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:

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


Recall the definition of the exponential as a power series:

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


Then, starting from the right hand side:

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

$\blacksquare$


Proof 2

Recall Euler's Formula:

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


Then, starting from the right hand side:

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

$\blacksquare$


Proof 3

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

$\blacksquare$


Proof 4

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 $(1)$.

\(\displaystyle y\) \(=\) \(\displaystyle \cos x\)
\(\displaystyle D^2_x y\) \(=\) \(\displaystyle D^2_x \cos x\) taking second derivative of both sides
\(\displaystyle \) \(=\) \(\displaystyle D_x \left({- \sin x}\right)\) Derivative of Cosine Function
\(\displaystyle \) \(=\) \(\displaystyle - D_x \left({\sin x}\right)\) Derivative of Constant Multiple
\(\displaystyle \) \(=\) \(\displaystyle -\cos x\) Derivative of Sine Function
\(\displaystyle \) \(=\) \(\displaystyle -y\)

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


Then from Cosine of Zero is One:

$\cos 0 = 1$

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


Then:

\(\displaystyle D_x \cos 0\) \(=\) \(\displaystyle - \sin 0\) Derivative of Cosine Function
\(\displaystyle \) \(=\) \(\displaystyle 0\) Sine of Zero is Zero

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

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

$\Box$


Step 2

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

\(\displaystyle z\) \(=\) \(\displaystyle \frac {e^{ix} + e^{-ix} } 2\)
\(\displaystyle D^2_x z\) \(=\) \(\displaystyle D^2_x \left({\frac {e^{i x} + e^{-i x} } 2}\right)\) taking second derivative of both sides
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({D^2_x e^{i x} + D^2_x e^{-i x} }\right)\) Linear Combination of Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({i D_x e^{i x} - i D_x e^{-i x} }\right)\) Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({i^2 e^{i x} - i \left({-i}\right) e^{-i x} }\right)\) Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({- e^{i x} - e^{-i x} }\right)\) $i^2 = -1$
\(\displaystyle \) \(=\) \(\displaystyle -\frac {e^{i x} + e^{-i x} }2\)
\(\displaystyle \) \(=\) \(\displaystyle -z\)

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


Then:

\(\displaystyle \frac {e^{i \times 0} + e^{-i \times 0} } 2\) \(=\) \(\displaystyle \frac {1 + 1} 2\) Exponential of Zero
\(\displaystyle \) \(=\) \(\displaystyle 1\)

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


Then:

\(\displaystyle \left.{D_x \frac {e^{i x} + e^{-i x} } 2}\right\vert_{x \mathop = 0}\) \(=\) \(\displaystyle \left.{\frac {i e^{i x} - i e^{-i x} } 2}\right\vert_{x \mathop = 0}\) Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {i - i} 2\) Exponential of Zero
\(\displaystyle \) \(=\) \(\displaystyle 0\)

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

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

$\Box$


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:

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

$\blacksquare$


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