Euler formula and relationship between sinx , cosx and exp(x)

Euler formula is given by

eix = cos x + i sin x

A straightforward proof of Euler's formula can be derived using the Taylor power series (My previous article on Taylor's theorem)  

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$

and

$$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots,sinx=x−3!x3​+5!x5​−⋯,

so

$$\begin{aligned} \cos{x} + i \sin{x} &= 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots \\ &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots \\ &= e^{ix}. \end{aligned}cosx+isinx​=1+ix−2!x2​−i3!x3​+4!x4​−⋯=1+ix+2!(ix)2​+3!(ix)3​+4!(ix)4​+⋯=eix.​

$$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots

Adding above two formulas

eix+e−ix=2cosx

$$\cos x=\frac{e^{ix}+e^{-ix}}{2},$$

and similarly

$$\sin{x} = \frac{e^{ix} - e^{-ix}}{2i}

For Hyperbolic functions

What is the relationship between sin x and sinh x?

sinx=eix−e−ix2i 

⟹isinx=eix−e−ix2 

⟹isin(ix)=ei(ix)−e−i(ix)2 

⟹isin(ix)=e−x−ex2 

⟹−isin(ix)=ex−e−x2 

⟹−isin(ix)=sinhx

Similarly cosh x = cos(ix)

De Moivre's Theorem
An important corollary of Euler's theorem is de Moivre's theorem.

(cosx+isinx)ϕ=cosϕx+isinϕx