Possible complex angles with no imaginary periodicity

[Some_dude91]

Trying the already known equation, sin^2(x) + cos^2(x) = 1 i wondered what would happen
if i took that either sin(x) or cos(x) squared equalled a number greater than 1, so when i plugged in sin(x) as 5/3 i got cos(x) 4i/3 ,
went to euler's equation and added the results,
then put the aggregate to the logarithm and got
angle = 2kp + pi/2 ± 1.0986i. Solving for cos(x) as well i got an imaginary angle equal to
angle = ±1.0986i,
but as i went on having the number grow larger, the imaginary part would grow proportionately. Now, based on the results i got, i saw such angles only poccessed periodicity on the real, unchanged, part whereas the imaginary ones poccessed none. The plus or minus part was due to the fact that for 1 possible cosine (or sine), the resulting sine (or cosine) was either positive or negative, on the equation
sin^2(x) + cos^2(x) = 1

I don't know if this is wrong but it yields an interesting result that might be telling something about a different, outside the usual trigonometry approach. However, seeing as i treated cosine and sine as continuous functions, and knowing that they can be used only to satisfy an equation, just numerically, i might be wrong, but based on the modification on the exponential function by euler,
e^x = cosh(x) + sinh(x) converted to e^ix = cos(x) + isin(x) i did the inverse, and got that for imaginary angles, cos(x) becomes cosh(x) and sin(x) becomes i*sinh(x), based on an approach
by the taylor series.

Also by using the trigonometrical identities sin(x+y) and cos(x+y) i got that
sin(x+iy) = sin(x)cosh(y) + i*cos(x)sinh(y) and
cos(x+iy) = cos(x)cosh(x) - i*sin(x)sinh(y) which in fact do satisfy the equation i did.
Could there really be such angles? Could this approach be right?

 

[mathman]

The term angle usually means a real number. However sin(z) or cos(z) where z is a complex number is a well defined function. When z is purely imaginary, then hyperbolic functions will be equivalent.

cos(ix) = cosh(x), sin(ix) = isinh(x)