Gravitational Potential Energy MDOF system

Question

I'm trying to find the Equation of motion for a MDOF system using the energy method. Using Lagrange's equation, I can get almost everything except for the potential energy of the bar. Using newton's method, and assuming small angle, $\sin(\theta)$ = $\theta$, and $\cos(\theta)$ = 1. I tried that in Lagrange's equation and there is no $\theta$ or u term to take a derivative of. This is a textbook example so I Find that they are using $\sin(\theta)$=$\theta$ and $\cos(\theta)$ = (1-$\frac{1}{2}\theta^2$). Can someone please explain to me how in the world are they getting (1-$\frac{1}{2}\theta^2$).Worked Example

Answer 1

The MacLaurin series for $\sin \theta$ and $\cos \theta$ are

$$\begin{align} \sin \theta &= \theta -\frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \\ \\ \cos \theta &= 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \end{align}$$

so for small $\theta$ we get

$$ \sin \theta \approx \theta \quad \text{and} \quad \cos \theta \approx 1 - \frac{\theta^2}{2}$$