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I am aware that according to Lindemann–Weierstrass theorem:

1) $\sin(a),\;\cos(a),\;\tan(a)$, and their multiplicative inverses $\csc(a),\;\sec(a),$ and $\cot(a)$, for any nonzero algebraic number $a$, the result will be transcendental.

2) $\ln(a)$ if $a$ is algebraic and not equal to $0$ or $1$ will also be transcendental.

However, if you put any value (except $1$) for $x$ in $\cos(\ln(x))$, according to Wolfram Alpha, the result will be transcendental.

I would like to know why that is so. Does anyone have a proof that every value of that function will be transcendental? (if that is the case)

Rushabh Mehta
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GL RM
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    You ask, "does anyone have a proof that every value of that function will be transcendental?". You might want to refine this question. The function $x \mapsto \cos(\ln x)$ is continuous, $\lim_{\varepsilon \rightarrow 0} \cos(\ln \varepsilon) = -1$, $\cos(\ln 1)$ is positive, so the function has a zero on the interval $(0,1)$. Zero is not transcendental. – Z. A. K. Jan 17 '20 at 03:12
  • Obviously $\cos(\ln(z)) = 0$ where $z = e^{\pi/2}$. I hope the question is whether $\cos(\ln(x))$ is transcendental where $x$ is algebraic and not $1$. – Robert Israel Jan 17 '20 at 03:37
  • I see those points, I will edit the question. My point, though, is that if x=5, for example, we have that, from rule 2, the value will be transcendental. However, when the cosine of this value is taken (cos(ln5)), this result is also transcendental. My doubt is that rule 1 makes no mention of these cases (since this rule only mentiones algebraic numbers), which happen with almost every value of x. – GL RM Jan 17 '20 at 03:43

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$\exp(i \ln(x)) = x^i$ (for a suitable branch). Since $i$ is algebraic and not rational, the Gelfond-Schneider theorem says that $\exp(i \ln(x))$ is transcendental whenever $x$ is an algebraic number other than $0$ or $1$. And from that you can get that $\cos(\ln(x))$, $\sin(\ln(x))$, $\tan(\ln(x))$ are also transcendental.

Robert Israel
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