# Intuitive Explanation of the graph $y = \sin x$ [duplicate]

Possible Duplicate:
Intuition for graphing Sine/Cosine

We've all seen the graph of $y = \sin x$
(I can't post an image because of reputation so I posted a link to a graph)

Sine Graph Picture

As off now, the only definitions of Sine I understand is the "ratio in a triangle" definition, and the "unit circle" definition. So I hope it's possible to answer my question using one of these definitions.

From these definition it's easy to see why at ($0.5\pi$) radians the value of $y$ is $1$, it's easy to see why at Pi radians that value of $y$ is $0$, and I can certainly see why the function repeats itself every $2 \pi$ radians, using the unit circle definition.

What I don't understand is why it has this exact shape, why does it look like it does between 0 and 0.5 Pi, why does it have this exact concavity?

1. Does this have an intuitive explanation?
2. How did the first mathematicians draw this function, did they actually measure the Sine of all the angles with a ruler and than drew the graph?

You ask why the graph of $\sin$ looks as it is on $[0,\pi/2]$. You can easily(with the help of some trig formulas) calculate the values of $\sin x$ for $x=0,\pi/6,\pi/3,\pi/4,\pi/2, \pi/12,\pi/5$. Plot these values.

Keep in mind that $\sin$ is continuous as a ratio of two continuous functions (opposite side/hypothenuse). It is increasing, since if the angle increases then the sine increases. Try now and plot $\sin$ knowing just the above mentioned facts. You'll see that this approximation of a graph seems to be the approximation of a concave function. Of course, one can prove concavity using trigonometrical formula. You asked for something more intuitive.

[May I ask why did you ask this question? I'm just curious. Don't take this the wrong way, but asking why is $\sin$ concave between $0$ and $\pi/2$ is just like asking why the letter $A$ looks like it looks. It is one of the elementary blocks of mathematics. There is no correct answer as to why the graph looks like that. It looks like that because of the properties of the $\sin$ function.]

• I'll try to explain, if I look at the graph y = x^2, I can write y = x*x, see the coefficient of x is in fact x, and say that the coefficient itself grows so of course the slope grows, this way I can intuitively understand many functions, but the sine function doesn't really fall in this category. – fiftyeight Dec 17 '11 at 23:10
• You'll be interested to find that you can do something like that for the sine, @fiftyeight; but you'll have to wait a bit for infinite series... – J. M. isn't a mathematician Dec 18 '11 at 1:52
• I don't wanna wait :) Can you tell me what you mean? – fiftyeight Dec 18 '11 at 18:54

Think of a point traveling on the unit circle, starting at $(1,0)$ and moving counterclockwise. The $y$ coordinate of the point is $\sin \theta$, where $\theta$ is the angle made by the point, the origin, and the point $(1,0)$ (and, of course, measured positive in the counterclockwise direction with the positive $x$-axis corresponding to $\theta=0$).

Now as the point moves from $(1,0)$ to $(0,1)$, what happens to the $y$ coordinate of the point? Well, it increases from 0 to 1. So, the graph of $\sin\theta$ (with $\theta$ as the horizontal axis) increases over $[0,\pi/2]$ from 0 to 1.

If you imagine the point traversing the entire unit circle, you should be able to convince your self that at least in terms of "increasing or decreasing" and in terms of where the zeroes and max/mins are, the graph of $\sin$ is as it is.

Edit: For the concavity, see my comment below.

• The shape of the concavity is actually what I don't understand, I rephrased my question, sorry if I wasn't clear enough – fiftyeight Dec 17 '11 at 21:58
• @fiftyeight I can almost convince myself that the rate of increase of the $y$ value of the point is decreasing from $(1,0)$ to $(0,1)$ by just considering its travel on the unit circle. This would give the graph its shape. Of course, just plotting values will reveal the shape (in a non-rigorous way). – David Mitra Dec 17 '11 at 22:12
• @fiftyeight Although you want an intuitive explanation I will say that the concavity may be explained in Calculus by the sign of the second derivative of $\sin x$, which is $-\sin x$. And $-\sin x<0$, for $0<x<\pi$. – Américo Tavares Dec 17 '11 at 22:21
• ... in the interval $[0,2\pi]$. – Américo Tavares Dec 17 '11 at 22:28

This section has the beginnings of an explanation, but it's not detailed. But it should nonetheless convey the idea that the methods were quite different from measuring with a ruler.

For some values, such as $30^\circ$ and $60^\circ$, it's easy to see by using elementary geometry why the values of the sine must be exactly what they are ($1/2$ and $\sqrt{3}/2$. For others, you have to get subtler.

1.

I have learned the sin function as a swinging object. You would see $y$ as the amount of distance from the center, and $x$ as the passed time. Notice, the swinging object stays on the side for the longest time ($x = k*Ï€$, $y = 0$), and moves quickly through the center($x = (k+1/2)*Ï€$, $y = 1$).

This also explains the derivative of $sin x$, which is $cos x$. The derivative of the position is the speed, which shows in the graph of $cos x$. The object passes through the center with the highest possible speed, and stops for a very short time at it furthest point from the center.

You should try it yourself, just swing anything (like a clock on a small chain), and imagine the $sin x$ function.

2.

(Purely speculation, I have no source)

Personally I would guess that early mathematicians would pull a 'paper' under a swinging object that leaves behind a track. You will end up with a sine function, if you pull the paper with a regular speed. Maybe they didnt have any paper or pen, but this is the most natural way to draw such a graph.

Function $\sin$ and $\cos$ are result of function that does exponential projection of line (number line $\mathbb{R}$) to the circle with radius $r=1$. Here's the definition of function:

$E: \mathbb{R} \mapsto k$

for $t \in [0, 2\pi\rangle, E(t)=T$ so that $|IT|$(this is arch)=t and that arch $IT$ is positively oriented (counter clockwise)

for $t \in \mathbb{R}\backslash [0, 2\pi\rangle, E(t)=E(t_0)$ where $t=t_0 +2k\pi,k\in \mathbb{Z},t_{0}=[0, 2\pi\rangle$

Point I is located on $(1,0)$. From this definition you can now clearly see that what you are basically doing is, informally speaking, "drawing" $\mathbb{R}$ to the circle, and circle can only have values in rage of $t[0, 2\pi\rangle$. Here's something interesting happening. Because you are representing numbers on circle it is easier for you to determinate where's $\pi$ (transcedent number) located than number $1$ e.g. . Therefor, because it is impossible to construct number $\pi$ it is impossible to construct the graph of functions $sin$ and $cos$, there are only aproximations. So the mathematician never draw this function they only aproximated it. It has shape of sinusoid because of projection done by function $E$.

EDIT:

Here are definitons of $\sin$ and $\cos$:

$E(t)(x_{E(t)},y_{E(t)})$

$\cos t:=x_{E(t)}$

$\sin t:=y_{E(t)}$

$\sin t\in[-1,1]$

$\cos t\in[-1,1]$