# what is sine of a real number

I never understand what the trigonometric function sine is..

We had a table that has values of sine for different angles, we by hearted it and applied to some problems and there ends the matter. Till then, sine function is related to triangles, angles.

Then comes the graph. We have been told that the figure below is the graph of the function sine. This function takes angles and gives numbers between $-1$ and $1$ and we have been told that it is a continuous function as it is clear from the graph.

Then comes taylor expansion of sine and we have $$\sin (x)=x-\frac{x^3}{3!}+\cdots$$

I know that for any infinitely differentiable function, we have taylor expansion. But how do we define differentiability of the function sine?

We define differentiability of a function from real numbers to real numbers..

But sine is a function that takes angles and gives real numbers..

Then how do we define differentiability of such a function? are we saying the real number $1$ is the degree 1?

I am confused.. Help me..

Above content is a copy paste of a mail i received from my friend, a 1 st year undergraduate. I could answer some things vaguely i am not happy with my own answers. So, I am posting it here.. Help us (me and my friend) to understand sine function in a better way.

• $\sin(1)$ is not the sine of $1$ degree, it's the sine of the angle of $1$ radian. – 5xum Jul 14 '16 at 5:39
• Spoiler alert: $\sin$ can even take complex number... – 3x89g2 Jul 14 '16 at 5:46
• It is viable to take $\sin(z)=\sum_{n\geq 0}\frac{(-1)^n z^{2n+1}}{(2n+1)!}$ as the definition of the sine function and simply ignore geometry. With such a definition, its regularity is trivial, but its periodicity or the sum formulas are a bit less. – Jack D'Aurizio Jul 14 '16 at 5:50
• I simply said it is viable, but I am not actually suggesting it. Another possible definition of the sine function is The solution of the Cauchy problem $y''+y=0$ with the constraints $y(0)=0$ and $y'(0)=1$, for instance. Thousand ways to skin a cat. – Jack D'Aurizio Jul 14 '16 at 6:01
• You don't define the differentiability of the sine function. You prove it from a definition. If you define the sine function using a Taylor series, it's easy. If you define it somehow else, you need to do the necessary work and show that the function you defined satisfies the needed properties to have the wanted Taylor series. – Asaf Karagila Jul 14 '16 at 8:37

The sine function doesn't actually operate on angles, it's a function from the real numbers to the interval [-1, 1] (or from the complex numbers to the complex numbers).

However, it just so happens that it's a very useful function when the input you give it relates to angles. In particular, if you express an angle as a number in radians (in other words, on a scale where an angle of $2\pi$ corresponds to a full circle), it gives you a value that relates to the ratio of two sides of a right-angled triangle that has that angle in one corner.

If that explanation doesn't satisfy you, then you can look at it another way - if you take it that the sine function does take an angle as input and outputs a number, then the differentiability of it relates to how its output changes as you change the angle slightly. If you go far enough in calculus, you'll learn about functions whose inputs and outputs are bizarre multi-dimensional concepts, and as long as the space of bizarre multi-dimensional concepts has the right properties, you can calculate derivatives in a meaningful sense, and if you can get your head around that then differentiating a function of an angle is small fry.

• It seems to take some time for me to understand this.. I will ask if i have any questions after understanding this – user311526 Jul 14 '16 at 6:03
• The one who asked me the question is happy with this answer.. I am happy with some other answers including this though.. Thanks.. – user311526 Jul 19 '16 at 13:51

Imagine the unit circle in the usual Cartesian plane: the set of pairs $(x, y)$ where $x$ and $y$ are real numbers. The unit circle is the set of all such pairs a distance of exactly $1$ from the origin.

Imagine a point moving around the circle. As it travels around the circle, it makes an angle of $t$ radians (not degrees!) with the positive $x$-axis. From now on we call the $x$ coordinate the cosine of $t$; and the $y$ coordinate the sine of $t$.

It's as simple as that. If you only remember this one fact you can figure everything else out: the definitions of the trig functions in terms of triangles, the shape of the graphs of the functions, and everything else.

To repeat: $\cos(t)$ and $\sin(t)$ are the $x$ and $y$ coordinates, respectively, of a point on the unit circle that makes an angle of $t$ radians with the origin and the positive $x$-axis.

There's a picture here ... https://en.wikipedia.org/wiki/Unit_circle

• This seems to be okay. I will ask if i have any further questions.. – user311526 Jul 14 '16 at 5:59
• You might want to mention that the point moves at unit speed. – filipos Jul 14 '16 at 13:10
• @filipos Speed of point makes no difference at all to the definition I gave. Only the x-y coords and the angle matter. – user4894 Jul 14 '16 at 15:48
• Alternatively, mention that $t$ is the distance travelled. – filipos Jul 14 '16 at 16:49
• @filipos Good point, especially to motivate concept of radians. – user4894 Jul 15 '16 at 0:11

How do we get derivative of $\sin(x)$?

1. Define $\sin(x)$ to be the series $A(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$
2. Define $\cos(x)$ to be the series $B(x) = 1-\frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$
3. Some analysis going on. First, you have to convince yourself that these two series do converge. Second, you have to convince yourself that you can differentiate these two series. This part is actually hard.
4. Finally, you just differentiate it, and you will see that $\frac{d}{dx}\sin(x) = \cos(x)$.

Notice that there is no cyclic reasoning here. The only trouble is that you have to convince yourself that defining $\sin$ and $\cos$ in this way actually makes sense.

In case you don't like this proof, there is another more elementary proof which can also be found on ProofWiki. It only uses some facts about limits.

• This seems to be okay. I will ask if i have any further questions.. – user311526 Jul 14 '16 at 6:04
• You need to get from that definition to the fact that $\sin$ and $\cos$ describes the lengths of the legs of a hypotenuse-$1$ right triangle, though. That requires some trickery. – Arthur Jul 14 '16 at 6:09
• @Arthur That is true and I admit that I'm not sure how to do that yet... – 3x89g2 Jul 14 '16 at 6:10
• @Misakov I think this is how I would do it: Define $s(x)$ and $c(x)$ as the legs of the triangle, derive the relations for $s(x + y)$ and $c(x + y)$, show that $\lim_{x \to 0}\frac{s(x)}{x} = 1$ and $\lim_{x \to 0}\frac{c(x) - 1}{x} = 0$, and finally that $s'(x) = c(x)$ and $c'(x) = -s(x)$. Now you can show that $s(x)$ has the same Taylor series as $\sin(x)$ and $c(x)$ coincides with $\cos(x)$. – Arthur Jul 14 '16 at 6:13
• @Arthur: My answer here describes a geometric interpretation of the power series. – Blue Jul 14 '16 at 9:35

An excellent question!

A nice thing about maths is that there are often plenty of different ways you can define something, but they turn out to be equivalent. It's really useful in practice to define the trig functions as a Taylor series. This way, it's completely obvious that the functions are differentiable and you can easily compute the derivatives.

Yet I'm not really a fan of this definition. Polynomials are nice as a local discription of all kinds of functions (i.e. for small angles), but for the big picture this is oblique at best, erroneous at worst. In fact, what you said

I know that for any infinitely differentiable function, we have taylor expansion.

is not correct: for any infinitely diff'able function, you can define some Taylor series, but only for analytic functions will this series actually describe the original function. There are many important functions that are infinitely differentiable yet not analytic, e.g. the test functions. But I digress.

An intuitive way to derive the trig functions is to actually look at physical oscillation motions, e.g. plot the height $h(t)$ of a point on a rotation wheel against time. If you want triangles, you can draw them in the wheel:

This is a continuous motion (the wheel must go through infinitesimally small angles in the course of its motion), and you have the derivatives “built in”. It's quite easy to find out that $$\frac{\mathrm{d}^2}{\mathrm{d}t^2} h(t) \propto - h(t)$$ must be fulfilled. ($\propto$ means proportional; there's actually a factor corresponding to the angle velocity.) Well, now you can look at the special case that the proportion is $1$, and use $$\frac{\mathrm{d}^2}{\mathrm{d}t^2} (\sin t) = - \sin t$$ as the defining equation of the sine function!

But wait a moment, is this actually well-defined? You need to prove it. It's not quite so easy as if you start from the Taylor series, but you can do it: the Picard-Lindelöf theorem tells you that, with suitable initial condition, such an ordinary differential equation gives a unique solution. It even tells you a power series expansion, which of course turns out to be equal to the Taylor series definition.

• It is much harder to prove the Picard-Lindelof theorem than to prove that the Taylor series for the exponential function is its own derivative; the latter takes only a few lines and Cauchy convergence... – user21820 Jul 14 '16 at 12:11
• Yes, as I said myself. My point is that it's conceptually nicer to start with the ODE. For instance because it generalises better: differential equations give the exact discription for many problems whose solution can not be written as a power series. – leftaroundabout Jul 14 '16 at 12:45
• Actually I also like starting with the ODE. From the ODE we can iteratively obtain polynomial approximations and then guess/hope that the 'limit' exists and is the actual solution we desire. Then we can switch over to the series definition to prove everything rigorously yet elementarily. The advantage is not just that it is pedagogically simpler and more intuitive but also that the proof that the series satisfies the differential equation works without change in the complex numbers, whereas the Picard-Lindelof theorem is for real ODE as far as I know. – user21820 Jul 14 '16 at 12:55
• Thanks for your answer.. It helped me to understand it better – user311526 Jul 19 '16 at 13:54

It is great that you asked this question (and +1 for putting your concerns very clearly in the question). Frankly speaking, the way these functions are introduced to the students as a part of a course in Trigonometry (meaning students of age 14 years) they don't make much sense apart from some set of fancy symbols satisfying certain identities. Moreover the way these functions are linked to angles and triangles makes them a systematic tool to utilize the concept of similarity of triangles. This is one of the approaches to trigonometry and it is nothing more than some algebra combined with properties of similar triangle.

The other approach to these functions is linked very intimately with the rectification and quadrature of a circle and almost always this part is never highlighted in a course of Trigonometry because it transcends the power of algebra and geometry. Not only a different approach but a different name "circular functions" is used when we link them to a circle. It is this approach which perfectly answers the question: If $x$ is a real then what is meant by symbol $\sin x$?

Let us then consider the unit circle $x^{2} + y^{2} = 1$ so that its center is origin and radius is $1$. Using a rigorous definition of length of a curve and definition of area of plane regions it is possible to establish that the unit circle has a well defined circumference and the plane region formed by the circle has a well defined area. This is the part which requires the tools of analysis. Going further it can be established that an arc of a circle has a length and the corresponding sector of the circle has an area.

Let $A = (1, 0)$ be the point where unit circle meets positive half of $x$-axis. Now let $x$ be a given real number and first let us suppose that $x$ is non-negative. Let us start from point $A$ and move on the circle in counterclockwise direction such that after covering $x$ units of distance we reach point $P$. Note that if $x$ is large we may have to move on the circle a number of times to cover the distance $x$. The coordinates of the point $P$ are defined to be $(\cos x, \sin x)$. If $x$ is negative then we have to start from $A$ and move on the circle in counterclockwise direction to cover a distance $|x|$ units and reach point $P$ and by definition again $P$ has the coordinates $(\cos x, \sin x)$.

By the above definition we see that since $P$ lies on unit circle we automatically get $$\cos^{2}x + \sin^{2}x = 1$$ You see that in this definition mentioned above there is no talk of angles. Given a real $x$ we get the value of $\cos x, \sin x$ as coordinates of a suitable point $P$ which depends on $x$.

Using the definition we get some obvious facts like $\cos 0 = 1, \sin 0 = 0$. Further the periodic properties are obvious, but to get them in their usual form we need to introduce the number $\pi$. And we don't need anything new here because $\pi$ is historically defined to be ratio of circumference and diameter of a circle and thus for unit circle the circumference comes out to be $2\pi$. Now consider any number $x$ and suppose we have covered $|x|$ distance on unit circle starting from $A$ and ending at $P$ (moving counterclockwise if $x \geq 0$ and clockwise if $x < 0$). Now starting from $P$ if we cover another distance of $2\pi$ in the counterclockwise direction we reach the same point $P$ again. Hence we see that $$\cos (x + 2\pi) = \cos x, \sin (x + 2\pi) = \sin x$$ for all $x$.

I have given the definition of circular functions using idea of length of an arc of circle. It is possible (and simpler from the point of view of analysis) to define them using areas of corresponding sectors. Using integral calculus it can be proved very easily (without any knowledge of $\sin x, \cos x$) that if $AP$ is an arc of unit circle with length $x$ and $AOP$ is corresponding sector then area of sector $AOP$ is $x/2$. As in case of arcs we can make convention that area of sector is to be treated positive when we move counter-clockwise from $A$ to $P$ and if we move from $A$ to $P$ in clockwise direction then area of sector $AOP$ is negative. Thus given any real number $x$ we move from $A$ to $P$ on circle such that area of sector $AOP$ is $x/2$ and define the coordinates of point $P$ as $(\cos x, \sin x)$.

Because of the relation between length of arc and area of corresponding sector mentioned in last paragraph it is clear that the definition of circular function based on length of arcs is equivalent to their definition based on area of sectors.

Using the definition based on areas it is possible to prove easily that if $0 < x < \pi/2$ then $$\sin x < x < \dfrac{\sin x}{\cos x}$$ which leads to the first and most important analytical property of $\sin x$ namely that $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ And combined with addition formulas for circular functions you can get the derivatives of $\sin x, \cos x$ and then justify the graph of $\sin x$. You can further find its Taylor's series and in fact any analytical property of $\sin x, \cos x$ can be derived from this point onward.

In case you were wondering about angles and their link to circular functions it is better to first understand the link between angles and circles (the use of semi circular protractor to measure angles gives us a hint that there is a deep link between angles and circles).

• Thanks for the answer... :) – user311526 Jul 19 '16 at 13:48

Try this answer by first just looking at the images (middle-click to enlarge in new tab).
If those aren't enough by themselves, then try reading the description.

## Where the input is in radians:

Imagine a circle, like the black one below: For convenience, let's call the up direction on this diagram "north."
Imagine you're sitting on the circle at the red dot, in a vehicle that has a very precise odometer.
You can imagine the vehicle is a train car, and the black circle is the track, if that helps.
Now imagine traveling along the circle and tracing out the orange arc.
Suppose you stop at some point. The reading on the odometer is x.
(You can imagine the units to be miles, kilometers, megameters, or other generic units that someone decided to call "radians.")

The "sine" of x is the distance you'd have to travel south, to reach your original latitude.
Stated another way, the "sine" of x is how far north you are of the horizontal blue line.

If you've gone more than halfway around the circle, and are in the "southern" half, that distance is going to be a negative number.
If you are exactly half way around, or all the way around, it'll be 0.
If you are one quarter of the way around, it'll be 1.
If you are three quarters of the way around, it'll be -1.

A circle's radius is the distance between the center and any point on its edge. (The fact that this is constant is what makes it a circle.)
This circle is centered where the blue lines cross, and the distance from that center to the red dot (a point on the circle's outside edge) is 1.
This means the radius of this circle is 1.
(Any circle with radius 1 gets the special name "unit circle," but you don't actually have to know that to understand this explanation.)

A circle's diameter (the distance across the circle through the middle) is always twice the radius.
Therefore, the diameter of this circle is 2.

$\pi$ is defined as the ratio between a circle's diameter and its circumference (the distance around the outside edge of a circle). Since $\pi$ = circumference / diameter, circumference = diameter * $\pi$, and the circumference of this circle is 2$\pi$.

Once the odometer reads 2$\pi$, you'll be back to exactly where you started. If you keep going, you'll be tracing out the same path, and for every reading on the odometer, the distance you'd have to travel south to reach the same latitude you started at is exactly the same as when your odometer read 2$\pi$ less than it does now. In other words, the sine value will be exactly the same as it was last time you were there.
This is why the sine function is periodic (meaning, it repeats itself).
$sin(x) = sin(x-2\pi) = sin(x+2\pi)$.

## Where the input is in degrees:

Imagine a circle, like the black one below:
For convenience, let's call the up direction on this diagram "north."
Imagine there is an obelisk at the point where the blue lines meet, which is visible from everywhere on the circle.
Imagine you're sitting on the circle at the red dot, one unit east of the obelisk.
Imagine the line between that obelisk and your starting point is permanently marked, by what I'll call the "positive horizontal axis."
Now imagine traveling along the circle and tracing out the orange arc.
Suppose you stop at some point, and make a purple line between where you are and the obelisk. Let's call the angle that sweeps from the positive horizontal axis to that line, $\theta$.

The "sine" of $\theta$ is the distance you'd have to travel south, to reach your original latitude.
Stated another way, the "sine" of $\theta$ is how far north you are of the horizontal blue line.

If you've gone more than halfway around the circle, and are in the "southern" half, that distance is going to be a negative number.
If you are exactly half way around, or all the way around, it'll be 0.
If you are one quarter of the way around, it'll be 1.
If you are three quarters of the way around, it'll be -1.

Once you get back to where you started, $\theta$ will be 360 degrees, because there are 360 degrees in a circle (by definition).
If you keep going, you'll be tracing out the same path, and for point you reach again, the sine value will be exactly the same as it was last time you were there.
This is why the sine function is periodic (meaning, it repeats itself).
$sin(\theta) = sin(\theta-360°) = sin(\theta+360°)$.

## Bonus explanation: Cosine

In both examples, the cosine of x, or $\theta$, is the distance you'd have to travel "west" to reach the vertical blue line.
Stated another way, it's how far "east" you are of the vertical blue line.
If you drew a horizontal line between the end of the arc and the vertical blue line, it's the length of that line.
It's the length of the blue side of the triangle in the second diagram above (where the length of the arc is still x).
At the starting point, it's 1.
In the "western" half of the circle, it's a negative number.
If you are exactly half way around, it'll be -1.
If you are one quarter of the way around, or three quarters of the way around, it'll be 0.
Cosine is periodic in the same way sine is; you could write "cos" instead of "sin" consistently in the formula lines above (that have two = signs) and each would still be correct.

Mathematicians call the location of the obelisk the "origin," but I chose not to use that term here to avoid confusion with a more common definition of "origin," namely "where you started," because that refers to a different location in the narrative underlying this answer. I also made an alternate version of the second diagram which uses "origin" instead.

This posting also uses the word "line" in multiple places where "line segment" would be more specifically correct; I think the meaning is clear in context and the language used better facilitates understanding here.

What is an angle and what isn't is just our interpretation of a number. Just like "3" in our equation can mean "3 apples" or "3 rotations", it just so happens that trigonometric function are useful when your number represents an angle. So... $\sin \pi$ is just a function of a real number. That $\pi$ may represent an angle ($180^\circ$ in this example), or it might represent a half-period of an oscillatory motion in time, or some factor in some formula that has nothing to do with angles - for instance, in statistics. Similar to the fact that $\pi$ appears in formulas that have nothing to do with a circle.

My point is... all functions take pure numbers (the domain may vary: analytical functions take any complex number as their arguments). No units are allowed. Unless you are dealing with abstract math, the numbers we put in these function usually mean something for us, but it depends on the context and it doesn't affect the way the function works, it's just nice for us to keep track what we're calculating. So, your assumptions that trigonometric function only work on angles, or that angles aren't real numbers, are wrong.

Allow me, please, say something about it based on my personal experience.

The first time we know something of the sine function we learned the following formula in a right triangle: $$\sin(x)=\frac{\text{opposite leg to the angle x}}{\text{hypotenuse}}$$ This definition, which can seem something "rudimentary", has served and continues to serve much engineers and craftsmen. At that time I thought that, by definition, did not apply to angles greater than $90^{\circ}$.

But soon came the lesson of the trigonometric circle so I learned that this "function" was periodic and that the definition extended to any number of "degrees". That was a "breakthrough" for me. On more than one occasion I asked some people (and many students) about a definition of the sine function and they have responded me very happy with the first "definition" above (although some have doubted quite what, I think, indicated that did not satisfy them about legs and hypotenuse).

Spend some time before I learned that the radian and not the degree is the most adequate variable and that the function properly is the infinite series well defined on $\mathbb R$ and taking values on $[-1,1]$ (the latter I knew from my first approach to the function!...) $$\sin(z)=\sum_{n= 0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}$$ But later I learned that in mathematics, equivalences give rise to different definitions all equivalent of course; (for example the absolute value of a real number defined by $|x|= +\sqrt{x^2}$ or $|x|=\max\{x,-x\})$ and in particular that the sine function can also be defined as the infinite product $$\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(n\pi)^2}\right)$$ and even as an integral inversion $$y=\int_0^{\sin x}\frac{dt}{\sqrt{1-t^2}}$$ There are several other (equivalent) definitions. I wanted to limit myself to the real numbers. I apologize for the bad English.