# Cosine Exponential Formulation/Real Domain/Proof 4

## Theorem

$\cos z = \dfrac {\map \exp {i z} + \map \exp {-i z} } 2$

## Proof

Consider the differential equation:

$(1): \quad D^2_x \map f x = -\map f x$

subject to the initial conditions:

$(2): \quad \map f 0 = 1$
$(3): \quad D_x \map f 0 = 0$

### Step 1

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

 $\ds y$ $=$ $\ds \cos x$ $\ds D^2_x y$ $=$ $\ds D^2_x \cos x$ taking second derivative of both sides $\ds$ $=$ $\ds \map {D_x} {-\sin x}$ Derivative of Cosine Function $\ds$ $=$ $\ds -\map {D_x} {\sin x}$ Derivative of Constant Multiple $\ds$ $=$ $\ds -\cos x$ Derivative of Sine Function $\ds$ $=$ $\ds -y$

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

Then from Cosine of Zero is One:

$\cos 0 = 1$

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

Then:

 $\ds D_x \cos 0$ $=$ $\ds -\sin 0$ Derivative of Cosine Function $\ds$ $=$ $\ds 0$ Sine of Zero is Zero

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

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

$\Box$

### Step 2

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

 $\ds z$ $=$ $\ds \frac {e^{i x} + e^{-i x} } 2$ $\ds D^2_x z$ $=$ $\ds \map {D^2_x} {\frac {e^{i x} + e^{-i x} } 2}$ taking second derivative of both sides $\ds$ $=$ $\ds \frac 1 2 \paren {D^2_x e^{i x} + D^2_x e^{-i x} }$ Linear Combination of Derivatives $\ds$ $=$ $\ds \frac 1 2 \paren {i D_x e^{i x} - i D_x e^{-i x} }$ Derivative of Exponential Function $\ds$ $=$ $\ds \frac 1 2 \paren {i^2 e^{i x} - i \paren {-i} e^{-i x} }$ Derivative of Exponential Function $\ds$ $=$ $\ds \frac 1 2 \paren {- e^{i x} - e^{-i x} }$ $i^2 = -1$ $\ds$ $=$ $\ds -\frac {e^{i x} + e^{-i x} }2$ $\ds$ $=$ $\ds -z$

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

Then:

 $\ds \frac {e^{i \times 0} + e^{-i \times 0} } 2$ $=$ $\ds \frac {1 + 1} 2$ Exponential of Zero $\ds$ $=$ $\ds 1$

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

Then:

 $\ds \intlimits {D_x \frac {e^{i x} + e^{-i x} } 2} {x \mathop = 0} {}$ $=$ $\ds \intlimits {\frac {i e^{i x} - i e^{-i x} } 2} {x \mathop = 0} {}$ Derivative of Exponential Function $\ds$ $=$ $\ds \frac {i - i} 2$ Exponential of Zero $\ds$ $=$ $\ds 0$

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

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

$\Box$

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

But a particular solution to a differential equation is unique.

Therefore $y = z$.

That is:

$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$

$\blacksquare$