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I was just testing out the Lagrangian on the simple pendulum and noticed that I got different results based on how I defined theta.

When defining $\theta$ from the vertical, we have $x=r\sin\theta$ and $y=-r\cos\theta$. Then, \begin{align} L&=T-U \\ &=\frac{1}{2}mr^2\dot\theta^2-mgr\cos\theta \end{align} Then the Euler-Lagrange equations give, $$ \frac{g}{r}\sin\theta+\ddot\theta=0\tag{1} $$

But when defining $\theta$ from the horizontal, we end up with $$ \frac{g}{r}\cos\theta+\ddot\theta=0\tag{2} $$

Since (1) and (2) are not the same when considering the small angle approximation, what am I missing here?

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The reason for the difference is because of the two different coordinate systems. When defining $\theta$ from the vertical axis, the stable stationary point is when $\theta=0$ which enables the small angle approximation. However, when defining $\theta$ from the horizontal axis, the stable stationary point is at $\theta=90^\circ$ and we cannot use the small angle approximation. Instead, we would have to do a slightly different small angle approximation: $$ \cos\left(\pi/2\pm\theta\right)\approx\cos(\pi/2)\mp\theta\sin(\pi/2)\simeq\mp\theta $$ which seems to give you the same solution as when considering $\theta$ from vertical axis.

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