show cosh and cos

**Dgphru** · March 14th 2010, 06:33 PM

Hi, can someone please help me prove this:

$cos(iy) = cosh (y)$

I used eulers formula for complex numbers but i ended up with
$cos (y) = cosh (iy)$

please help or provide any guidance

thanks\ you!!

**tonio** · March 14th 2010, 06:59 PM

Originally Posted by Dgphru

Hi, can someone please help me prove this:

$cos(iy) = cosh (y)$

I used eulers formula for complex numbers but i ended up with
$cos (y) = cosh (iy)$

please help or provide any guidance

thanks\ you!!

Ok, so Euler it is: for $x\in\mathbb{R}\,,\,\,e^{ix}=\cos x+i\sin x\,,\,\,e^{-ix}=\cos x-i\sin x$ . Now sum up both these equations:

$2\cos x=e^{ix}+e^{-ix}\Longrightarrow \cos x=\frac{e^{ix}+e^{-ix}}{2}$ . OTOH, the definition of the hyperbolic cosine is:

$\cosh x=\frac{e^x+e^{-x}}{2}$ . Well, now just put $ix$ intead of $x$ and compare with the above...

Tonio

**drumist** · March 14th 2010, 10:22 PM

Let $y=ix$

Then:

$\cos(y)=\cosh(iy) ~\implies~ \cos(ix)=\cosh(i^2x)=\cosh(-x)=\cosh(x)$

because cosh is an even function.

So basically, since both cos and cosh are even functions, we have that:

$\cos(x) = \cos(-x) = \cosh(ix) = \cosh(-ix)$

and

$\cos(ix) = \cos(-ix) = \cosh(x) = \cosh(-x)$

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