Cosine Exponential Formulation/Real Domain

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Theorem

For any real number $x \in \R$:

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

where:

$e^{i x}$ denotes the exponential function
$\cos x$ denotes the real cosine function
$i$ denotes the inaginary unit.


Proof 1

Recall the definition of the real cosine function:

\(\ds \cos x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n!} }\)
\(\ds \) \(=\) \(\ds 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots + \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + \cdots\)


Recall the definition of the exponential as a power series:

\(\ds e^x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)
\(\ds \) \(=\) \(\ds 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots + \frac {x^n} {n!} + \cdots\)


Then, starting from the right hand side:

\(\ds \frac {e^{i x} + e^{-i x} } 2\) \(=\) \(\ds \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n + \paren {-i x}^n} {n!} }\) Cosine Function is Absolutely Convergent
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} + \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }\) split into even and odd $n$
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!}\) $\paren {-i x}^{2 n + 1} = -\paren {i x}^{2 n + 1}$
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n} } {\paren {2 n}!}\) $\left({ -1 }\right)^{2n} = 1$
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} } {\paren {2 n}!}\) cancel $2$
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}\) $i^{2 n} = \paren {-1}^n$
\(\ds \) \(=\) \(\ds \cos x\)

$\blacksquare$


Proof 2

Recall Euler's Formula:

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


Then, starting from the right hand side:

\(\ds \frac {e^{i x} + e^{-i x} } 2\) \(=\) \(\ds \frac {\cos x + i \sin x + \cos \paren {-x} + i \sin \paren {-x} } 2\)
\(\ds \) \(=\) \(\ds \frac {\cos x + \cos \paren {-x} } 2\) Sine Function is Odd
\(\ds \) \(=\) \(\ds \frac {2 \cos x} 2\) Cosine Function is Even
\(\ds \) \(=\) \(\ds \cos x\)

$\blacksquare$


Proof 3

\(\text {(1)}: \quad\) \(\ds e^{i x}\) \(=\) \(\ds \cos x + i \sin x\) Euler's Formula
\(\text {(2)}: \quad\) \(\ds e^{-i x}\) \(=\) \(\ds \cos x - i \sin x\) Euler's Formula: Corollary
\(\ds \leadsto \ \ \) \(\ds e^{i x} + e^{-i x}\) \(=\) \(\ds \paren {\cos x + i \sin x} + \paren {\cos x - i \sin x}\) $(1) + (2)$
\(\ds \) \(=\) \(\ds 2 \cos x\) simplifying
\(\ds \leadsto \ \ \) \(\ds \frac {e^{i x} + e^{-i x} } 2\) \(=\) \(\ds \cos x\)

$\blacksquare$


Proof 4

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:

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

$\blacksquare$


Also presented as

This result can also be presented as:

$\cos x = \dfrac 1 2 \paren {e^{-i x} + e^{i x} }$


Also see


Sources

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