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}
 A: It's a Pythagorean identity and comes from $$\sin^2 x + \cos ^2 x = 1$$
Just subtract $\cos ^2 x$ from both sides and you have your answer. Now, as to where $\sin^2 x + \cos ^2 x = 1$ comes from:
Let's say you have a right triangle with legs $a$ and $b$. By the Pythagorean theorem, the hypotenuse is $$\sqrt {a^2 + b^2}$$ 
Next, $\sin x$ is defined as $\dfrac{opposite}{hypotenuse}$, and $\cos x$ is defined as $\dfrac{adjacent}{hypotenuse}$ 
So in your triangle you have $\sin x = \dfrac{a}{\sqrt {a^2 + b^2}}$ and $\cos x = \dfrac{b}{\sqrt {a^2 + b^2}}$
$\sin^2 x = \dfrac{a^2}{a^2 + b^2}$
$\cos^2 x = \dfrac{b^2}{a^2 + b^2}$
$\sin^2 x + \cos^2 x = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$
A: This is just a basic property in Trigonometry.
There are many ways to prove it . Here is one way
$\sin^2x + \cos^2x= \sin x \sin x+ \cos x \cos x = \cos(x-x) = \cos 0 = 1$
A: Here is another simple way to prove the trig. identity, we know $i^2=-1$ so we have 
$$\sin^2 x+\cos^2x$$$$=\cos^2x-i^2\sin^2x$$ $$=(\cos x+i\sin x)(\cos x-i\sin x)$$
using Euler's theorem, $$=(e^{ix})(e^{-ix})=e^{0}=1$$
A: 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
$$\sin ^2 x + \cos ^ 2 x=1$$
A: 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.
