When proving them the "modern" way (from first principles) it seems impossible to get around proving the identities $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$ and the related $\cos$ limit. This itself requires some geometric reasoning and uses the definition of radians that we have today.

But surely people knew that $(\sin \theta)'=\cos \theta$ and so forth long before limits came about. Is this just obvious? How would they have justified it?

  • $\begingroup$ Think about a triangle like $\angle$ and what $\sin$ actually is, the shape should be pretty obvious. The rate of change is easily visualised. What you're talking about is more real-analysis and there's a bit of background before that really. However you can use "small angle approximations" and note that when $x$ is small $\sin(x)$ is basically $x$, then you get $\frac{x}{x}=1$ - this is A-level style, not formal. As I said for formal you need real analysis and mathematical experience. $\endgroup$
    – Alec Teal
    Dec 5, 2015 at 16:25
  • $\begingroup$ But how would they have justified the small angle approximation? Did they know and use it? Did they even use radians? I know how to formally prove it, that's not what I'm asking. I'm asking how it was originally discovered. $\endgroup$
    – user3339
    Dec 5, 2015 at 16:29
  • $\begingroup$ By thinking about a picture of a triangle in their heads. $\endgroup$
    – Alec Teal
    Dec 5, 2015 at 16:32
  • $\begingroup$ I'm sort of looking for historical references :) $\endgroup$
    – user3339
    Dec 5, 2015 at 16:35
  • 2
    $\begingroup$ If instead of radians we use some other units equal to $k$ radians, we just get $(\sin kx)' = k(\cos kx)'$, i.e. the instantaneous increase in the sine at a particular angle is a constant multiple of the cosine at that angle. And this specific fact was known to Manjula (around 932) and Aryabhata II (around 950) and explicitly given with geometric reasoning by Bhaskara II (around 1150) even before the general calculus or power series of Madhava (around 1500). See Indian Journal of the History of Science 19 (2): 95–104 (1984), "Use of Calculus in Hindu Mathematics" by Datta/Singh/Shukla. $\endgroup$ Feb 21, 2016 at 18:22

4 Answers 4


Did you try looking in any books on the history of calculus? The following is taken from "The Historical Development of the Calculus" by C. H. Edwards (p. 205 ff).

The inverse sine function (for radian angles in the first quadrant) can be related to an area under an arc of the unit circle, which is $y = \sqrt{1-x^2}$ in the first quadrant. Newton knew from his discovery of the binomial expansion for rational exponents how to write $\sqrt{1-x^2}$ as a power series, which he could then integrate termwise. In this way he found the power series for $\arcsin x$. Then he inverted that series until he recognized the pattern to "establish" the power series of $\sin x$, from which he could find the power series for $\cos x$ as the series for $\sqrt{1-\sin^2x}$ with constant term $1$. From the power series expansions of $\sin x$ and $\cos x$ it is clear that $\sin'x = \cos x$.

The power series for $\sin x$ and $\cos x$ were known in India long before they were found by Newton. See https://en.m.wikipedia.org/wiki/Madhava_series.

While the power series formulas reveal the derivatives of the sine and cosine to us, it is not the case that Newton or Leibniz in the 1600s took this step. The trigonometric functions were not systematically thought of as functions to be graphed and have a derivative. Euler, in the 1700s, was the one who first incorporated the trigonometric functions into calculus. See "The calculus of the trigonometric functions" by Victor Katz, which is available online at http://ac.els-cdn.com/0315086087900644/1-s2.0-0315086087900644-main.pdf?_tid=a15b6d9e-9c37-11e5-b17b-00000aacb35e&acdnat=1449420025_6f4ed309a5aed67668cf37b14c276280.


Knowledge of the specific fact that $(\sin x)' = \cos x$ actually predates the general knowledge of calculus and derivatives. It was known in the following form: that for very small $\Delta x$, when you increase $x$ to $x + \Delta x$, the increase in value of the sine, from $\sin x$ to $\sin (x + \Delta x)$, is proportional to $\Delta x$ times $\cos x$. In other words, that $$\frac{\sin (x + \Delta x) - \sin x}{\Delta x} \approx \cos x$$ The approximation being exact in the limit as $\Delta x \to 0$ is of course the modern definition of the derivative.

This happened historically in Indian mathematics, where Muñjala (around 932), Āryabhaṭa II (around 950), Prashastidhara (around 958) all give the above rule for calculating $\sin(x + \Delta x)$, and an explicit geometric reasoning / justification is given by Bhāskara II (around 1150) in his Siddhanta Shiromani. I have not found a perfectly good reference to these, but you can start with the following article:

  • Use of Calculus in Hindu Mathematics, by Bibhutibhusan Datta and Awadhesh Narayan Singh, revised by Kripa Shankar Shukla, Indian Journal of the History of Science, 19 (2): 95–104 (1984). (PDF)

It was first pointed out by Bapu Deva Shastri in Bhaskara's knowledge of the Differential Calculus, Journal of the Asiatic Society of Bengal, Volume 27, 1858, pp. 213–6.

  • 1
    $\begingroup$ I remember having seen a paper from Pascal explaining the concept of "triangle inassignable" (in english, approximately, "vanishing triangle"), which was more or less a rectangular triangle with sides dx, dy (I use notations that will appear some 60 years later than his writing). See on the web the paper (in French):La «géométrie calculante» de Pascal, dans le traité des sinus du quart de cercle et dans le traité des trilignes rectangles (by Merker, Université de Franche-Comté). $\endgroup$ Feb 24, 2016 at 12:58
  • $\begingroup$ @JeanMarieBecker Thank you, that's intereresting (epiphymaths.univ-fcomte.fr/publications/…). It will take me a bit of effort to go through the French, but it is not surprising that this sort of geometric reasoning for trigonometry predates (even by centuries) a general calculus of arbitrary functions. $\endgroup$ Feb 24, 2016 at 17:02
  • $\begingroup$ The URL for the lecture by Claude Merker has changed: now at epiphymaths.univ-fcomte.fr/seminaire/publications/… $\endgroup$
    – terry-s
    Nov 25, 2018 at 10:10

Formally, $(\sin x)'=\cos x$ is true only if you measure x in radians. So you need to use radians for this fact anyway. If you will to ignore this, you can say that Euclid knew that the limit of $\frac{\sin x}{x}$ is 1, at least when $x=2\pi/n$ and $n\rightarrow \infty$. Of course Euclid did not use radians, sin and lim, but he proved that the difference between the area of a circle and the area of an inscribed right polygon will become arbitrarily small as number of sides becomes large (by method of exhaustion). The area of right n-gon inscribed in a circle of radius 1 is $\frac 12 n\sin (2\pi/n)$. So using our notation, we can say that Euclid proved that $$\lim_{ n\rightarrow \infty} \frac 12 n\sin (2\pi/n) = \pi.$$ This is equivalent to $$\lim_{n\rightarrow \infty}\frac{ \sin (2\pi/n)}{2\pi/n} = 1.$$ Btw, that was used by Archimed to estimate $\pi$: he approximated $\pi$ by area of 96-gon. That is, he calculated $\sin(2\pi/96)$ and used that $2\pi/96 >\sin(2\pi/96)$. And to prove the upper bound he used $x< \tan x$.


I guess the initial discovery and subsequent justification may have been done just by trying some calculations, like below. Nowadays we can use Excel for this kind of investigation.

enter image description here

  • 1
    $\begingroup$ No, it is other way around. To do this calculation you need to know the value of $\pi$. But the value of $\pi$ was calculated using the formula $\sin(\pi/n)\approx \pi/n$ for some large $n$. See my answer. $\endgroup$ Jul 9, 2020 at 17:52
  • $\begingroup$ Alexei, so your point is that in order to calculate Sin(x), we need to use the value of pi. But Sin(x) has been used historically to calculate pi thus making the definitions of Sin(x) and pi circular. Is that right? $\endgroup$
    – ramana_k
    Jul 9, 2020 at 23:04
  • $\begingroup$ No, you don't need to calculate pi to calculate six x. For example, you can calculate sin(30°) or sin(15°) without knowing what pi is. But if you want translate 15° to radians you need pi: 15°=pi/12. And you need to use radians to estimate that sin(x)/x ≈1. $\endgroup$ Jul 9, 2020 at 23:48

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