I am working on a simple pendulum problem. The $y$ direction is vertical and the $x$ direction is horizontal. Displacement in the $x$ direction is taken to be much less than the length of the string, $L$.

One of the small angle approximations given for this problem was $${\theta \over 2} \approx {y \over x}. $$ where $y$ and $x$ represents the coordinates of the pendulum.

Why is this true? One of the small angle approximations I know is $$\tan \theta \approx \theta, $$ giving $$\frac{x}{y}\approx\theta.$$

Where did the factor of two in the first equation come from?

enter image description here

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    $\begingroup$ It would help to know exactly where you are applying the small angle approximation. I would guess to Newton's laws of motion, right? $\endgroup$ – ggcg Dec 17 '18 at 12:58
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    $\begingroup$ Where is your origin? That matters a lot too. $\endgroup$ – QuIcKmAtHs Dec 17 '18 at 13:34
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    $\begingroup$ $sin(\theta/2)=\theta/2-1/3!((\theta)/2)^3+....$ so for small angles like 6 1/3!((\theta)/2)^3=1.9*10^-4 $\endgroup$ – Abdelrhman Fawzy Dec 17 '18 at 13:38
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    $\begingroup$ Is this the original diagram that was used? Maybe, in the original diagram, theta, y, and x were defined differently? $\endgroup$ – Chet Miller Dec 17 '18 at 13:49
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    $\begingroup$ If you look carefully the geometric scheme you should realize that, with the notation uset there, $tan \theta = y/x$. Of course, for small angles $tan \theta$ and $\theta$ can be identified. $\endgroup$ – GiorgioP Dec 17 '18 at 13:50

It seems like the origin is not P, but the point where the pendulum intersects the vertical axis. As a result the coordinates are:

$x = L \sin(\theta)$

$y = L - L \cos(\theta)$

Which gives us:

$\frac{y}{x} = \frac{1 - \cos(\theta)}{\sin(\theta)} = \tan(\frac{\theta}{2}) \approx \frac{\theta}{2}$

Another way of finding the same result is to calculate (geometrically) the angle between the pendulum, the origin, and the horizontal axis: it is equal to $\frac{\theta}{2}$.

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