De Moivre's Theorem

The De Moivre's formula is given by

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

Let us prove for n = 2

(sin(θ)+cos(θ))2= cos (2θ) + i sin (2θ)

(sin(θ)+cos(θ))2= (cos θ + i sin θ) (cos θ + i sin θ)

                            = cos θ cos θ - sin θ sin θ + 2i cos θ sin θ

                            = cos2θ - sin2θ  + 2i cos θ sin θ

We know

sin (A + B) = sin A cos B + cos A sin B      ................(1)

cos (A + B) = cos A cos B - sin A sin B       ................(2)

if A = B

the above equations can be rewritten as 

sin(2A) = 2 sin A cos A

cos(2A) = Cos2 A − Sin2A

Using above equations 

(sin(θ)+cos(θ))2= cos (2θ) + i sin (2θ)

Proof of De Moivre's Theorem

Assume it is true for n=k

Let (cos θ + i sin θ)^k = (cos kθ + i sin kθ) is True  

Now, prove it is true for "k+1"

 (cos θ + i sin θ)^k+1 = cos(k+1)θ + i sin(k+1)θ 

The left hand side can be written as  

                 (cos θ + i sin θ) ^k  (cos θ + i sin θ)

                 (cos kθ + i sin kθ)  (cos θ + i sin θ)

                 cos kθ(cos θ + i sin θ) + i sin kθ(cos θ + i sin θ)

                 cos kθ cos θ + i cos kθ sin θ + isin kθ cos θ + i² sin kθ sin θ

cos kθ cos θ + i cos kθ sin θ + isin kθ cos θ − sin kθ sin θ

cos kθ cos θ − sin kθ sin θ + i(cos kθ sin θ + sin kθ cos θ)

 Using equations 1 and 2

cos(kθ+θ) + i sin(kθ+θ)

cos((k+1)θ) + i sin((k+1)θ)

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