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I know that $\sin\theta=\frac{y}{r}$ and $\cos\theta=\frac{x}{r}$.

My question is: is $\sin$ a function of $\theta$, as in $\sin (\theta$)?

If yes, why is there no $\theta$ on the right hand side of the equation? For example, $f(x) = mx + c$. Here $f$ is a function of $x$.

If not, what is the relationship between the symbols $\sin$ and $\theta$?

Another question is: why can we write $\theta = \sin ^{-1} (\frac{y}{r})$ and what does it actually mean?

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    $\begingroup$ Fixing $r=1$ as is traditional, then any angle $\theta$ uniquely identifies a point on the unit circle. Hence, $\theta$ specifies a point, which specifies that point's $y$-coordinate. That is to say, the right hand side does depend on $\theta$. This dependence just isn't reflected by the notation. $\endgroup$ – pjs36 Jul 25 '15 at 19:34
  • $\begingroup$ It is a function. A function is a rule for turning a number into another number (roughly). If you give me a number $\theta$, I will draw a triangle with angle $\theta$ and divide the opposite side by the hypotenuse and call the answer $\sin(\theta)$. That's a rule for turning a number into another number, so it's a function. $\endgroup$ – Akiva Weinberger Jul 26 '15 at 6:06
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    $\begingroup$ $\sin$ is not a function of $\theta$. $\sin \theta$ is a function of $\theta$. $x,y,r,\theta$ are collectively dependent variables, so the statement $\sin \theta = y/r$ isn't as trivial as, say, defining $f(t) = \pi$, and by geometry it turns out not to be ill-defined like trying to define $g(x) = y$ would be in a context where $x$ and $y$ are independent (and so, e.g., it would require $g(0) = 0$ and $g(0) = 1$ as $y$ varies while keeping $x$ fixed)). $\endgroup$ – Hurkyl Jul 26 '15 at 7:00
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    $\begingroup$ $\sin$ is a function, but $\sin \theta = \frac{y}{r}$ isn't a strictly correct way of defining it. $\endgroup$ – immibis Jul 26 '15 at 22:59

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I think the historical reason for the confusion stems from graphing trigonometric functions in polar form versus rectangular form. In rectangular form, the following statement below is true. $$ \theta \quad =\quad x $$ (Where the meaning of this equality is that we let the measure of the angle on the unit circle be representative of the rectangular distance along the x-axis, or the domain, of the rectangular function.) But in polar form, this is not true, and $\theta$ means something different, namely, it is used to define the reference angle from zero radians, not the distance along the $x$-axis. Thus, the meaning of $\theta$ or $x$, depends on which way you are graphing in the two-dimensional plane, i.e., polar or rectangular plotting. In the polar method, the meaning of the rectangular coordinate pair $$(x,y)$$ is that it is an analogous location to the coordinate pair defined by the directed distance from the origin for a certain measure of $\theta$, with the abscissa, or $x$-value of the rectangular coordinate pair, defined by $$x=r\cdot \cos(\theta )$$ and the ordinate, or $y$-value of the rectangular coordinate pair, defined by $$y=r\cdot \sin(\theta )$$

In polar method, we therefore represent the coordinate pair, or point on the plane, using a different meaning of $\theta$, namely, as an input value for the angle from the origin, in terms of $r$ defined as a function of $\theta$, and thus we commonly refer to each coordinate pair on the plane with: $$(r,\theta )$$

If, instead, we intend to graph on the two-dimensional plane in rectangular form, then we let: $$ \theta \quad =\quad x $$ (Where $\theta$ is now being understood to be the rectangular distance along the $x$-axis, or the domain of the rectangular function.) Thus, instead of representing the angle measure from the zero radians, the angle theta now represents the distance along the $x$-axis from the origin. Additionally, in the case of the parent function derived from the unit circe, since: $$r=1$$ we can construct the following function machine of a single variable for rectangular graphing: $$ r\cdot \sin(\theta )=y $$ And this one is intuitive because the ordinate, or y variable, winds up conveniently isolated. And since we are very accustomed to y being a function of x, it makes perfect sense to write this as: $$[r\cdot \sin(\theta )=y=f(x)=\sin(x)]\quad \Longleftrightarrow \quad \theta =x\quad \cap \quad r=1$$ But where it gets less intuitively obvious is when we construct the function machine for rectangular graphing for cosine, which when we multiply both sides in your example by r: $$r\cdot \cos(\theta )=x$$ And this is strange because even though rectangular and polar graphing are referring to the same two-dimensional planes, this statement below is using x in two different ways. First, let's consider the logically equivalent statement: $$ [r\cdot \cos(\theta )=x=f(x)=\cos(x)]\quad \Longleftrightarrow \quad \theta =x\quad \cap \quad r=1 $$ The problem is that in the polar / rectangular conversion formula, we are defining $x$ as the abscissa value for the polar representation of a coordinate, but now that we are switching to rectangular graphing form, we are using $x$ as the input variable to represent theta. In short, we are using $x$ in two different ways. (Similarly, we are also using $y$ in two different ways.) Basically, we are replacing the polar usage of $x$ with the rectangular ordinate, or $y$, and we are replacing the polar usage of theta in the above statement with the rectangular usage of $x$ as an independent variable. This ambiguity can be removed if we replace our common way of writing the ratios in polar form, with dummy variables. This is why some teachers prefer using opposite, adjacent, etc., instead of $x$ and $y$ because we are using them in different ways when we construct the function machines. To remove the conflation, simply use phi for the polar abscissa formula, yielding: $$ [r\cdot \cos(\theta )=\varphi =f(x)= \cos(x)]\quad \Longleftrightarrow \quad \theta =x\quad \cap \quad r=1 $$ And thus both trigonometric ratios have been transformed into functions of $x$. Let's be clear, which $x$ do we mean!? Well, in this case, we mean $x$ as the independent variable and not the abscissa output of the polar representation's directed distance. The same can be mathematically deduced for all of the other functions. Again, it might be helpful to replace your original ratios with dummy variables that are defined as the abscissa and ordinate so as not to conflate the use of $x$ in the first case, which is used as an abscissa, with $x$ in the function machine, which, for cosine, has two different usages as described above. By so using dummy variables, the apparent multiple usages and meanings of $x$ are removed. This answers your first three questions. Understanding this properly, also sheds light on the fact that even in the construction of sine, and the other functions, we are still committing the same mathematical conceit of switching between uses of the variables $x$ and $y$ in the polar method that differ from uses of the variables $x$ and $y$ in the rectangular method. To be clear, these are uses, and we are still graphing in the two-dimensional coordinate plane in both cases.

Your next question is what is the relation of sine and $x$, and cosine and $x$, etc. There is a reason that trigonometric functions are called transcendentals. The relationship is transcendental. That means one cannot use basic arithmetic, i.e., addition, subtraction, multiplication, or division, to solve, for example: $$ \cos(30) $$ Now, in antiquity, this was painstakingly measured by iterations around the unit circle. As a result, tables were developed, notably in the case of Ptolemaic astronomy. Later, when the Calculus was developed, infinite series were used to approximate these transcendentals with more accuracy, and these are now collectively referred to as Taylor Series, which are MacLaurin Series that are not centered at 0. The relationship is therefore symbolic, and is that sine "of an angle" is measured by an ordinate on the unit triangle of a specific theta, or angle. Likewise, cosine "of an angle" is measured by the abscissa of the unit triangle of a specific theta, or angle. And so on for the others ...

You then additionally ask what the inverse function means and why we can write it. First off, in response to why we can write it, it should be noted that each trigonometric function machine requires a different domain and range restriction in order to be a function in the first place. This is why some camps of thought prefer to write "arc" before the function as opposed to using the negative one superscript which, to some, implies there is a perfect inverse function without restriction. Now, if you are asking more broadly why one can construct inverse functions, I would either defer to a theorist, or suggest more simply that sometimes one knows the ratio, but needs the angle, and sometimes one knows the angle, but needs the ratio. Consider early analysis of planetary movement, hot air balloon, calculating the distance of a kite string, determining the distance of a trajectory using parametric trig functions, and so on, of course all limited to two dimensions at present.

Then, the answer to your question about the meaning of the inverse function, is simply the logical converse, or arcsine "of a ratio (with $r=1$)" is measured by a specific theta on the unit triangle of a specific ordinate, or y-value. Likewise, arccosine "of a ratio (with $r=1$)" is measured by a specific theta on the unit triangle of a specific abscissa, or $x$-value. It is simply the logical converse function and meaning of the above trigonometric function machines (including the necessary restricted domains and ranges). Additionally, they are both transcendental and require either iterations or measurement, resulting in tables, or require approximations of transcendental values through the use of so-called Taylor Series, which are MacLaurin Series that are not centered at 0. This last part answers what the inverse function means.

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    $\begingroup$ This is of course assuming you wanted the deeper meaning / answer to your questions. $\endgroup$ – oemb1905 Jul 25 '15 at 20:07
  • $\begingroup$ is there a way you could reword the first half? I am very comfortable with sines and cosines, but I have absolutely no idea what is going on in the first half. I am guessing that $\theta=x$ at the beginning is something profound, but for the life of me I cannot figure out why one would wish to do that besides simply graphing a sin or arcsine. I'm also not sure how we can declare "Additionally, since r=1..." without any explanation (I think you intended to assume r=1 without loss of generality, rather than just remidning us that r=1) $\endgroup$ – Cort Ammon Jul 26 '15 at 16:27
  • $\begingroup$ I'll try to reword it for you later, but while on my phone let me attempt to explain. Theta is referring to an angle, and that tradition comes from polar graphing. X also refers to the angle but comes from rectangular graphing. As for r = 1 that is simply because all functions start with "parents," or primary versions that are special cases of transformations that can be developed afterwards. In the special case, r = 1, because the ways of graphing each method are based on the "unit circle." Remember, r is simply the radius value either for a polar theta, or for a rectangular x. $\endgroup$ – oemb1905 Jul 26 '15 at 16:34
  • $\begingroup$ "This is why some camps of thought prefer to write "arc" before the function as opposed to using the negative one superscript which, to some, implies there is a perfect inverse function without restriction." I've never heard this before, it makes so much sense! $\endgroup$ – Kik Jul 27 '15 at 14:11
  • $\begingroup$ Yes, and even the domain and range restrictions were not generally agreed on until the late 60s. I had an old Calculus book that introduced two different ways of restricting arc secant, for example. Additionally, some camps capitalize the first letter to indicate the inverse. $\endgroup$ – oemb1905 Jul 27 '15 at 14:27
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There is a $\theta$ on the right hand side. The definition of $\sin(\theta)$ is not just $y/r$; instead it is something like:

$y/r$, after you've drawn a right triangle with $\theta$ as an angle, and where $y$ is the length of the side opposite $\theta$ and $r$ is the length of the hypotenuse.

As you can see, that full definition does in fact contain a $\theta$, no less than $mx+c$ contains an $x$.

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    $\begingroup$ You didn't answer all his questions. $\endgroup$ – Jane Smith Jul 25 '15 at 20:01
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Looking at your profile, you post a lot on StackOverflow, so maybe a programming analogy will help.

Say you want to make a Point class. There's a lot of ways you could do it. You could have member variables p.x and p.y, for the $x$ and $y$ coordinates (duh). That's the most common way. But in theory, you could also write it in terms of p.rad and p.theta. Why not? Or if you were feeling really perverse, you could use some kind of bipolar coordinate system. There's a lot of choices, and you don't want to have to re-write your code if you change your mind.

So, as programmers are wont to do, you introduce a layer of abstraction. All Point objects, however they're implemented, have the following methods: p.get_x(), p.get_y(), p.get_rad(), and p.get_theta(). So now you can safely forget about implementation details*. A Point doesn't necessarily have an x or a y anymore, no more so than it has a rad or a theta.

Now let's go back to math-land. Instead of p.get_x(), we have $x(p)$, and your formula becomes: $$ \sin \left( \theta(p) \right) = \frac{y(p)}{r(p)} $$

But since we're lazy, and the $p$s everywhere kinda clutter things up, we drop the $p$ from the notation. So whenever you see these kinds of statements, remember that there is an implicit "for any point $p$" in them. Hopefully this makes more sense now.

$\newcommand{\RR}{\mathbb{R}}$ EDIT: I guess I didn't really address what "sine" really means. It's just a function from $\RR$ to $\RR$ that's got certain properties. Nothing special about applying it to $\theta$.


*Other than performance concerns, of course. :)

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    $\begingroup$ "Looking at your profile, you post a lot on StackOverflow, so maybe a programming analogy will help." --- Hilarious and frightening!! $\endgroup$ – user6704 Jul 25 '15 at 21:21
  • $\begingroup$ As a programmer, THANK YOU. I wish all math questions could be answered in programmer notation rather than symbols, since I find it vastly more intelligible. $\endgroup$ – Dewi Morgan Jul 27 '15 at 1:05
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    $\begingroup$ “bipolar coordinate system”?  As in manic-depressive?  … … … … … … … … … … … … … … … … …  But seriously, as long as you’re going to bring up programming, Point_Cartesian($x, y$) == Point_Cartesian($0, 0$) $\implies x=0$ and $y=0$, but whether Point_polar($42, 0$) == Point_polar($42, 2\pi$), and whether Point_polar($0, 0$) == Point_polar($0, any\ \theta$), may be less straightforward. $\endgroup$ – Scott Jul 27 '15 at 18:40
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I guess it depends on how you define $\sin\theta$. One possible definition is $f(\theta)\equiv \theta- \frac{\theta^3}{3!}+\frac{\theta^5}{5!}...$ This series converges for all $\theta$ and is called $\sin\theta$. For more info look up Maclaurin series.

More generally I think you are confused about what a function is. The technical definition is daunting (at least for me) but what you put in can be different to what you get out. Consider a machine that assigns a number to the letters of the alphabet. $A\to 1$ here the domain of the function are letters whilst the range numbers.

The inverse function $\sin^{-1}$ will essentially assign the limit of the convergent series (or ratio) to an angle. The role of domain and range reversed. Note though that it is necessary to restrict the value of $\theta$ as multiple values converge to the same limit.

I hope this helps.

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Yes $\sin$ is just a function on the real (or complex) numbers. People often write $\sin(\theta)$ or $\sin\theta$ because the argument of the $\sin$ function is often an angle in physical applications, and $\theta$ is often used to denote angles.

For your follow-up: $\sin^{−1}$ is the inverse function of the $\sin$ function. Similar to how $\log$ is the inverse function for the exponential function. Note that since $\sin$ is not bijective, $\sin^{-1}$ is rather 'arbitrarily' chosen partial inverse.

The $\theta$ symbol doesn't appear on the rhs because it is usually included in a diagram like this. So you have to determine from the context what the $\theta$ refers to.

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It depends on how you want to define $\sin$ and $\cos$. I suspect you're looking for an explicit definition in terms of things you already know (such as polynomials), in which case @Karl's answer (the MacLaurin series definition) is what you're looking for.

However, others like me find it more elegant to define them implicitly as bases for the the set of solutions to

\begin{align*} \ddot{r}(\theta) &= -r(\theta) \end{align*}

that satisfy \begin{align*} &\dot{r}(0) = 1 && \dot{r}(0) = 0 \\ &r(0) = 0 && r(0) = 1 \\ &\text{(for $r \equiv \sin$)} && \text{(for $r \equiv \cos$)} \end{align*}

The reason some people prefer this kind of definition is that it displays the relationship of sinusoids to exponentials and hyperbolic sinusoids. For example, we can define $e^\theta \equiv \exp(\theta)$ to be the solution to

\begin{align*} \dot{r}(\theta) &= +r(\theta) \\ r(0) &= 1 \end{align*}

and, similarly to the case of $\sin$ and $\cos$, we can define $\sinh(\theta)$ and $\cosh(\theta)$ to be the solutions to

\begin{align*} \ddot{r}(\theta) &= +r(\theta) \end{align*}

that satisfy

\begin{align*} &\dot{r}(0) = 1 && \dot{r}(0) = 0 \\ &r(0) = 0 && r(0) = 1 \\ &\text{(for $r \equiv \sinh$)} && \text{(for $r \equiv \cosh$)} \end{align*}

Defining them as such lets us remember that these functions are different faces of the same coin.

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  • $\begingroup$ (+1) I like the idea that the functions are linked by their properties $\endgroup$ – Karl Jul 25 '15 at 21:47
  • $\begingroup$ @Karl: Thanks! I added a couple more too :) $\endgroup$ – Mehrdad Jul 25 '15 at 23:03
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The trigonometric functions relate an angle to a length. More precisely, in a rectangle triangle, the sine, cosine and tangent of an angle are the ratio of two of the sides. enter image description here

$$\cos(\theta)=\frac xr,\ \sin(\theta)=\frac yr,\ \tan(\theta)=\frac yx.$$

For convenience, we will now assume that $r=1$, as this doesn't change the proportions.

$$\color{green}{\cos}(\theta)=\color{green}x,\ \color{blue}{\sin}(\theta)=\color{blue}y,\ \color{red}{\tan}(\theta)=\frac{\color{blue}y}{\color{green}x}.$$

As you vary $\theta$, the opposite vertex describes a circle, and $x$ and $y$ vary accordingly. So the trigonometric functions are indeed... functions. Here is their plot.

enter image description here

In the expression $$\sin(\theta)=y,$$

you don't see $\theta$ explicitly in the right-hand side, but it appears implicitly in $y$. This is a definition of the sine resulting from a "geometric computation". Other expressions exist, such as the Taylor's development which allows you to compute numerically $$\sin(\theta)=\theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}-\frac{\theta^7}{5040}+\frac{\theta^9}{362880}\cdots$$ where $\theta$ must be measured in radians (a full turn, $360°$, is $2\pi$ radians).

The sine function transforms an angle to a ratio of lengths. You can invert the relation and ask what is the angle that gives some ratio. This inverse function is called the arcsine, and is denoted $\arcsin$ or $\sin^{-1}$.

$$\sin(\theta)=y\longleftrightarrow\theta=\arcsin(y).$$

Similarly, you can evaluate with a Taylors' development

$$\arcsin(y)=y + \frac{y^3}{6} + \frac{3y^5}{40} + \frac{5y^7}{112} +\frac{35y^9}{1152}\cdots $$

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Like others said, it does depend on the context—but here's the way I intuitively answer your questions:


What are the trigonometric functions?

$\sin$, $\cos$, and $\tan$ are the ratios of the length of the sides of a right triangle

They are functions of $\theta$, but all $\theta$ is doing is telling you which sides to look at. As an example, let's consider the below triangle.

enter image description here

Now, define the trigonometric functions like this:

$\sin{\theta}$ - Take the ratio of the side opposite to $\theta$ and the triangle's hypotenuse.

$\cos{\theta}$ - Take the ratio of the side adjacent to $\theta$ and the triangle's hypotenuse.

$\tan{\theta}$ - Take the ratio of the side opposite to $\theta$ and the side adjacent to $\theta$.

If we build a triangle like the above, and want the $\sin{30^\circ}$, all we have to do is take the ratio of $A$ and $C$. The ratio $\frac{A}{C}$ will always give the correct value of $\sin{30^\circ}$.

I think this is pretty cool. It doesn't matter what the length of the sides are. These functions depend only on the structure of a right triangle.


Why can we write $\theta = \sin^{-1}{\left(\frac{A}{C}\right)}$ and what does it mean?

I'll answer what it means first: $\sin^{-1}{\left(\frac{A}{C}\right)}$ is asking "If we have a right triangle whose ratio of sides is $\frac{A}{C}$, what angle does this have to correspond to?" That angle is $\theta$. Therefore, $\theta = \sin^{-1}{\left(\frac{A}{C}\right)}$. "Why can we write it?" is difficult to answer by itself, but maybe my answers to your other questions will help.


Is sin a function of $\theta$, like $\sin{\theta}$? If yes, why are there no $\theta$ on the right hand side of the equation?

Yes! It is a function of $\theta$. You can think of a function like this:

A set of inputs which each have exactly one output

Your example, $f(x) = mx + c$ is certainly a function. Every input of $x$ has exactly one output of $mx + c$. On the other hand, $f(x) = y$ is also a function—for every input of $x$, there is exactly one output. The output doesn't necessarily have to depend on the input!

Now, let's look at the trigonometric functions as we've defined them above. $\sin{\theta}$ definitely has an input, $\theta$. And, it definitely has an output: "The ratio of the side opposite to theta and the triangle's hypotenuse."

Back to your other question, "Why can we write this?" Well... we've set up rules for how we write functions, and, as we've seen, $\sin$ is definitely a function—so we can write this.

Viola! Trigonometry!

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It means different things in different contexts. However, it is function notation. $\sin(\theta)$ is the function that takes an input angle $\theta$ and outputs the ratio of the opposite side and hypotenuse for a right triangle.

$\sin^{-1}(y/r)=\theta$ is the inverse function. It takes as input a ratio of opposite to hypotenuse and outputs the angle to which that ratio corresponds.

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$\sin(\theta) = \sin \theta$, without brackets is just a short hand notation. And yes, it means that $\sin$ depends on $\theta$ as its argument like a function $f(\theta)$. In this sense your expression $\theta = \sin^{-1}(\frac y r)$ means that you use the inverse of $\sin$ (actually the inverse of $\sin$ is $\arcsin$) to get the related $\theta$ to the given ratio of $y$ and $r$.

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If you wish to think about it from the perspective of "why isn't $\theta$ on the right hand side," it may be helpful to approach sin and cos backwards - start with $\sin^{-1}(x)$ and work our way backwards. Why? Because doing so makes it look more like a typical definition of a variable, with a variable on one side and some expression on the other side.

$\theta = \sin^{-1}(\frac{x}{r})$

Now, this looks more like what you would expect. However, we still don't seem to have something that looks quite like what you expect. Accordingly, I'm going to do two nonstandard things:

  • I'm going to replace $\theta$ with a function, $\Theta$, simply so that it looks more like a function in the sense you are used to seeing.
  • I'm going to replace $\sin^{-1}$ with f. All I'm doing is renaming it to make it look more like the simple function you intuitively want to see, without all those interesting looking exponents and such. No meaning is changed, just the symbol.

$\Theta(x, r) = f(\frac{x}{r})$

This looks more like what you would expect for a typical functional notation. it shows that there is a relationship between $\Theta(x, r)$ and the ratio of x and r. This specific relationship ($f$ or $sin^{-1}$) is exatly the correct relationship to convert from the ratio of a leg of a right triangle and its hypotenuse to an angle within the triangle (specifically the angle between the leg and hypotenuse). Why is this ratio what it is? The answer is rather boring: because it is defined that way. When the Greeks were working their way through trigonometry, they defined sine and cosine to be these relationships.

But we're still in a decidedly nonstandard notation. Let's see if we can work our way back. Let's start by bringing $sin^{-1}$ back into the picture

$\Theta(x, r) = \sin^{-1}(\frac{x}{r})$

Now, $\sin^{-1}$ is what we call a bijection. In laymans terms, each input maps to one output and each output maps to one imput. This means $\sin^{-1}$ has an inverse, known unsprisingly as $sin$.

$\sin(Theta(x,r)) = \sin(\sin^{-1}(\frac{x}{r})) = \frac{x}{r}$

(Purists will note that I'm not dealing with the domain of sin and arcsin just yet. It will come later)

Now, this is technically all correct, but we usually don't bother writing theta as a function of x and r. There's too any cases where it is more convenient to just think of a specific \theta, remembering that $\sin$ is the function that matters.

$\sin(\theta) = \frac{x}{r}$

So now we've taken a round trip, simply to make the formula look the way you'd expect, then return back to the formula which is traditionally shown.

There's a good reason why they define $\sin$ with this equation rather than starting from $\sin^{-1}$ like I did. Sine is what we call a periodic function. Consider a 30 degree angle. Now consider a 390 degree angle (30 + 360). They go in the same direction right? If you were to draw right triangle, as we did earlier, they'd be the same right triangle. Thus they should have the same sine. Well they do. Sine is a function that repeats every 360 degrees (or rather every $2\pi$ radians... might as well get used to thinking in radians soon!). By making the left hand side of the equation $\sin\theta$ instead of just $theta$, all of that periodic behavior gets taken care of with the sine function, rather than having to put together some complex piecewise expression for $\theta$ directly.

And you'll find later that this periodic function is the basis for a remarkably large number of things in physics, so it's worth getting the hang of it now!

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I know that, $\sin\theta=\frac{y}{r}$, and, $\cos\theta=\frac{x}{r}$.

Probably you mean the "trigonometric functions as quantities computed from a right triangle" definition. To make it clearer that everything is well defined it is better to use the more modern "point on a fixed circle of radius $r$" definition, but this answer will use the presumed definition using right triangles.

My question is, Is $ \sin$ a function of $\theta$, like $\sin (\theta$)?

Yes. In the "compute sin, cos, tan from right triangle" definition, $x$, $y$, and $r$ are not functions of $\theta$, because there are many different (and mutually similar) triangles with angle $\theta$. But the ratios like $\frac{x}{y}$ and $y/r$ and $\frac{x}{x+y}$ all depend only on the shape of the triangle, not its scale, and so are uniquely determined by $\theta$. Therefore they can be considered as functions of $\theta$.

If yes, why are there no $\theta$ on the right hand side of the equation? For example, $f(x) = mx + c$. Here, $f$ is a function of $x$.

There is an implicit $\theta$ on the right hand side in the recipe "given $\theta$, construct a right triangle with that angle, and let $\sin (\theta)$ be the ratio of the side opposite the angle equal to $\theta$, to the longest side". There is also an implicit verification that this does not depend on the scale or position of the triangle so constructed, so that the function depends on $\theta$ alone and not the choice of triangle.

If not, what is the relationship between the symbols $\sin$ and $\theta$?

It doesn't hurt to write it as $\sin (\theta)$ with parentheses, but it's not the notation that makes it a function, it's the fact that $\sin$ is uniquely determined by $\theta$.

Another question is, why can we write $\theta = \sin ^{-1} (\frac{y}{r}) $ and what does it actually mean?

The -1 superscript denotes the inverse function, and has the same meaning and limitation as inverting any other function; the inverse is not unique without some restrictions, since $\sin$ is not a one-to-one function. Restricting to positive values of $\sin$ and $\theta$ between $0$ and a right angle, the inverse is unique and a function.

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