User:Swissmaster 5000

Test
$$R=\{A\mid A\not\in A\}.$$ 

Derivation of Euler's Formula
On a whim, I took the derivative of this function, which is the reverse of Euler's formula.
 * $$ Let f(x) = (\cos x + i\sin x)\ ( e^{-ix} )\ $$


 * $$\begin{align}

f'(x) &{}= (\cos x + i\sin x)(-i e^{-ix}) + (-\sin x + i\cos x)\ ( e^{-ix} ) \\ &{}= (-i\cos x - i^2\sin x)\ ( e^{-ix} ) + (-\sin x + i\cos x)\ ( e^{-ix} ) \\ &{}= (-i\cos x + \sin x - \sin x + i\cos x)\ ( e^{-ix} ) \\ &{}= 0. \end{align} $$ This showed me that the function is constant, because its derivative is zero. DAMN! Therefore, I could do the following: If

\begin{align} f(0)=(\cos x + i\sin x)\ ( e^{-ix} ) = (\cos 0 + i\sin0)\ ( e^{-i0} ) = 1 \\ \end{align} $$ and
 * $$ e^{-ix} \cdot e^{ix} = e^0 \,, $$

and
 * $$ e^0=1 \,, $$

then, multiplying by $e^{ix}\,$ gives you


 * $$ e^{ix} = \cos x + i \sin x \ .$$