Why aren't trigonometric functions dimensionless regardless of the argument?

Consider this equation :-

$$y = a\sin kt$$ where $a$ is amplitude, $y$ is displacement, $t$ is time and $k$ is some dimensionless constant.

My instructor said this equation is dimensionally incorrect because the dimension of $[kt] = [\text{T}^1]$ and since $\text{angles}$ are dimensionless, we can conclude that it is dimensionally incorrect.

I don't understand why it is so. Why do we need to check the dimension homogeneity of the term inside the $\sin$ to conclude whether the equation is dimensionally correct or not?

Why isn't the whole sine function is dimensionless $(\sin kt = \text{[T}^0])$ regardless of the dimension of the argument inside as the range of sine function is $[-1, 1]$.

• What is $\sin(1\ \mathrm{s})$? Is is the same as $\sin(1000\ \mathrm{ms})$? – Javier Aug 4 '16 at 15:36
• $k$ is not dimensionless: it has units of radians per second and is the angular frequency – Henry Aug 4 '16 at 21:11
• @Henry Angles in radians are considered to be dimensionless. Don't ask why. – user253751 Aug 5 '16 at 0:50
• @immibis: It's angles/second. The /second cancels out t which makes the argument dimensionless again. That's what I think Henry is saying – slebetman Aug 5 '16 at 6:17
• @ritwiksinha "How we can prove that the argument inside sine must be dimensionless?" We don't need to prove that because we define $sin$ as a function of dimensionless argument. The argument is dimensionless by definition. – Kamil Maciorowski Aug 5 '16 at 10:36

One definition of the sinus function (in fact, the one probably used by your calculator) is $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ This definition is dimensionally inconsistent unless $x$ is dimensionless. The proof that this gives you the familiar trigonometric sine if $x$ is the ratio of two carefully chosen lengths takes you through a surprisingly interesting swath of mathematics.

• I know taylor series but haven't learnt it formally. So, if i only consider circle definition of sine , is $\sin x = [T^0]$, can you show me the contradiction with circle definition ? – A---B Aug 4 '16 at 17:44
• I'm not entirely clear whether your follow-up question is about the units for the argument of the sine function, or the units for the result of the sine function. In regular trigonometry the result of the sine function is defined as the ratio of lengths of two sides of a triangle, which is also dimensionless. I'm not sure that the dimension of the argument of the sine function is constrained until you start using calculus, but it must be some sort of an angle or a ratio rather than, say, a length or a mass. – rob Aug 4 '16 at 19:07
• For what it's worth (I didn't know this until about four years ago), calculators don't use Taylor series but rather the CORDIC algorithm which stores the angles whose tangents are of the form $2^{-n}$, decomposes an angle into the sum of these special angles (to within desired approximation) and then imparts successive rotation matrices - which by construction only involve left / right shifts of binary numbers for multiplications. Most software libraries, however, use a version of Taylor series. See qc.cuny.edu/Academics/Degrees/DMNS/Faculty%20Documents/… – WetSavannaAnimal Aug 5 '16 at 1:18
• "... the sinus function ..." I thought it was the sine function? Or have I wandered into one of the strange parts of physics again? – Οurous Aug 5 '16 at 4:58
• I'd be very surprised if a calculator used that definition to compute a sine. It's slow, doesn't work well for big $x$ and has trouble achieve high precision results. – CodesInChaos Aug 5 '16 at 7:33

The sine function $\sin(\theta)$ is defined as a function of an angle $\theta$ (measured in radians). So the question amounts to asking why an angle must be dimensionless. (Of course, the answers invoking properties of Taylor series are also correct.) The angle $\theta$ between two directions is defined as the ratio $\theta = a/r$, where $a$ is the arc length connecting the endpoints of a pair of lines of equal length $r$ pointing in those directions (see diagram below). An angle is therefore a ratio of two lengths, making the angle dimensionless. In particular, it doesn't matter if you measure $a$ and $r$ in metres, centimetres, inches or parsecs, the answer for their ratio $\theta$ will always be the same.

• thanks for such a nice answer. Now i am stuck between which answer to choose, yours or rob's.:) – A---B Aug 5 '16 at 6:03
• On the other hand, angle itself makes a perfectly good dimension in an appropriate coordinate system. – bright-star Aug 5 '16 at 16:32

The quantity $kt$ is dimensionless.
If $t$ has the dimensions of time then the dimensions of $k$ must be $\text{time}^{-1}$ so your series expansion works.

You will meet this idea again and again in Physics and checking the dimensions often is a good way of checking a derivation.
The charging and discharging of a capacitor $C$ through a resistor $R$ has a term $e ^{-\frac {t}{RC}}$ and so if $t$ is a time then $RC$ will also have the dimensions of time so that $\frac{t}{RC}$ will be dimensionless which would not be case if the derived expression contained $\frac{tC}{R}$.

• You're making an assumption that $k$ has dimensions of $\text{time}^{-1}$. This is not stated in the question, so where did that assumption come from? – DanielSank Aug 4 '16 at 15:54
• It is my understanding that you must make sure that equations a dimensionally consistent. So if one takes the sine of an expression the expression must be dimensionless even though it might have a value in radians or degrees. – Farcher Aug 4 '16 at 16:00
• Uh, yeah but the whole point of the original question is that the OP's instructor said that $kt$ has dimensions of $\text{time}$, which strongly suggests that in this particular case $k$ is a dimensionless constant. You are explicitly contradicting the main premise of the post and ignoring the main question, which is why the argument of the trig function need to be dimensionless in order for the equation to be dimensionally sound. – DanielSank Aug 4 '16 at 16:06
• As was pointed out in one of the comments above if $k$ was just a number then the same time but measured in different units would give different values to the trig function which would in effect mean that the trig function was indeterminate. – Farcher Aug 4 '16 at 16:16
• I understand your answer, I'm just trying to point out that it doesn't answer OP's question. – DanielSank Aug 4 '16 at 17:19

All maths functions can only be used with dimensionless arguments. The reason is quite boringly that these functions are only defined for real numbers, or perhaps integers, complex numbers, real vectors. But time is none of these.

The only exception you can make are homogeneous functions, especially linear functions. A linear function allows to you pull physical units (which are basically just multipliers) in and out of the argument, so you can use an $\mathbb{R}\to\mathbb{R}$ function as well as a function mapping times to times (or, at least, time-differences to time-differences). With a homogeneous function, you can do the same, but may pick up some power on the unit in the process.

• From my understanding, any homogeneous function (and not just linear ones) can be used with dimensional units. For example, we can use the function CircleArea(r) with any unit and have it return that unit squared. – yoniLavi Aug 5 '16 at 0:46
• Quite right you are! – leftaroundabout Aug 5 '16 at 0:59

In Physics you do not calculate with numbers you calculate with physical quantities which have a magnitude and a dimension/unit. So in order for an equation to make any sense it must be dimensionaly correct: you can only add and substract quantities with the same dimensions and function arguments must be dimensionless: Log(x), Sin(x), Exp(x)... and so on only make sense if x is a dimensionless quantity. @Rob gave an explanation for that. It makes no sense to add for example one meter and one second, that is just not defined.

To your example: in order to be dimensionaly correct your function argument needs to be dimensionless. Assuming $t$ has the dimension of a time it needs to be multiplied with something that has the dimension of an inverse time. This would be a frequency to be specific actually an angular frequency often denoted by $\omega$. $k$ often is used as a symbol for a wave vektor which has the dimension of an inverse length.

A typical one-dimensional traveling wave looks like:

$\Psi(x,t)=A*\sin(\vec{k}\cdot\vec{x}-\omega t + \phi)$.

It has a wave vector $\vec{k}$ with an norm $|\vec{k}|=k=2\pi/\lambda$, an angular frequency of $\omega=2\pi/T$, an amplitude $A$ and a phase $\phi$, where $\lambda$ is the wave length and $T$ is the period. The amplitude can have an arbitray dimension which determines the dimension of the wave.

Your instructor is correct. The argument of sine (ie $kt$) should be dimensionless - eg an angle in degrees or a fraction of $2\pi$.

However if $k$ is dimensionless and $t$ is time then $kt$ has units of time and is not dimensionless. So this combination of units cannot be correct.

protected by Qmechanic♦Aug 4 '16 at 19:38

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