Cosine Exponential Formulation/Real Domain/Proof 4

Proof
Consider the differential equation:
 * $(1): \quad D^2_x f \left({x}\right) = - f\left({x}\right)$

subject to the initial conditions
 * $(2): \quad f \left({0}\right) = 1$


 * $(3): \quad D_x f \left({0}\right) = 0$

Step 1
We will prove that $y = \cos x$ is a specific solution of $(1)$.

Thus $y = \cos x$ fulfils $(1)$.

Then from Cosine of Zero is One:
 * $\cos 0 = 1$

Thus $y = \cos x$ fulfils $(2)$.

Then:

Thus $y = \cos x$ fulfils $(3)$.

So $y = \cos x$ is a specific solution of $(1)$.

Step 2
We will prove that $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a specific solution of $(1)$.

Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(1)$.

Then:

Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(2)$.

Then:

Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(3)$.

So $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a specific solution of $(1)$.

We have shown that $y$ and $z$ are both specific specific solutions of $(1)$

But a specific solution to a differential equation is unique.

Therefore $y = z$.

That is:
 * $\cos x = \dfrac {e^{i x} + e^{-i x} } 2$