Euler Formula for Cosine Function

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Theorem

\(\ds \cos x\) \(=\) \(\ds \prod_{n \mathop = 0}^\infty \paren {1 - \frac {4 x^2} {\paren {2 n + 1}^2 \pi^2} }\)
\(\ds \) \(=\) \(\ds \paren {1 - \dfrac {4 x^2} {\pi^2} } \paren {1 - \dfrac {4 x^2} {9 \pi^2} } \paren {1 - \dfrac {4 x^2} {25 \pi^2} } \dotsm\)


Proof

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

$\cos x = 1 - \dfrac {x^2} {2!} + \dfrac {x^4} {4!} - \dfrac {x^6} {6!} + \dotsb$

The roots of cosine are the numbers $\paren {k + \dfrac 1 2} \pi$, where $k$ is any integer.

From the Polynomial Factor Theorem, the following (after simplification) might be true:

$\ds \cos x = A \prod \paren {1 - \frac {2 x} {\paren {2 k + 1} \pi} }$

where the product is taken over all $k \in \Z$, and $A$ is some constant.

The intuition is as follows.

\(\ds \cos x\) \(=\) \(\ds \ldots \paren {1 - \frac {2 x} {5 \pi} } \paren {1 - \frac {2 x} {3 \pi} } \paren {1 - \frac {2 x} \pi} A \paren {1 + \frac {2 x} \pi} \paren {1 + \frac {2 x} {3 \pi} } \paren {1 + \frac {2 x} {5 \pi} } \cdots\)
\(\ds \) \(=\) \(\ds A \paren {1 - \frac {4 x^2} {\pi^2} } \paren {1 - \frac {4 x^2} {3^2 \pi^2} } \paren {1 - \frac {4 x^2} {5^2 \pi^2} } \cdots\)

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

It remains to formalize the above claims.


Also see


Sources