A defining feature of radians as a unit of measurement is that if an angle $\theta$ is expressed in radians, the height of a point on the unit circle at this angle is $\sin(\theta)$.

Is it possible to choose a different unit of measurement for the angle so that we get something other than a sinusoid?

Why doesn't the height of a point on the unit circle as a function of the angle give us a well-shaped half-circle on each half-period rather than a sinusoid? Wouldn't such a circle shape be more fundamental than a sinusoid?

If not, why must the coordinates of a point on the unit circle be sinusoidal as a function of the angle?

I may have a very basic error in my thought, maybe sines were not taught properly to me, but when I learned them they were something that suddenly came out of nowhere. I know it's like the movement of a wriggling snake or the time diagram of a spring that is bouncing, but I still have this question in my head. I would be thankful if I hear your input on it.

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    $\begingroup$ Only with the radian as unit is true that $sin' = cos$... $\endgroup$ Jun 15, 2019 at 8:43
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    $\begingroup$ Changing the unit of angles used only stretches the sine curve parallel to the $x$-axis. You cannot choose a unit such that the stretched sine wave becomes a periodic semi-circular wave. $\endgroup$ Jun 15, 2019 at 8:44
  • $\begingroup$ You always get a sinusoidal shape when you measure the height of a point in a circle with respect to an angle -- it doesn't matter how quickly you go around the circle (and any different choice of unit for "angle" would basically just be scaling radians up or down to represent going around the circle "faster" or "slower"), you still get a sinusoidal one, just with a tighter or flatter periodic shape. Consider the height of a point in the circle relative to an angle in degree (rather than radians), for example! It's still sinusoidal, but "slowed down" by a ratio of $\frac{\pi}{180}$. $\endgroup$ Jun 15, 2019 at 8:44
  • $\begingroup$ With any other unit of angle, you will always have a sine-like shape for the sine function. Note that the unit of angle is about angles, so we want $\angle AOB+\angle BOC=\angle AOC$ to hold for example (and of course equalaity of congruent angles) $\endgroup$ Jun 15, 2019 at 8:45
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    $\begingroup$ these are interesting points..., can you also provide a proof (or an intuitional explanation) that the shape of it will always be sinosuidal? $\endgroup$
    – aderchox
    Jun 15, 2019 at 8:48

2 Answers 2


Suppose $\theta$ is an angle in radians. Let's create a new unit for angles called the $\mathrm{Moytaba}$, or $\mathrm{Moy}$ for short. Let us define it by $1 \,\,\mathrm{Moy} = c\,\,\mathrm{rad}$ where $c$ is some constant scaling factor (of course, any transformation between units of the same dimension has to be a constant scaling factor, for obvious reasons).

Then we know that $\theta \,\,\mathrm{rad}$ is just $\frac{\theta}{c} \,\,\mathrm{Moy}$. Since the height of a point on a circle wrt the angle in radians gives us a height of $\sin(\theta)$, then the height of a point on the circle in $\mathrm{Moy}$s must be $\sin(\frac{\theta}{c})$. Clearly this is still sinusoidal! So this is your proof that any other units are still sinusoidal (and just represent scaling the period of the sine wave).

But for the real intuitions, as I noted in my comments above:

I think this animated GIF (which handles the angle in a purely unit-less fashion) will give you the intuition you're after: sine and cosine with respect to angle http://en.wikipedia.org/wiki/File:Circle_cos_sin.gif


In fact, the reason why sine describes the time diagram of a bouncing spring is because the Kinetic E + Potential E = Total Energy (constant) equation actually parametrises a circle, where we end up getting momentum/velocity along one axis and position along the other. As time passes, we just move around this circle, with energy flowing back and forth between kinetic and potential. When you graph the position with respect to time, you're just getting the height of the point on the circle, just like we are here! Oscillation of mass in phase space http://en.wikipedia.org/wiki/File:Simple_Harmonic_Motion_Orbit.gif

Edit: After talking a little more with the OP, I couldn’t resist citing one last, brilliant animation: https://www.deviantart.com/woodmath/art/Euler-s-formula-3d-visualization-268936785

  • $\begingroup$ Nice answer! But any specific reason for choosing Moytaba? $\endgroup$ Jun 15, 2019 at 9:28
  • $\begingroup$ @MohammadZuhairKhan That is the name of the person who asked the question, so I figured if anybody deserved to have the hypothetical unit named after them, it would be them. ;) $\endgroup$ Jun 15, 2019 at 9:30
  • $\begingroup$ Oh, I didn't notice that. I thought it was a famous mathematician who worked on this subject. Though OP certainly did work on this subject... $\endgroup$ Jun 15, 2019 at 9:31
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    $\begingroup$ (Sorry for reacting with delay, something urgent came up.) I laughed so much when I saw you used Moy, because at the same time I was trying to find a proof and I had used a same name. :)) thank you. $\endgroup$
    – aderchox
    Jun 15, 2019 at 9:54

NOTICE: This answer is NOT CORRECT. It's only not deleted because it might be a useful mistake to inform unaware people that sinusoids and cycloids are not the same things. what the answer below describes is indeed a cycloid and NOT a sinusoid.

We can rotate the circle itself too! Say it's a wheel and there's a marker on some point of its circumference and this wheel is rotating near a wall(so that the marker will draw the path taken by the point for us), now the change in the angle between the horizontal radius of the circle and that point is what we simulate by rotation of the wheel!.

I've drawn different states of the wheel in the picture below, we can see that even if the circle moves an epsilon(assuming epsilon is an infinitesimally small number) forward, the new location of the point can not be on the same previous circle obviously, and that means it's impossible to produce a movement path that consists of half circles for a fixed point on a rotating circle. Instead, they will be in a shape which we call "sinusoidal"(they are indeed cycloids).

Moving Circle With Fixed Point On It

  • $\begingroup$ I'm not quite sure this is right -- if I'm understanding your example correctly, this isn't actually sinusoidal, it's a cycloid! See the intro section of this video: youtube.com/watch?v=S8EWKm2FwBo (and if you're interested in some of the deeper properties of the cycloid, check out youtube.com/watch?v=Cld0p3a43fU) $\endgroup$ Jun 15, 2019 at 14:05
  • $\begingroup$ Oh that's right, this is very interesting! I never knew cycloids exist... I will search about them. BTW, should I delete this answer or edit and explain it is wrong? $\endgroup$
    – aderchox
    Jun 15, 2019 at 14:39
  • $\begingroup$ Either of those options sound fine. Just make it clear so people don’t get confused ;) Maybe you need to build yourself a toy model to play with in Mathematica or Desmos or something if you’re still not quite feeling the sinusoid. But yeah, as far as I’ve learned it, I have always considered the fact that sine gives us the height of a point on a circle given an angle to be the defining feature of sine and everything else about it to be a conclusion of that. It really goes deep into the nature of the circle. That’s why we end up using it in trigonometry so much! $\endgroup$ Jun 15, 2019 at 14:44
  • $\begingroup$ I’m sure you’ve seen that $x^2 + y^2=1$ is the equation defining a circle; consider that $\sin^2(\theta)+\cos^2(\theta) = 1$ is basically the first law you learn about sine and cosine! They really serve the same purpose here: $\sin$ and $\cos$ get the coordinates of a point on the circle expressed in two orthogonal basis directions in the same way that $x$ and $y$ do! $\endgroup$ Jun 15, 2019 at 14:49
  • $\begingroup$ I just thought of the perfect thing! Here, I think the illustrations on this page might demonstrate how sinusoids emerge from the height of a point on the circle even better: en.m.wikipedia.org/wiki/Circular_polarization $\endgroup$ Jun 15, 2019 at 15:03

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