𝑒^𝑖𝑥 = 𝑐𝑜𝑠(𝑥) + 𝑖·𝑠𝑖𝑛(𝑥)

derivations

Derivation #0 - via Power/Taylor/MacLaurin Series Expansion Taylor series expansion of 𝑒𝑖𝑥:

• $e^{ix}=\frac{x^0}{0!}+\frac{i{x}^1}{1!}-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}-\frac{x^6}{6!}...$

Grouping the real and imaginary terms together yields:

• $e^{ix}=(\frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)+i(\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-...)$

Let’s consider the Taylor series of sine and the Taylor series of cosine:

• $sin(x)=\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$
• $cos(x)=\frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}...$

Thus:

• $e^{ix}=cos(x)+i\cdot sin(x)$

Let’s consider the

Derivation #2 - via trigonometry and complex numbers Making Sense of Euler's Formula with Trigonometry & Complex Numbers

see: 

Derivation #3 - via Group Theory Euler's Formula - Intuition via Group Theory

see: 

𝑒^{±𝑖𝑘𝑥}= 𝑐𝑜𝑠(𝑘𝑥) ± 𝑖·𝑠𝑖𝑛(𝑘𝑥)

where 𝑥=𝑘𝑥

𝑒^𝑖𝜋 = −1

where 𝑥=𝜋 (i.e. Euler’s identity)