Cosine Exponential Formulation/Real Domain/Proof 4

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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.


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Therefore $y = z$.

That is:

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

$\blacksquare$

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