 s\tag{halve both sides}$$
</s>