What is a rational trigonometric function? Is $\cos x$ rational? I am reading Trigonometry by Gelfand and Saul. On p.140 they discuss rational trigonometric functions and define one as:

A rational trigonometric function is a function you can get by taking the sine and cosine of various angles, together with all the constant functions, and adding, subtracting, multiplying or dividing them.

I want to check my understanding of what exactly is meant by a rational trigonometric function.
Is $\tan x$  rational because $\tan x = \dfrac{\sin x}{\cos x}$? (You take the sine and cosine of $x$ and divide)
$\sin(x+y)$ is rational because $$\sin(x+y) = \sin x \cos y + \cos x \sin y$$ (You take the sine and cosine of $x$ and $y$ and there is multiplication and addition).
$\sin x$ is not rational (because you are just taking the sine of $x$, and there is no multiplication, division, addition or subtraction).
$\cos x$ is not rational for the same reason.
$\sqrt{2} \sin\alpha$ is not rational either.
Are my thoughts about rational trigonometric functions along the right lines?
 A: I think you are simply misreading the definition. They perhaps should have used the words "starting with" rather than "taking." That is:


*

*$\sin x, \cos x$ and constant functions are rational trig functions

*If $p(x),q(x)$ are rational trig functions, then $p(x)\cdot q(x), p(x)+q(x),p(x)-q(x)$ are rational trig functions. If $q(x)\neq 0$ for some $x$, then also $p(x)/q(x)$ is a rational trig function.


Even with your reading of the text, since $\sin^2 x + \cos^2 x=1$, we can get $$\cos x =\frac{\cos x}{\sin^2 x + \cos^2 x}$$ But I suspect that it was not the author's intent for the paragraph to be interpreted that way. 
Some care will need to be taken if you want to also include multiple variable rational trig functions.
You aren't explicitly allowed to take square roots, but that doesn't mean that $\sqrt{\sin x}$ is not a rational trig function. I suspect that the authors just meant to make it clear that they didn't include some operations in allowing you to create rational trig functions.
For example, while $\sqrt{\sin x}$ is not rational, $\sqrt{2+2\sin x-\cos^2 x}$ is rational, since it happens to be equal to $1+\sin x$. Proving that $\sqrt{\sin x}$ can't be written that way is actually some work, and probably most easily done with complex analysis. Essentially, we can show that a rational trig function can only be undefined for finitely many $x\in[0,2\pi]$, and when it is defined at $x$, it is differentiable at $x$. This breaks for $\sqrt{\sin x}$. 
But again, I suspect the authors don't want you to go that far into it, and instead just note, "we only allow these operation, and square root wasn't one of them."
A: $\sqrt{2}\sin x$ is rational because $\sqrt{2}$ is a constant function of $x$, and you find constant functions mentioned in your definition.   $\sin x$ and $\cos x$ are both rational functions.  You can look at that in any of several ways:


*

*The functions you listed are the first rational functions, and the ones you can get from them by adding, subtracting, multiplying, and dividing other rational functions;

*$\sin x$ is $1\cdot\sin x$, so you're multiplying two of the functions you initially listed: a constant function and $\sin x$;

*The number of things you multiply can be $1$.  So you get $\sin x$.


$\tan x$ is a rational function for precisely the reason you mention, and similarly $\cos x\sin y + \sin x\cos y$.
I should add that when I say "rational functions", I mean rational functions of sine and cosine.  The term "rational function" without that modifier would mean just what you get by starting with constants and variables and adding, subtracting, multiplying, or dividing them.
A: A rational function is the quotient of two polynomials (with the lower one non-zero of course) so something like $$f(x)=\cfrac{\sum\limits_{k=0}^na_kx^k}{\sum\limits_{k=0}^mb_kx^k}$$
And so a trigonometric rational function is the quotient of two polynomials in $\sin x$ and $\cos x$
$$f(x)=\cfrac{\sum\limits_{i=0}^n\sum\limits_{j=0}^ma_{i,j}\cos^i(x)\sin^j(x)}{\sum\limits_{i=0}^p\sum\limits_{j=0}^qb_{i,j}\cos^i(x)\sin^j(x)}$$
