We measure an angle to be -55 degrees (2 sigfigs). Let us take the sine of this angle. We get sin(-55) (2 sigfigs). Now, since sin(-55)=sin(305), then we can also take the sin(305) (3sigfigs). Now, since sine(305) is just sin(3905), then we can also take sin(3905) (4sigfigs) and so on. In this manner we can get a precise sine with arbirary number of sigfigs from an imprecise measurement. What is wrong with this example?
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1$\begingroup$ The error is that $f(x)$ does not have the same significance as $x$. For a trivial example, the function $f(x) = 2$ has infinite sigfigs irrespective of the significance of the input. $\endgroup$– Marius Ladegård MeyerMar 27 at 9:05
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$\begingroup$ What do you recommend I should do with a physics textbook that thinks otherwise? $\endgroup$– wosMar 27 at 9:13
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2$\begingroup$ Please mention title, author and/or ISBN of your textbook, so we can have a look. $\endgroup$– MS-SPOMar 27 at 9:36
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$\begingroup$ Does the book take into account that sine is a cyclic function? $\endgroup$– MS-SPOMar 27 at 9:37
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$\begingroup$ University Physics with Modern Physics in SI Units by Young and Freedman pearson link. See Example 1.6 in the first chapter. When computing Dx, the x component of vector D. The cos(-45) is multiplied by 3.00 meters (3sigfigs) to get 2.1 meters (2sigfigs) which assumes that cos(-45) has 2 sigfigs. $\endgroup$– wosMar 28 at 6:17
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1 Answer
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It is important to understand that significant figures are not used in actual scientific publications. Significant figures are essentially "training wheels" to get new students to think about measurement uncertainty in a simple way that doesn't interfere with learning other topics. Like actual training wheels, significant figures are flimsy and prone to break. Here you have found one way to break them.
In actual scientific publications the expression of the uncertainty in a measurement is done explicitly through the "standard uncertainty", which is the estimated standard deviation of a series of repeated measurements of the same quantity. The calculation and reporting of uncertainty in scientific publications is described by the BIPM's "Guide to the expression of uncertainty in measurement" (GUM).
If you have measured an angle to be $-55^\circ$ what that actually means is that you are confident that the angle is somewhere between $-54.5^\circ$ and $-55.5^\circ$. If we assume that your uncertainty is normally distributed and that range represents a 95% confidence interval, then according to the GUM (section 7.2.2) we would report that as "$\theta=-55.00^\circ$ with a standard uncertainty of $u(\theta)=0.25^\circ$" or more concisely as "$\theta=-55.00(25)^\circ$ where the number in parentheses is the numerical value of the standard uncertainty, $u(\theta)$ referred to the corresponding last digits of the quoted result" (I will use this last format for the remainder of this post).
Now, to find the standard uncertainty of $\sin(\theta)$ where $\theta$ is measured in degrees we follow the propagation of uncertainty procedure given in GUM 5.1.2 and elsewhere. $$u(\sin(\theta))=\sqrt{\left(\frac{\partial \sin(\theta)}{\partial \theta}\right)^2 u(\theta)^2}=\left|\frac{\pi}{180}\cos(\theta)\right| u(\theta)$$ Now, for $\theta=55.^\circ$ and $u(\theta)=0.25^\circ$ we have $\sin(55^\circ)=0.8192$ and $u(\sin(\theta))=0.0025$ so we would write $$\sin(55.00(25)^\circ)=0.8192(25)$$This is how it would appear in an actual scientific publication.
Now, your idea is to add $n \ 360^\circ$ to $\theta$ to increase the significant figures where $n \ 360^\circ$ is not a measured value with uncertainty but an exact constant. However, $$u(\sin(\theta+n \ 360^\circ))=\left|\frac{\pi}{180}\cos(\theta+n \ 360^\circ)\right| u(\theta+n \ 360^\circ) =\left|\frac{\pi}{180}\cos(\theta)\right| u(\theta) = u(\sin(\theta))$$ so doing that does not change the standard uncertainty at all. We still have $$\sin(55.00(25)^\circ+ n \ 360^\circ)=0.8192(25)$$
Note that $0.8192(25)$ means that we expect the value of $\sin(\theta)$ to be somewhere between $0.8142$ and $0.8242$ and if we are using significant figures $0.82$ means that we expect the value to be somewhere between $0.815$ and $0.825$.
So if you don't deliberately try to break the significant figures approach, it does give roughly the correct result in this case. The "training wheels" are not bad, they are just prone to break when misused. And they are not used professionally.
