# What are the units of the $\sin(10^\circ)$? [closed]

Say I measure the angle of incidence to be $10^\circ$, what is the unit for $\sin(10^\circ)$?

## closed as off-topic by Hanul Jeon, Xander Henderson, Chris Custer, Claude Leibovici, GNUSupporter 8964民主女神 地下教會Apr 9 '18 at 12:46

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• $\mathrm {sin}\ \theta = \frac {\mathrm {perpendicular}\ \mathrm {w.r.t.}\ \theta} { \mathrm {hypotenuse}\ \mathrm {w.r.t.}\ \theta}$. Both perpendicular and hypotenuse are lengths so they have the same dimensions and hence their ratio has no dimension. – Arnab Chatterjee. Apr 8 '18 at 10:23
• You should also be aware that "radians" are also unitless. While they are the measure of an angle, as are degrees, it's incorrect to say an angle is x radians. – JoeTaxpayer Apr 8 '18 at 14:11
• if you consider the series expansion of $\sin$ you will see all terms must be of the same dimensionality in order to be consistent and the leading order term is equal to $1$ which is dimensionless. Therefore all terms are dimensionless and so is the result of the $\sin$ function. – nluigi Apr 8 '18 at 14:20
• @nluigi And if you consider the series expansion of $\sqrt{x} = 1- (1-x)/2 - (1-x)^2/8 + \dots$ analogously you see that everything you take a square root of has to be dimensionless? Not a sound argument... – Federico Poloni Apr 8 '18 at 17:05
• @Narasimham So velocity $v=\frac{s}{t}$ is dimensionless? – Federico Poloni Apr 8 '18 at 17:15

The short answer has already been given: Because the sine function is defined as the ratio between the opposite leg and the hypotenuse, its dimension is $\frac{\mathrm{length}}{\mathrm{length}} = 1$ (i.e. it is dimensionless).

As was pointed out by nluigi in the comments, the Taylor expansion $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ (for $x$ given in radians) shows that $x$ and $\sin(x)$ both have to be dimensionless, since all terms must have the same dimension. Indeed, the angle $x$ is defined as the ratio between the radius of a circular arc with angle $x$ and its arclength. In this spirit, $^\circ$ is not a real unit, but just a convenient shorthand for the numerical constant $\frac{2\pi}{360}$, and $\mathrm{rad}$ is just an alias for the constant $1$.

• It was very clarifying for me when I realized that a radian, rather than being strictly dimensionless, actually is a unit of $\text{arclength}\over\text{radial distance}$. – hBy2Py Apr 8 '18 at 17:51
• @hBy2Py But how do we reconcile that with the Taylor expansion? – Ovi Apr 8 '18 at 18:32
• @Ovi A similar "unitless units" difficulty arises in the $2\pi$ factor needed to convert between angular and cyclic frequencies; see here. – hBy2Py Apr 8 '18 at 18:41
• @DanRobertson How can $1 + x$ be dimensionful if $1$ is dimensionless? – Elias Riedel Gårding Apr 8 '18 at 20:22
• I am sorry, but I believe the heuristic argument with the series expansion to be hopelessly flawed. See my comments to the main question for details. – Federico Poloni Apr 9 '18 at 6:09

Sine function is a ratio of lengths and has no unit.

sine is definitely dimensionless - for a given angle (no matter what units are used to express it) you get the same sine. The sine of a right angle is always 1, for example.

Interesting question about whether an angle has dimension. It definitely has units: an angle of 1 degree is nothing like an angle of 1 radian; but does that mean it has dimension?

@Federico Poloni - velocity definitely does have dimension: it's length/time. Just because you divide x by y does not make something dimensionless - you must keep in mind the units of x and y, and if they are different, you do not get a dimensionless value.

There are already good answers but I would like to add a note on the Taylor series in the case of dimensional arguments, a question that has come up a few times in the comments on this page.

A Taylor series can be used to expand a function whose value is dimensional and/or whose argument is dimensional. It is easy to see how by noticing that each term of the Taylor series of $f(x)$ around the expansion point $x_0$ has the form:

$$\frac{1}{n!} \left. \frac{\mathrm{d}^n f}{\mathrm{d}x^n} \right|_{x_0} (x - x_0)^n.$$

Since the dimension of $\left.\mathrm{d}x^n\right|_{x_0}$ and that of $(x - x_0)^n$ are the same, all terms have the dimension of $\mathrm{d}^n f$, which is the same as that of $f$. In other words all the dimensions sort out nicely by themselves. Please also see the answers to this other question.

In the case of the function "$\sin$", as already said by others, both the argument and the value are dimensionless by the very definition of $\sin$. Nevertheless, I could ask myself what happens if I still want to explicitly consider the units. I could define $\sin$ as a function from radians to the ratio of lengths: $[\mathrm{rad}] \rightarrow [\frac{\mathrm{m}}{\mathrm{m}}]$ (I know it's redundant but let's continue anyway). If I do so, I cannot simply use the conventional Taylor expansion for $\sin$ but I have to use one where the value of $f$ and of its derivatives were computed using the same units: $$\sin(x) = 0\,\frac{\mathrm{m}}{\mathrm{m}} + 1\,\frac{1\,\mathrm{m}}{\mathrm{m}\cdot\mathrm{rad}} \, (x-0\,\mathrm{rad}) + \frac12\, \frac{0\,\mathrm{m}}{\mathrm{m}\cdot\mathrm{rad}^2} \, (x-0\,\mathrm{rad})^2 + \dots \quad \text{with x in [\mathrm{rad}]}.$$ This also works nicely if one uses degrees instead of radians: the derivatives are not 1, 0, -1 any more but are powers of $\pi/180^{\circ}$ and the $^{\circ}$ "unit" in these coefficients cancels out with that of the powers of $x$.

Of course I am not advocating the use of this needlessly overcomplicated representation but I just wanted to show that everything would work nicely anyway.

• Does this not just reduce to “Taylor series give us a function with the right dimensions when the right dimensions are whatever we choose them to be?” In such a case you can say “it’s fine to look at the dimensions of the Taylor series because the dimensions of the Taylor series are the same as whatever we choose the dimensions of the function to be. So one can say “why is sin a function of dimensionless to dimensionless?” and get the answer “well sin is a function of dimensionless to dimensionless so it’s Taylor series is dimensionless to dimensionless so you look at the series and ... – Dan Robertson Apr 9 '18 at 12:49
• @DanRobertson I am not sure I fully understand what you are saying. If this is your objection, I wouldn't answer the question “why is sin a function of dimensionless to dimensionless?” by using the Taylor argument (as I showed it is not as strong because choosing a different expansion point, one with units, makes it possible to consider units). I would simply say that the sin is dimensionless because its argument is a ratio of lengths (arc length divided by radius) and its value is a ratio of lengths (opposite cathetus divided by hypotenuse). – Luca Citi Apr 9 '18 at 13:44