Cosine Exponential Formulation/Real Domain/Proof 1

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Theorem

$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$


Proof

Recall the definition of the real cosine function:

\(\displaystyle \cos x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n!} }\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots + \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + \cdots\)


Recall the definition of the exponential as a power series:

\(\displaystyle e^x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots + \frac {x^n} {n!} + \cdots\)


Then, starting from the right hand side:

\(\displaystyle \frac {e^{i x} + e^{-i x} } 2\) \(=\) \(\displaystyle \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n + \paren {-i x}^n} {n!} }\) Cosine Function is Absolutely Convergent
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} + \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }\) split into even and odd $n$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!}\) $\paren {-i x}^{2 n + 1} = -\paren {i x}^{2 n + 1}$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n} } {\paren {2 n}!}\) $\left({ -1 }\right)^{2n} = 1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} } {\paren {2 n}!}\) cancel $2$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}\) $i^{2 n} = \paren {-1}^n$
\(\displaystyle \) \(=\) \(\displaystyle \cos x\)

$\blacksquare$