So I was playing with some sine and cosine functions, after my exams to see from where do they originate and I got introduced to the unit circle and so, but I noticed something weird; for a given sector in the unit circle the sum between the are of the triangle whose base is the cosine of the angle and height is the sine of the angle and the quarter sector whose two radii are the sine of the angle and (1 - cosine the angle) did not equal to the area of the sector of the circle.


I am only concerned about the sector of the circle formed by the angle theta.

Areas of the triangle and quarter ellipse

By using simple area rules, triangle area is $\frac12\sin(\theta)cos(\theta)$ And the area of the quarter ellipse is $\frac\pi4\sin(\theta)(1-\cos(\theta))$

Area of sector

Assuming that theta is in radians the area equals $\frac12r^2\theta$ as the radius is 1 so this can be simplified to $\frac12\theta$

Well, I am pretty sure I missed something up here because all of the next paragraphs talks about the weird result of this, yet I could not identify whether I made a mistake or there is something in core maths that restricts doing this.

As mentioned earlier I expect that summation of areas of the triangle and the quarter ellipse equals the area of the sector circle. $$ \frac12\theta = \frac12\sin(\theta)cos(\theta) + \frac\pi4\sin(\theta)(1-\cos(\theta)) $$ But that didn't work, I tried with some values of theta and only $\frac\pi2$ and $0$ worked, any other value gave either a near value or a very wrong value. I thought well, what about making an equation in one variable the sine function of the above equation! (I mean making sine theta the variable) $$ f\left(x\right)=x\sqrt{1-x^{2}}+\frac{\pi}{2}x\left(1-\sqrt{1-x^{2}}\right) $$ I just simply applied the famous identity $\sin(x)^2+cos(x)^2=1$ to have a function that supposedly equals $arcsin()$ function, if my suggestion of the whole question was correct.

Arcsine vs theory

And it actually kinda looks like it, I by no means understand what's going on, and to my surprise, I discovered that actually sine, cosine, their inverses and so are calculated using very complex algorithms using what's called Taylor's series or complex plane rotating and stuff like that. Not an easy equation to do. Desmos.com Calculator Graph for above image

So what's I am missing up here? and why the graph looks nearly identical to an $arcsin()$ function?

  • $\begingroup$ How did you come up with the idea that the area of the sector is a triangle plus a quarter ellipse? $\endgroup$ Commented May 19, 2021 at 19:43
  • 2
    $\begingroup$ That's not quarter of an ellipse. If you you connect two circular segment with the same base and even if cut out from the same circle, it does not make an ellipse. For starter, it is not smooth at two opposite ends. $\endgroup$
    – Math Lover
    Commented May 19, 2021 at 19:44
  • $\begingroup$ please note that its equation is still $x^2+y^2 = 1$ and equation of an ellipse with center on x-axis would be $\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2} = 1$ where $a \ne b$. $\endgroup$
    – Math Lover
    Commented May 19, 2021 at 19:49

1 Answer 1


As indicated by the comments, the red area in your figure is (almost-never) a quarter-ellipse. A quarter-ellipse (almost-always) bulges a bit beyond the circle, due to the horizontal tangent at the top:

enter image description here

That your graph approximates arcsine can be attributed to the fact that the discrepancy with the quarter-ellipse is relatively small compared to the area of the sector.


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