Cosine Exponential Formulation/Real Domain/Proof 1

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 $2$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ $i^{2 n} = \paren {-1}^n$ $\displaystyle$ $=$ $\displaystyle \cos x$

$\blacksquare$