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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?

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    $\begingroup$ $\cos(x) = 1 \cdot \cos(x)$ which would make it rational... $\endgroup$ – gt6989b Jul 8 '13 at 19:41
  • $\begingroup$ I've improved your question's formatting; apologies if I changed your meaning. You can see here how I edited your question. Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ – Zev Chonoles Jul 8 '13 at 19:41
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    $\begingroup$ Since $\cos x = \dfrac{\cos x}{\sin^2x +\cos^2 x}$, it is a rational trig function even when requiring division. :) $\endgroup$ – Thomas Andrews Jul 8 '13 at 20:01
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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."

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  • $\begingroup$ +1 for the nice comment about $\sqrt{2+2\sin x-\cos^2 x}$. $\endgroup$ – Dave L. Renfro Jul 8 '13 at 21:22
  • $\begingroup$ Thank you. It is correct that I found the author's definition hard to understand. $\endgroup$ – mikoyan Jul 9 '13 at 16:33
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$\sqrt{2}\sin x$ 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.

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  • $\begingroup$ Thank you, that makes sense. $\sqrt{\sin x}$ is not a rational function in the book. Why not? $\endgroup$ – mikoyan Jul 8 '13 at 20:06
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    $\begingroup$ Well, proving $\sqrt{\sin x}$ is not a rational trig function might be hard, but in the definition, you aren't allowed to take square roots. It still could be the case that it is also representable in the above format somehow, using some trig identity. The only way I can think of to prove this can't happen is using complex analysis. @mikoyan $\endgroup$ – Thomas Andrews Jul 8 '13 at 20:15
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    $\begingroup$ The short answer for why, ultimately, $\sqrt{\sin x}$ can't be written as a rational trig function is that we can prove that if $f(x)$ is rational trig, then $f(x)$ can only be undefined for finitely many points in $[0,2\pi]$ and when $f(x)$ is defined, $f$ is differentiable at $x$. $\sqrt{\sin x}$ has differentiable problems at $x=0$. $\endgroup$ – Thomas Andrews Jul 8 '13 at 20:51
  • $\begingroup$ Notice that $\sqrt{\sin x}$ is not defined when $\sin x$ is negative. That won't happen with rational functions of sine and cosine. One way of showing that $\sqrt{\sin x}$ is not a rational function of sine and cosine is to show that its graph has vertical tangent lines at some points, and also that that can't happen with rational functions of sine and cosine. I think that second part will take more work than the first part. $\endgroup$ – Michael Hardy Jul 9 '13 at 0:48
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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 $\cos x$

$$f(x)=\cfrac{\sum\limits_{i=0}^n\sum\limits_{j=0}^ma_{i,j}\cos^i(x)\sin^j(x)}{\sum\limits_{i=0}^p\sum\limits_{j=0}^qb_{i,j}\cos^i(x)\sin^j(x)}$$

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    $\begingroup$ This answer is technically correct but does not directly address the OP's incorrect beliefs that $\sin x$ and $\cos x$ are not rational trigonometric functions. $\endgroup$ – Zev Chonoles Jul 8 '13 at 19:46
  • $\begingroup$ Also, he has rational across several variables. Basically, a rational function of $e^{ix_1},e^{ix_2},\dots,e^{ix_n}$. $\endgroup$ – Thomas Andrews Jul 8 '13 at 19:54
  • $\begingroup$ Ok, so $\cos x$ and $\sin x$ are rational trigonometric functions. Why is this so, particularly in terms of Gelfand and Saul's definition? Why is $sqrt\sin x$ not rational? $\endgroup$ – mikoyan Jul 8 '13 at 19:57

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