Trigonometric values of complex Numbers Is it sensible and mathematically correct to extend the definitions of the trigonometric ratios into the complex plane?  If yes, then how And why?
Also, does the following set of equations hold true?
$$e^{ix} = \cos(x) + i\sin(x)$$
If we let $x=i$, then we have
\begin{align}
e^{i^{2}} &= \cos(i) + i\sin(i)\\
e^{-1} &= \cos(i) + i\sin(i)\\
1&=e(\cos(i) + i\sin(i))
\end{align}
Also ,what is the need to define trigonometric ratios in the complex plane??
What are it's practical implementations?
 A: Yes, this is all correct.
From the basic formulas $e^{\pm it}=\cos t \pm i\sin t$ one has
$$\cos t = \frac{e^{it}+e^{-it}}{2}$$ $$\sin t = \frac{e^{it}-e^{-it}}{2i}$$
It's also easy to show that
$$\cos it = \cosh t$$ $$\sin it = i\sinh t$$ which, with the usual addition formulas, gives
$$\cos (x+iy) = \cos x \cosh y - i\sin x\sinh y$$
$$\sin(x+iy) = \sin x \cosh y + i \cos x\sinh y$$
All of the other trig functions are defined in the obvious way using these.
Hyperbolic functions, and inverse trig/hyperbolic functions work the same way.
A: 
Is it sensible and mathematically correct to extend the definitions of trigonometric ratios in the complex plane?

Yes

If yes, then how?

Starting from $e^{ix} = \cos(x) + i\sin(x)$, we can substitute $x := -x$ to get $e^{-ix} = \cos(-x) + i\sin(-x)$, and then by standard properties we get $e^{-ix} = \cos(x) - i\sin(x)$. Therefore $$\cos(x) = \frac{e^{ix} + e^{-ix}}2 \\ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

And why?

One option is to define the three functions by their Taylor expansions: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \\
\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \\
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
