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