You don't magically remove the i, but you are on the right track. Input ix into cos. So you get cos(ix). This will give you cosh(x). So the relationship between cos and cosh is cos(ix) = cosh(x).
I'd actually write it the other way: cosh(ix) = cos(x) and sinh(ix) = isin(x). In general, exp(x) = cosh(x) + sinh(x), so exp(ix) = cosh(ix) + sinh(ix) = cos(x) + isin(x). I'd also like to note that i doesn't need to be _the_ imaginary unit, it can be _anything_ as long as its square is -1, so the formula also works for any normalised pure quaternion. What happens if the input is multiplied by some unit with a square other than -1? Well there are really only two other cases to consider: j² = 1 and ε² = 0. exp(jx) = sinh(x) + jcosh(x). This is the Split-complex Euler's formula. On the other hand, exp(εx) = cosh(εx) + sinh(εx) = 1 + εx. This is the dual number Euler's formula, and yes, the cosine equivalent really is a constant 1 and the sine equivalent is the identity function.
You don't magically remove the i, but you are on the right track. Input ix into cos. So you get cos(ix). This will give you cosh(x). So the relationship between cos and cosh is cos(ix) = cosh(x).
I'd actually write it the other way: cosh(ix) = cos(x) and sinh(ix) = isin(x). In general, exp(x) = cosh(x) + sinh(x), so exp(ix) = cosh(ix) + sinh(ix) = cos(x) + isin(x).
I'd also like to note that i doesn't need to be _the_ imaginary unit, it can be _anything_ as long as its square is -1, so the formula also works for any normalised pure quaternion.
What happens if the input is multiplied by some unit with a square other than -1? Well there are really only two other cases to consider: j² = 1 and ε² = 0.
exp(jx) = sinh(x) + jcosh(x). This is the Split-complex Euler's formula. On the other hand, exp(εx) = cosh(εx) + sinh(εx) = 1 + εx. This is the dual number Euler's formula, and yes, the cosine equivalent really is a constant 1 and the sine equivalent is the identity function.