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}

If you define $\cos$ and $\sin$ as ratios of legs to hypotenuse of a right triangle, then $\cos^2 (x) +\sin^2 (x) =1$ follows directly from the Pythagorean identity.
