Cosine of Complex Number/Proof 2
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Theorem
Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
- $\cos \left({a + b i}\right) = \cos a \cosh b - i \sin a \sinh b$
where:
- $\cos$ denotes the cosine function (real and complex)
- $\sin$ denotes the real sine function
- $\sinh$ denotes the hyperbolic sine function
- $\cosh$ denotes the hyperbolic cosine function
Proof
\(\ds \cos a \cosh b - i \sin a \sinh b\) | \(=\) | \(\ds \frac {e^{i a} + e^{-i a} } 2 \frac {e^b + e^{-b} } 2 - i \frac {e^{i a} - e^{-i a} } {2 i} \frac {e^b - e^{-b} } 2\) | Euler's Cosine Identity, Definition of Hyperbolic Cosine, Euler's Sine Identity, Definition of Hyperbolic Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{b + i a} + e^{-b + i a} + e^{b - i a} + e^{-b - i a} - e^{b + i a} + e^{-b + i a} + e^{b - i a} - e^{-b - i a} } 4\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{-b + i a} + e^{b - i a} } 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{i \paren {a + b i} } + e^{-i \paren {a + b i} } } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \cos {a + b i}\) | Euler's Cosine Identity |
$\blacksquare$