why $ 1 - \cos^2x = \sin^2x $? I'm trying to prove this result $$\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0$$ In this process I have come across an identity $1-\cos^2x=\sin^2x$. Why should this hold ? Here are a few steps of my working:
\begin{array}\\
 \lim_{x\to 0} \dfrac{1 - \cos(x)}{x}\\ = \lim_{x\to 0} \left[\dfrac{1 - \cos(x)}{x} \times    \dfrac{1 + \cos(x)}{1 + \cos(x)}\right] \\
=\lim_{x\to 0} \left[\dfrac{1 - \cos^2(x)}{x(1+\cos(x))}\right]  \\
=\lim_{x\to 0} \left[\dfrac{\sin^2(x)}{x(1+\cos(x))}\right]
\end{array}

Let $F(x)=\sin ^2 x + \cos ^ 2 x$.
$$F'(x)=2 \sin x\cos x- 2 \cos x\sin x=0$$
Since $F(0)=1$ and $F$ is constant, we get
$$\sin ^2 x + \cos ^ 2 x=1$$
