What is the value of $\sin(\sqrt{-1})$ ? What does it signify?

Question

I used the formula $e^{ix} = \cos(x) + i\sin(x)$, substituting $x = i$, we get $1/e = \cos(i) + i\sin(i)$, as real part on LHS $= 0$, $\sin(i) = 0$, is this correct? I don't really understand trigonometric formulas with imaginary inputs.

Answer 1

No, it is not correct. It would be correct if $\cos(i)$ and $\sin(i)$ were real numbers, but they are not. Actually,$$\cos(i)=\frac{e+\frac1e}{2}\text{ and }\sin(i)=\frac{e-\frac1e}2i.$$

Answer 2

$$e^{i(x+iy)}=e^{-y}(\cos x+i\sin x)$$ and $$e^{-i(x+iy)}=e^{y}(\cos x-i\sin x).$$

From this,

$$\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$$ and $$\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y.$$

Answer 3

$\sin i=\sum_{n=0}^\infty (-1)^n\dfrac{i^{2n+1}}{(2n+1)!}=\dfrac{e^{-1}-e}{2i}$. The first equality comes from the Taylor series; the second can be gotten from Euler's formula.