what is cosix? how can an angle be imaginary?
In the complex plane, we define $\displaystyle \cos z=\frac{e^{iz}+e^{-iz}}{2}$.
Letting $\displaystyle z=ix$, we have $\displaystyle \cos ix=\frac{e^{i(ix)}+e^{-i(ix)}}{2}=\frac{e^{i^2x}+e^{-i^2x}}{2}=\frac{e^{-x}+e^{-(-x)}}{2}=\frac{e^x+e^{-x}}{2}=\cosh x$
Thus, when evaluating $\displaystyle \cos ix$ at an angle $\displaystyle x$, its the same as evaluating $\displaystyle \cosh x$.
An angle can't be imaginary! But the "trig functions", sine, cosine, etc. are not always related to angles. Sine and cosine, in particular are the "ideal" periodic functions and are often used to model periodic phenomena that have nothing to do with angles (although engineers will still insist upon talking about the "phase angle" of an electrical current where there is no actual angle involved).
The easiest thing to do with complex numbers is to multiply and add them so for more complicated functions it is best to work with power series expansions:
$\displaystyle e^x= 1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot$
$\displaystyle cos(x)= 1- \frac{1}{2!}x^2+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n)!}x^{2n}+ \cdot\cdot\cdot$
$\displaystyle sin(x)= x- \frac{1}{31}x^3+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}x^{2n+1}$
If you replace x by "ix" in the first, you get
$\displaystyle e^{ix}= 1+ ix+ \frac{1}{2}(ix)^2+ \cdot\cdot\cdot+ \frac{1}{n!}(ix)^n+ \cdot\cdot\cdot$
But $\displaystyle i^2= -1$, $\displaystyle i^3= (i^2)(i)= -i$, $\displaystyle i^4= (i^2)(i^2)= (-1)(-1)= 1$ and then $\displaystyle i^5= (i^4)i= i$, etc.
$\displaystyle e^{ix}= 1+ ix- \frac{1}{2}x^2- i\frac{1}{3!}x^3+ \frac{1}{4!}x^4+ \cdot\cdot\cdot$
so all of the "odd" terms have "i" while the "even" terms do not. Separating "real" and "imaginary" parts,
$\displaystyle e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4+ \cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5+ \cdot\cdot\cdot)$
or $\displaystyle e^{ix}= cos(x)+ i sin(x)$. Replacing x by -x, $\displaystyle e^{-ix}= cos(-x)+ i sin(-x)= cos(x)+ i sin(x)$, since cos(-x)= cos(x) and sin(-x)= -sin(x).
Adding those two equations, $\displaystyle e^{ix}+ e^{-ix}= cos(x)+ i sin(x)+ cos(x)- i sin(x)= 2 cos(x)$ so $\displaystyle cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$. In particular, if we now replace x by ix, we have $\displaystyle cos(ix)= \frac{e^{-x}+ e^{x}}{2}= cosh(x)$.
(Which is why you titled this "hyperbolics", right?)
It means
$\displaystyle e^{ix}= 1+ ix+ \frac{1}{2}(ix)^2+ \cdot\cdot\cdot+ \frac{1}{n!}(ix)^n+ \cdot\cdot\cdot$
just as I said above.
From those same formulas you can get $\displaystyle e^{ix}= cos(x)+ i sin(x)$, where, since x is now a real number, cos(x) and sin(x) have their usual significance.