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I have a set of results from an experiment I did recently in the lab where one array of results is for the variable $\theta$. In order to obtain a graph from which I can analyse the results and obtain a value for the electron mass I must take the cosine of this variable, to give me the following data:

$\cos \theta = [0.94, 0.77, 0.50, 0.17, -0.17, -0.50]$

Now, the errors on theta are plus/minus 0.5° and so using the standard error formula and assigning a variable y to the cosine of theta such that $y = \cos \theta$, I obtain the error on y (which will be the error on cos theta) as:

$\sigma (y) = \frac{\partial y}{\partial \theta} \, \sigma (\theta) = -\sigma (\theta) \sin (\theta)$

But when I plug the values of theta and the error on theta into said error formula I obtain the following errors:

$\sigma (\cos \theta) = \pm [0.17, 0.32, 0.43, 0.49, 0.49, 0.43]$

As you can see, these errors are pretty large (nearly 300% in some cases) and so I get the feeling I am doing something wrong here. So, am I being stupid and missing something blindingly obvious?

Edited by x(x-y)
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Seriously real results? Because they look awfully like the cosine of 20 40 60 80 100 and 120 degrees!

If you think what experimental uncertainty is you can work long hand to give your self an idea - and your errors clearly do not work out, as you have mentioned. When things seem to be banjaxed try adding and subtracting the error from your result and running each high and low through your calcs (in this case simple taking cosine) - does it work out as huge errors? In this case no. So where are your errors springing from ? Take a look at the formulation and work out exactly what it is telling you

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I suspect the problem stems from using 0.5º as $\sigma (\theta)$. You should be using radians.

Once you've made this mistake once, you tend not to forget it.

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I suspect the problem stems from using 0.5º as $\sigma (\theta)$. You should be using radians.

Once you've made this mistake once, you tend not to forget it.

Ah, yes, what an amateur mistake. That seems odd though, if I change everything to radians then the errors make sense - however surely if you were using degrees for theta then you'd stick with that convention all the way through your calculation? Well, obviously not I guess - I really should stop using degrees altogether, I work in radians most the time anyway.

Thanks.

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Ah, yes, what an amateur mistake. That seems odd though, if I change everything to radians then the errors make sense - however surely if you were using degrees for theta then you'd stick with that convention all the way through your calculation? Well, obviously not I guess - I really should stop using degrees altogether, I work in radians most the time anyway.

Thanks.

When taking a derivative of a sinusoidal function, if you're working in degrees then you pickup a factor of $\pi /180^{\circ}$. This is because the derivative is "naturally" in terms of radians. So if $\theta$ is in terms of degrees, then the equivalent angle in radians is $\pi \theta/180^{\circ}$.

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