Taylor series as an approximation for cos(θ)

Question

So I was watching one of Walter Lewin's lectures and he writes down that...

$L(1-\cos \theta)=L\frac{\theta ^2}{2}$

This was evaluated by using the first 2 terms in the taylor series for cosθ which makes sense however he then writes down

$L(1-\cos \theta)=\frac{x^2}{2L}$ (???How did he come to this conclusion???)

Here is a sketch I drew to help visualize the situation. It involves the motion of a pendulum, but since the specific problem has to do with the taylor approximation here seems like a suiting place for the problem. Thanks!

Answer 1

Note that $\sin\theta=x/L$. And when $\theta$ is small, we have $\sin\theta\approx\theta$. Hence $\theta\approx x/L$.

Substituting this into your first equation gives the second.

Answer 2

These are approximations, appropriate for small $\theta$, not equalities.

The Taylor series gives $$1-\cos\theta=\frac{\theta^2}2-\frac{\theta^4}{24}+\cdots.$$ For small $\theta$, $\theta^4/24$ etc., are much smaller than $\theta^2/2$ and so we have the approximation $$1-\cos\theta\approx\frac{\theta^2}2$$ so that $$L(1-\cos\theta)\approx L\frac{\theta^2}2.$$

In your triangle $x=L\sin\theta\approx L\theta$ using the series $$\sin\theta=\theta-\frac{\theta^3}6+\cdots.$$ Substituting $x/L$ for $\theta$ gives $$L(1-\cos\theta)\approx\frac{x^2}{2L}.$$

All of this is standard theory of the simple pendulum.