Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$? 
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?

Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$ 
When I look up trig identities, however, it says $\sin^2(x) = \frac{1-\cos(2x)}{2}$. 
Why is this?

From the angle addition formula, we have
$$\begin{align}
\cos(2x)&=\cos(x+x)\\\\
&=\cos(x)\cos(x)-\sin(x)\sin(x)\\\\
&=\cos^2(x)-\sin^2(x)\\\\
&=\left(1-\sin^2(x)\right)-\sin^2(x)\\\\
&=1-2\sin^2(x)\tag 1
\end{align}$$
Solving $(1)$ for $\sin^2(x)$ yields
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
as expected.
