Uncertainty of angle with known uncertainty of cosine of angle Quick question, how do I proceed if I managed to get (from some kind of a measurement) the value of $\cos(\phi)$ and its error $\sigma_{\cos\phi}$ and I'd like to find $\phi \pm \sigma_{\phi}$?
Would it be reasonable to just calculate $\arccos(\cos(\phi) \pm \sigma_{\cos\phi})$ to find the range the angle is likely in? Can I somehow use the error propagation formula to get the error directly? Or is there any other way this is commonly dealt with? Thank you.
 A: So you have the measurements in the form $cos\phi\pm \Delta(cos\phi)$ and need to find $\phi\pm\Delta\phi$
Note that $$\left| \frac{\Delta cos(\phi)}{\Delta\phi}\right|=sin(\phi)$$ where we assume $\{\phi\ne 0^\circ, 90^\circ\}$ so that $$\mid\Delta cos(\phi)\mid=sin(\phi)\Delta\phi$$ From what you say, it looks like you have a value for $\Delta cos(\phi)=\sigma_{\cos\phi}$
So the uncertainty in $\phi$ $$\Delta\phi=\frac{\sigma_{\cos\phi}}{sin(\phi)}$$ meaning you would write the range as $\phi\pm\Delta\phi$

As per a comment made by Mark H. for very small angles, the computed uncertainty will be large (since the the expression for $\Delta\phi$ becomes large if the denominator approaches zero) so an additional term to the $|\Delta \cos(\phi )|$ expression is added so that we now have the equation (Taylor series expansion about $\phi\approx 0$ - second term is the second derivative of original $cos$ term)  $$|\Delta\cos\phi| \approx \sin\phi\Delta\phi + \frac{1}{2}\cos\phi(\Delta\phi)^2$$ so that for small angles, i.e., $cos(\phi)\approx 1$, the quadratic term dominates.
A: Let $y = f(x)$ be a function of $x$.  Let $S_x$ be the standard deviation for $x$.  The standard deviation for $S_y$ for $y$ is $\sqrt{({\partial f \over \partial x})^2 S_x^2}$.  [See, for example, the text Data Analysis for Scientists and Engineers by Meyer.]
Let $x$ be $\theta$ and $y = cos(\theta)$.  $S_{cos(\theta)} = \sqrt{S_{\theta}^2 \space sin^2(\theta)}$; so $S_{\theta} = {S_{cos(\theta)} \over |sin(\theta)|}$.
The other answers address small $\theta$.
A: By definition, for small errors, the derivative is an approximation of the ratio of the dependent variable's error to the independent variable's error. Since the derivative of $\cos$ is $\sin$, that gives us that $\frac{\sigma_{\cos \phi}}{\sigma _{\phi}}=\sin\phi$. Rearranging gives $\sigma_{ \phi} =\frac{ \sigma_{\cos\phi}}{\sin \phi }$.
As Mark H points out in a comment to another answer, this goes to infinity as $\sin \phi$ goes to zero. Looking at a graph of $\cos$, you can see that near $\phi =0$, the curve is close to horizontal, meaning that small variations in $\cos \phi$ correspond to large variations in $\phi$. Your idea of taking $\arccos$ would also be problematic in this region, as $\cos$ is not 1-1 in the neighborhood around $0$.
However, $\cos \phi$ can, in the neighborhood around $0$, be approximated by $1-\frac{\phi^2}2$, thus giving $\phi = \pm \sqrt {2-2\cos \phi}$, so for $\cos \phi$ close to zero, you can plug $\cos \phi \pm\sigma_{\cos \phi}$ into that formula. With two $\pm$ (a \pm in the value to plug into the formula and a \pm in the formula), that will give you four different values, and you should take the widest spread.
And all of this, of course, applies only to small $\sigma_{\cos \phi}$.
