Euler's Cosine Identity

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Theorem

For any complex number $z \in \C$:

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

where:

$\exp z$ denotes the exponential function
$\cos z$ denotes the complex cosine function
$i$ denotes the imaginary unit.


Real Domain

This result is often presented and proved separately for arguments in the real domain:


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


Proof 1

Recall the definition of the cosine function:

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


Recall the definition of the exponential as a power series:

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


Then, starting from the right hand side:

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

$\blacksquare$


Proof 2

Recall Euler's Formula:

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


Then, starting from the right hand side:

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

$\blacksquare$


Proof 3

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

$\blacksquare$


Also presented as

Euler's Cosine Identity can also be presented as:

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


Also see


Source of Name

This entry was named for Leonhard Paul Euler.


Sources

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