Cosine Exponential Formulation

Theorem
Let $x \in \R$ be a real number.

Let $\sin x$ be the sine of $x$.

Let $\cos x$ be the cosine of $x$.

Let $i$ be such that $i^2 = -1$.

Then:
 * $ \displaystyle \sin x = \frac 1 2 i \left({ e^{-i x} - e^{i x} }\right)$


 * $ \displaystyle \cos x = \frac 1 2 \left({ e^{-i x} + e^{i x} }\right)$

Proof
Recall the definition of the sine function and cosine function:


 * $\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$


 * $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

Recall the definition of the exponential as a power series:


 * $\displaystyle e^x = \sum_{n=0}^\infty \frac {x^n}{n!} = 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots$

Then, starting from the RHS:

Similarly: