Hi, can someone please help me prove this:
$\displaystyle cos(iy) = cosh (y) $
I used eulers formula for complex numbers but i ended up with
$\displaystyle cos (y) = cosh (iy) $
please help or provide any guidance
thanks\ you!!
Ok, so Euler it is: for $\displaystyle x\in\mathbb{R}\,,\,\,e^{ix}=\cos x+i\sin x\,,\,\,e^{-ix}=\cos x-i\sin x$ . Now sum up both these equations:
$\displaystyle 2\cos x=e^{ix}+e^{-ix}\Longrightarrow \cos x=\frac{e^{ix}+e^{-ix}}{2}$ . OTOH, the definition of the hyperbolic cosine is:
$\displaystyle \cosh x=\frac{e^x+e^{-x}}{2}$ . Well, now just put $\displaystyle ix$ intead of $\displaystyle x$ and compare with the above...
Tonio
Let $\displaystyle y=ix$
Then:
$\displaystyle \cos(y)=\cosh(iy) ~\implies~ \cos(ix)=\cosh(i^2x)=\cosh(-x)=\cosh(x)$
because cosh is an even function.
So basically, since both cos and cosh are even functions, we have that:
$\displaystyle \cos(x) = \cos(-x) = \cosh(ix) = \cosh(-ix)$
and
$\displaystyle \cos(ix) = \cos(-ix) = \cosh(x) = \cosh(-x)$