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
$$\sin^2(\theta)=1-\cos^2(\theta)\tag{15}$$