Determination of the uncertainty of the cosine of the half angle of a measured angle

I am trying to determine the uncertainty of the cosine of the angle, $$\beta$$, when the angle that is measured is $$2\beta$$. If the uncertainty in the measurement of $$2\beta$$ is 1 degree, then is the uncertainty in $$\beta$$ one half of a degree?

• Use either $\cos(\beta^*-\epsilon/2)=\cos(\beta^*)\cos(\epsilon/2)+\sin(\beta^*)\sin(\epsilon/2)$ or $\cos(\beta^*-\epsilon/2)=\cos(\beta^*)+\sin(\beta^*)\epsilon/2+O(\epsilon^2)$ for small uncertainty $\epsilon.$ You might want to bear in mind that the measured angle is $2\beta^*=2\beta+\epsilon,$ not just $2\beta.$
– whuber
Commented Apr 5 at 17:27
• There are two questions „If the uncertainty in the measurement of $2\beta$ is 1 degree, then is the uncertainty in $\beta$ one half of a degree?” and „Determination of the uncertainty of the cosine of the half angle of a measured angle“ Is this question about the cosine or about the division by two? Commented Apr 7 at 16:15

Let $$2b = 2\beta+\epsilon$$ be the measurement of the true but unknown angle $$\beta$$ so that $$\epsilon$$ represents the measurement uncertainty as an additive discrepancy $$\epsilon = (2b) - (2\beta).$$ Naturally--and this has many theoretical justifications--you would estimate $$\beta$$ as half of your measurement,

$$\hat\beta = (2b)/2 = b.$$

An equally natural estimate of $$\cos\beta$$ is the "plug-in" estimate,

$$\widehat{\cos\beta} = \cos(b).$$

The uncertainty in this estimate, also expressed additively as the difference $$\widehat{\cos\beta}-\cos(\beta),$$ can be found in several ways. A quick and easy one, relying on the expectation that $$\epsilon$$ is small (one degree is less than $$0.02$$ radians) uses a Taylor expansion and simplifies it through repeated subsitution of the foregoing equalities:

\begin{aligned} \widehat{\cos\beta}-\cos(\beta) &= \cos(b)-\cos\beta\\ &= \cos(\beta+\epsilon/2)-\cos(\beta)\\ &= -\sin(\beta)\frac{\epsilon}{2}+O(\epsilon^2)\\ &= -\sin(b-\epsilon/2)\frac{\epsilon}{2}+O(\epsilon^2)\\ &= -(\sin(b) + O(\epsilon))\frac{\epsilon}{2}+O(\epsilon^2)\\ &=-\sin(b)\frac{\epsilon}{2} + O(\epsilon^2). \end{aligned}

Dropping $$O(\epsilon^2)$$ as negligible, the answer is

The error in the plug-in estimate of the cosine of $$\beta$$ is $$\sin(b)/2$$ times the measurement error inherent in the observation $$2b.$$

This supplies more information than the standard "delta method" that relies on a similar analysis, insofar as when we adopt a convention that the measurement $$2b$$ is between $$0$$ and $$360$$ degrees, the sine of $$b$$ is positive, showing that the error in the cosine of $$\beta$$ tends to be of the opposite sign as the error in the measurement. (This conclusion breaks down near $$\beta\approx 0,$$ where the neglected quadratic term matters.)

To illustrate and verify this result, here are the results of four independent simulations using errors $$\epsilon$$ with standard deviations of one degree. (I generated these errors asymmetrically with exponential tails to show this analysis does not require the errors to be "nicely" distributed.)

The near-perfect negative correlations for $$\beta\ne 0$$ illustrate the "opposite sign" conclusion, while the quadratic behavior of the errors in the cosine show the (vanishingly small) effects of the quadratic terms neglected in the Taylor expansion.

• Another illustration of a break down of the Taylor approximation when the first derivative is zero, is described here for the case of the Delta method is here: stats.stackexchange.com/questions/343187/… Commented Apr 7 at 16:31
• @Sextus Are you claiming the sines of angles between 90 and 180 degrees are negative? ;-) // The Taylor approximation does not "break down" here: it remains just as accurate as assured by the various forms of Taylor's theorem. But when the derivative is zero, the behavior is of higher order. That is evident in the tiny range of the vertical axis in the $\beta=0$ plot.
– whuber
Commented Apr 7 at 18:59
• I misread your post as assuming that 0 < b < 360, but I see now that you (indirectly) assumed 0 < b < 180. The comment about the breakdown is unrelated to that wrong reading (I agree that it is not the Taylor approximation that breaks down, I should have commented more specifically that the first order Taylor approximation, and conclusions based on it, breaks down) Commented Apr 7 at 19:48