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?
 A: Notice that $\cos^{2}(x):=(\cos(x))^{2}$ is not the same thing as $\cos(2x)$. It is indeed true that $\sin^{2}(x)=1-\cos^{2}(x)$ and that $\sin^{2}(x)=\frac{1-\cos(2x)}{2}$.
A: Both formulas are true, however, both are useful in different contexts (applications).


*

*You use $\sin^2(x) = \frac{1-\cos(2x)}{2}$ for integrating $\sin^2(x)$. 

*You use $\sin^2(x) = 1 - \cos^2(x)$, for example, when solving $\sin^2(x) = 2\cos(x)$. 
Note that it is just in some way more "natural" to write $\sin^2(x) + \cos^2(x)=1$, because this gives both $\sin^2(x) = 1 - \cos^2(x)$ and $\cos^2(x) = 1 - \sin^2(x)$ in one "natural looking" formula.
A: 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.
A: Let's address two things, the question in the post and the confusion others have pointed out.
Part 1: Proof
Here's a proof that $\cos^2(\theta)+\sin^2(\theta)=1$, from which you can show that $\sin^2(\theta)=1-\cos^2(\theta)$. It requires a bit of set up using the Euler identity and that $i=\sqrt{-1}$.  If you are not familiar with complex numbers, find yourself a video on YouTube, you'll get the hang of them in next to no time.
$$e^{i\theta}=\cos(\theta)+i\sin(\theta)\tag{1}$$
$$e^{-i\theta}=\cos(\theta)-i\sin(\theta)\tag{2}$$
Now, add (1) and (2).
$$e^{i\theta}+e^{-i\theta}=\cos(\theta)+i\sin(\theta)+\cos(\theta)-i\sin(\theta)=2\cos(\theta)\tag{3}$$
Subtract (2) from (1)
$$e^{i\theta}-e^{-i\theta}=\cos(\theta)+i\sin(\theta)-\cos(\theta)+i\sin(\theta)=2i\sin(\theta)\tag{4}$$
From (3) and (4) we get (5) and (6)
$$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}\tag{5}$$
$$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}\tag{6}$$
now, add the squares of (5) and (6)
$$\cos^2(\theta)=\frac{e^{2i\theta}+2e^{i\theta-i\theta}+e^{-2i\theta}}{4}=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}\tag{7}$$
$$\sin^2(\theta)=\frac{e^{2i\theta}-2e^{i\theta-i\theta}+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}\tag{8}$$
$$\cos^2(\theta)+sin^2(\theta)=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}+\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{4}\tag{9}$$
$$\cos^2(\theta)+sin^2(\theta)=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{4}=\frac{4}{4}=1\tag{10}$$
Part 2: Addressing the confusion of $\cos(2x)$ and $\cos^2(x)$
$$\cos(2\theta)=\frac{e^{2i\theta}+e^{-2i\theta}}{2}= \frac{\left(e^{i\theta}\right)^2+\left(e^{-i\theta}\right)^2}{2} = \frac{(\cos(\theta)+i\sin(\theta))^2+(\cos(\theta)-i\sin(\theta))^2}{2}\tag{11}$$
Processing this result further, we obtain
$$\cos(2\theta)=\frac{\cos^2(\theta)+2i\cos(\theta)\sin(\theta) - \sin^2(\theta) + \cos^2(\theta) -2i\cos(\theta)\sin(\theta)-\sin^2(\theta) }{2}=\frac{2\cos^2(\theta)-2\sin^2(\theta)}{2}$$
Therefore
$$\cos(2\theta)=\cos^2(\theta)-\sin^2(2\theta)\tag{12}$$.
Ok, you stated that $$\sin^2(x)=\frac{1-\cos(2x)}{2}\tag{13}$$
well, if you substitute the result from (12) into (13), you find that 
$$\sin^2(\theta)=\frac{1-\cos^2(\theta)+\sin^2(2\theta)}{2}\tag{14}$$
Rearrange this (multiply both sides by 2 and then subtract the $sin^2(\theta)$ from the right side to the left) to get 
$$\sin^2(\theta)=1-\cos^2(\theta)\tag{15}$$
