Complex Number Trigonometry

**Simplicity** · October 30th, 2008, 10:07 AM

I understand the fact:

$\frac{e^{x}+e^{-x}}{2} = \mathrm{cosh}x$

But why does:

$\frac{e^{jx}+e^{-jx}}{2} = \cos x$ ?

**Simplicity** · October 30th, 2008, 10:11 AM

Oops, just remembered:

$e^{jx} = \cos x + j \sin x$

**Chris L T521** · October 30th, 2008, 10:11 AM

Originally Posted by Air

I understand the fact:

$\frac{e^{x}+e^{-x}}{2} = \mathrm{cosh}x$

But why does:

$\frac{e^{jx}+e^{-jx}}{2} = \cos x$ ?

The reason is due to Euler's Formula:

$e^{jx}=\cos(x)+j\sin(x)$

$e^{-jx}=\cos(-x)+j\sin(-x)=\cos(x)-j\sin(x)$ [Due to the even and odd properties of sine and cosine]

Thus, $\frac{e^{jx}+e^{-jx}}{2}=\frac{\cos(x)+j\sin(x)+\cos(x)-j\sin(x)}{2}=\frac{2\cos(x)}{2}=\color{red}\boxed{ \cos(x)}$

Does this make sense?

--Chris

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