Euler Formula for Sine Function/Real Numbers/Proof 3

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Theorem

\(\ds \sin x\) \(=\) \(\ds x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }\)
\(\ds \) \(=\) \(\ds x \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \dotsm\)

for all $x \in \R$.


Proof

We have that $\sin x$ has a power series representation:

$\sin x = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots$

The roots of sine are the numbers $k \pi$, where $k$ is any integer.

From the Polynomial Factor Theorem, the following might be true:

$\ds \sin x = A x \prod \paren {1 - \frac x {k \pi} }$

where the product is taken over all $n \in \Z \setminus \set 0$, and $A$ is some constant.

The intuition is as follows.

\(\ds \sin x\) \(=\) \(\ds \ldots \paren {1 - \frac x {2 \pi} } \paren {1 - \frac x \pi} A x \paren {1 + \frac x \pi} \paren {1 + \frac x {2 \pi} } \cdots\)
\(\ds \) \(=\) \(\ds A x \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots\)
\(\ds \leadsto \ \ \) \(\ds \frac {\sin x} x\) \(=\) \(\ds A \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots\) for $x \ne 0$

From Limit of $\dfrac {\sin x} x$ at Zero:

$\dfrac {\sin x} x \to 1$ as $x \to 0$

Letting $x$ tend to $0$ in the above equation implies that $A = 1$.

We now formalize the above claims.


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