I was using the textbook A History in Mathematics by Victor J. Katz. I saw a theorem from Nasir al-Din al-Tusi. The way the theorem is written in the book is like this:

In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB:AC=sin(angle ACB):sin (angle ABC). [Note that since we are considering a ratio it is irrelevant whether we use Sines or sines.)

This theorem is about the law of sines. My question is about the last sentence in parenthesis.

What is the difference between Sine and sine?

  • $\begingroup$ Check the text, but some use Sine for usual sine restricted to $[-\pi,\pi]$. $\endgroup$ – coffeemath Dec 20 '20 at 16:43
  • $\begingroup$ mathwords.com/i/inverse_trigonometry.htm $\endgroup$ – saulspatz Dec 20 '20 at 16:45
  • 1
    $\begingroup$ @coffeemath I think the interval is meant to be $[-\pi/2,\pi/2]$. $\endgroup$ – Joe Dec 20 '20 at 16:46
  • 1
    $\begingroup$ @Joe is right. I went too quickly. $\endgroup$ – coffeemath Dec 20 '20 at 16:47
  • 1
    $\begingroup$ @saulspatz but such a distinction would not match the quoted text which is about ratios. I.e., it would match better if we had Sine = 1000 times sin or the like. $\endgroup$ – Hagen von Eitzen Dec 20 '20 at 17:07

Six pages earlier, at the beginning of 9.6.1, the first subsection on Islamic trigonometry, a parenthetical remark notes:

The Islamic sine of an arc, like that of the Hindus, was the length of a particular line in a circle of given radius $R$. We will keep to our notation of "Sine" to designate the Islamic sine function

Additionally, in 8.7.1 on Indian trigonometry, the book explains:

In what follows, we generally use the word "Sine" (with a capital S) to represent the length of the Indian half-chord, given that the half-chord is a line in a circle of radius $R$, where $R$ will always be stated. We reserve the word "sine" (with a small s) for the modern function (or, equivalently, when the radius of the circle is 1). Thus, $\mathrm{Sin}\,\theta=R\sin\theta$.

I have not seen this convention outside of this book. As noted in the comments, in modern mathematics, some might use a capital letter to denote a restriction of the domain of the sine function.


Edit: it appears that the notation used in the book does not correspond to how I have seen $\sin$ and $\mathrm{Sin}$ being used elsewhere. However, I will keep this answer for the benefit of other readers.

$\sin$ is a function that (generally speaking) maps real numbers to real numbers. Here is how it is defined on Wolfram Mathworld:

Let $\theta$ be an angle measured counterclockwise from the $x$-axis along an arc of the unit circle. Then $\sin(\theta)$ is the vertical coordinate of the arc endpoint.

The definition of sine

There are other equivalent definitions of sine, but however you choose to define it, it is not a one-to-one function:

Graph of sine

Notice that $\sin(0)$ equals $0$, but so does $\sin(\pi)$, $\sin(2\pi)$, etc. (I assume you are familiar with radians.) This means that the $\sin$ function does not have an inverse. To get around this, we often restrict the domain of $\sin(\theta)$ by requiring that $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$:

Graph of Sine

Notice that along this restricted domain, there are no two values of $\theta$ that produce the same output. This means that the inverse exists. However, because a function is defined not just by its values but by its domain, when we restrict the domain of $\sin$ it becomes a new function that we often denote $\DeclareMathOperator{\Sin}{Sin} \Sin$. The inverse of $\Sin$ is called $\Sin^{-1}$. You will often see $\Sin^{-1}$ being written simply as $\sin^{-1}$, but this is really an abuse of notation. The function $\sin$ does not have an inverse—only $\Sin$ does. A more popular alternative to $\Sin^{-1}$ is $\arcsin$. This notation has the advantage of being more 'correct' than $\sin^{-1}$, while also avoiding the $\sin$/$\Sin$ confusion.

  • 1
    $\begingroup$ Correction: $\sin\frac{\pi}{2} \ne 0$ $\endgroup$ – Tavish Dec 20 '20 at 19:19
  • $\begingroup$ @Tavish Thank you, I'll edit the post. $\endgroup$ – Joe Dec 20 '20 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.