关于证明sin(1)为超越数（非代数数）的方法问询及简化证明需求

Hey there, let's break this down step by step—first we'll cover the easier proofs that $\sin(1)$ isn't an integer or rational number, then dive into the trickier question of proving it's non-algebraic (transcendental), and address your request for a straightforward approach without overpowered theorems.

First, a quick recap of $\sin(1)$'s definition #

We start with the alternating Taylor series for $\sin(1)$:
$$ \sin(1) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} = \frac{1}{1!} - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \cdots $$

Proving $\sin(1)$ isn't an integer is trivial #

Using the properties of alternating series, we can bound $\sin(1)$ tightly between two values:
$$ \frac{1}{1!} - \frac{1}{3!} \leq \sin(1) \leq \frac{1}{1!} - \frac{1}{3!} + \frac{1}{5!} $$
Calculating these bounds gives us:

• Lower bound: $1 - \frac{1}{6} = \frac{5}{6} \approx 0.833$
• Upper bound: $\frac{5}{6} + \frac{1}{120} = \frac{101}{120} \approx 0.8417$

Since $\sin(1)$ lies strictly between 0 and 1, it can't possibly be an integer.

Proving $\sin(1)$ isn't rational takes a clever lemma and contradiction #

This requires a bit more work, but here's a solid argument:

Key Lemma #

Every rational number can be written uniquely as $\frac{p}{q!}$ where $q$ is odd, and $q$ is the smallest such integer for that rational number.

Proof of the lemma: Take any rational number $\frac{a}{b}$. We can rewrite it as $\frac{a \cdot \frac{(2b+1)!}{b}}{(2b+1)!}$. Then, repeatedly divide both the numerator and denominator by pairs of the form $(2k)(2k+1)$ (starting from $k=b$ and working downwards) until dividing further would make the numerator a non-integer. The resulting denominator will be $q!$ where $q$ is odd, and this $q$ is minimal by how we constructed it.

Contradiction Argument for Irrationality #

Suppose $\sin(1)$ is rational. By our lemma, we can write it as $\frac{p}{q!}$ where $q$ is odd and minimal. Multiply both sides of the Taylor series by $q!$:
$$ p = q!\left( \frac{1}{1!} - \frac{1}{3!} + \frac{1}{5!} - \cdots \pm \frac{1}{q!} \right) \mp q!\left( \frac{1}{(q+2)!} - \frac{1}{(q+4)!} + \cdots \right) $$

Let $X$ be the sum of the first set of terms—all of these are integers, since $q!$ is divisible by every $(2k+1)!$ for $2k+1 \leq q$. Let $R$ be the remainder term, so:
$$ p = X \mp R $$

Now we analyze $R$:

• If the first term of $R$ is positive: $R = \frac{1}{(q+1)(q+2)} - \frac{1}{(q+1)(q+2)(q+3)(q+4)} + \cdots$. This alternating series is strictly bounded between $\frac{1}{(q+1)(q+2)}$ and $\frac{q^2 -7q +11}{(q+1)(q+2)(q+3)(q+4)}$—both positive fractions less than 1.
• If the first term of $R$ is negative: $R = -\left( \frac{1}{(q+1)(q+2)} - \frac{1}{(q+1)(q+2)(q+3)(q+4)} + \cdots \right)$, so it's strictly bounded between $-\frac{1}{(q+1)(q+2)}$ and $-\frac{q^2 -7q +11}{(q+1)(q+2)(q+3)(q+4)}$—negative fractions with absolute value less than 1.

In either case, $R$ is not an integer. But $X$ is an integer, so $p = X \mp R$ would not be an integer—contradicting the fact that $p$ must be an integer (since $\frac{p}{q!}$ is rational). Thus our initial assumption is wrong: $\sin(1)$ is irrational.

The hard part: Proving $\sin(1)$ isn't algebraic (transcendental) #

This is significantly more challenging. The standard proof relies on the Lindemann-Weierstrass theorem, which states that if $\alpha$ is a non-zero algebraic number, then $e^\alpha$ is transcendental. Since $\sin(1) = \frac{e^i - e^{-i}}{2i}$, if $\sin(1)$ were algebraic, then $e^i$ would satisfy an algebraic equation—contradicting Lindemann-Weierstrass (since $i$ is algebraic and non-zero).

But you asked for a straightforward method without invoking "absurdly powerful" theorems. Unfortunately, there's no widely known elementary proof (like the rationality one) for the transcendence of $\sin(1)$. Most simpler approaches still rely on variants of Lindemann-Weierstrass or related results (like Hermite's proof of the transcendence of $e$), which are non-trivial but not "absurdly powerful" if you're familiar with the background.

If you want to avoid the full Lindemann-Weierstrass theorem, you could build up from Hermite's work: first prove $e$ is transcendental, then extend that result to $e^i$, then use the trigonometric identity linking $e^i$ to $\sin(1)$. Even this path requires heavy lifting with polynomials and integrals (Hermite's method uses contour integrals or recurrence relations for polynomial derivatives to derive contradictions), but it's a more incremental approach than invoking the full Lindemann-Weierstrass.

In short: There's no truly elementary proof for the transcendence of $\sin(1)$, but you can use slightly less general theorems if you're willing to work through the intermediate steps.

备注：内容来源于stack exchange，提问作者Greg Nisbet