Cosine Complexity

Note first the identity $\cosh(x) = \dfrac{e^{x} + e^{-x}}{2}$ . Then $\cosh(iy) = \dfrac{e^{iy} + e^{-iy}}{2} = \dfrac{(\cos(y) + i \sin(y)) + (\cos(-y) + i \sin(-y))}{2} = \dfrac{2\cos(y)}{2} = \cos(y)$ ,

where Euler's formula $e^{i\theta} = \cos(\theta) + i \sin(\theta)$ was used. We then see that

$\cos(\sqrt{-1}) = \cos(i) = \cosh(i \times i) = \cosh(-1) = \dfrac{e^{-1} + e^{1}}{2} = \dfrac{e^{2} + 1}{2e} \approx \boxed{1.543}$ .

From Euler’s formula: $e^{ix} = cos(x) + i\sin(x)$ ,

making $\cos(x)$ the subject and substituting $i$ , we get

$\cos(x)= \frac{e^{ix}+e^{-ix} }{2}$ .

Substituting $i$ ,

$\cos(i)= \frac{e+e^{-1} }{2}$ , which we know is $\cosh(1)$ , which is 1.54308.

e^(ix) = cos(x) + I sin(x). Letting x =I, then x = -I, e^(-1) cos(I) + I sin(I), and e^(-1) = cos(I) - I sin(I). adding, cos(I) = (1/2) 9e + 1/e)) = 1.543. Ed Gray