Euler's Cosine Identity/Proof 2
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Theorem
- $\cos z = \dfrac {e^{i z} + e^{-i z} } 2$
Proof
Recall Euler's Formula:
- $e^{i z} = \cos z + i \sin z$
Then, starting from the right hand side:
\(\ds \frac {e^{i z} + e^{-i z} } 2\) | \(=\) | \(\ds \frac {\cos z + i \sin z + \map \cos {-z} + i \map \sin {-z} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos z + \map \cos {-z} } 2\) | Sine Function is Odd | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \cos z} 2\) | Cosine Function is Even | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos z\) |
$\blacksquare$