Messing around with the $\cos$ & $\sin$ relation with $\pi$. Apparently got wrong.

Question

So the $\sin$ relation with $\pi$ is;

$$\pi =\lim \limits_{n \to \infty} n\cdot\sin(\frac{180}{n}) \tag{Eq. 01}\label{1} $$

And the $\cos$ one is;

$$\pi = \lim \limits_{n \to \infty} n\sqrt2\cdot\sqrt{1-\cos(\frac{180}{n})} \tag{Eq. 02}\label{2}$$

So from these two, we can get;

\begin{align} n*\sin(\frac{180}{n}) & = n\sqrt2\cdot\sqrt{1-\cos(\frac{180}{n})} \\ \sin(\frac{180}{n}) & = \sqrt2\cdot\sqrt{1-\cos(\frac{180}{n})} \\ \sin^2(\frac{180}{n}) & = 2(1-\cos(\frac{180}{n})) \\ \frac{\sin^2(\frac{180}{n})}{1-\cos(\frac{180}{n})} & = 2 \\ \frac{1-\cos^2(\frac{180}{n})}{1-\cos(\frac{180}{n})} & = 2 \\ 1+\cos(\frac{180}{n}) & = 2 \\ \cos(\frac{180}{n}) & = 1 \\ \end{align}

This isn't obviously true. So what's the problem here?

Answer 1

This is the same as the following argument $$\lim_{n\to\infty}\frac1n=\lim_{n\to\infty}\frac2n=0$$ $$\implies\frac1n=\frac2n$$ $$\implies1=2$$ Why would the first equality imply the second?