Euler's Cosine Identity/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:
\(\ds \cos x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n!} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots + \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + \cdots\) |
Recall the definition of the exponential as a power series:
\(\ds e^x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots + \frac {x^n} {n!} + \cdots\) |
Then, starting from the right hand side:
\(\ds \frac {e^{i x} + e^{-i x} } 2\) | \(=\) | \(\ds \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n + \paren {-i x}^n} {n!} }\) | Cosine Function is Absolutely Convergent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} + \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }\) | split into even and odd $n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!}\) | as $\paren {-i x}^{2 n + 1} = -\paren {i x}^{2 n + 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n} } {\paren {2 n}!}\) | as $\paren {-1}^{2 n} = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} } {\paren {2 n}!}\) | cancel $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}\) | as $i^{2 n} = \paren {-1}^n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos x\) |
$\blacksquare$