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The Euler's trigonometric formulas

$$ \forall x \in \hspace{0.04em} \mathbb{R},$$
$$ e^{ix} = \cos(x) + i \sin(x) \qquad \bigl(\text{Euler's formula}\bigr) $$
$$ \forall x \in \hspace{0.04em} \mathbb{R}, \enspace p \in \mathbb{Z}, $$
$$ \Biggl \{ \begin{gather*} 2 \cos(px) = e^{ipx} + e^{-ipx} \\ 2i \sin(px) = e^{ipx} - e^{-ipx} \end{gather*} \qquad \bigl(\text{Euler's trigonometric formulas}\bigr) $$

Proofs

Euler's formula: exponential form of a complex number

Let \( z \in \mathbb{C}\) a complex number having as a module \( |z|= 1\) under its trigononmetric form such as:

$$ z = \cos(\theta) + i \hspace{0.2em} \sin(\theta)$$

We know that the Taylor series of \(e^x\) at \(0\) is:

$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} \frac{x^n}{n!} + o \bigl(x^n\bigr) $$

Thus, if we develop a Taylor series of \(e^{ix}\) at \(0\), we do have:

$$ e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} \frac{(ix)^n}{n!} + o \Bigl((ix)^n\Bigr) $$
$$ e^{ix} = 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} \frac{(ix)^n}{n!} + o \Bigl((ix)^n\Bigr) $$

Now, we notice two known Taylor series , those of the \(\cos(x)\) and \(\sin(x)\) functions:

$$ \cos(x) = 1 -\frac{x^2 }{2!}+ \frac{x^4}{4!} + \ ... \ + \hspace{0.2em} (-1)^{n} \frac{x^{2n}}{(2n)!} + o \bigl(x^{2n}\bigr) $$
$$ \sin(x) = x -\frac{x^3 }{3!}+ \frac{x^5}{5!} - \frac{x^7}{7!} + \ ... \ + \hspace{0.2em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o \bigl(x^{2n}\bigr) $$
$$ e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots + (-1)^{n} \frac{x^{2n}}{(2n)!} \right) + ix - i \frac{x^3}{3!} + i \frac{x^5}{5!} - \dots + i(-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o\bigl(x^{2n+1}\bigr) $$
$$ e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots + (-1)^{n} \frac{x^{2n}}{(2n)!} \right) + i \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots + (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \right) + o\bigl(x^{2n+1}\bigr) $$

Assuming that all the remainder are tending towards \( 0 \) when \( n \to \infty \):

$$ e^{ix} = \underbrace { 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\hspace{0.1em} \hspace{0.1em} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n}}{(2n)!}} _{\cos(x)} \hspace{0.1em} + i \times \underbrace {\Biggl(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \Biggr)} _{\sin(x)} $$
$$ e^{ix} = \cos(x) + i \sin(x) $$

Thus, the complex number \( z \) having as a module \( |z|= 1\) can be written under its exponential form:

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

And finally,

$$ \forall x \in \hspace{0.04em} \mathbb{R},$$
$$ e^{ix} = \cos(x) + i \sin(x) \qquad \bigl(\text{Euler's formula}\bigr) $$

Now, whatever complex number \( z \) will be written as:

$$ z = x + iy \Longleftrightarrow z = |z|e^{i\theta} = |z| \bigl(\cos(\theta) + i \sin(\theta)\bigr) $$
$$ \text{with} \enspace \left \{ \begin{gather*} |z| = \sqrt{x^2 + y^2} \\ \cos(\theta) = \frac{x}{|z|} \\ \sin(\theta) = \frac{y}{|z|} \end{gather*} \right \}$$

Be careful not to get mixed up with the "\( x \)" of \( e^{ix} \) of the Euler's formula with the one in a complex number written\( z = x + iy \). In the Euler's formula, le "\( x \)" represents an angle, or even the argument of a complex .

Euler's trigonometric formulas

Let \( z \in \mathbb{C}\) be a complex number having as a module under its trigonometric form, and its conjugate \( \overline{z} \) such as:

$$ \Biggl \{ \begin{gather*} z = \cos(\theta) + i \hspace{0.2em} \sin(\theta) \\ \overline{z} = \cos(\theta) - i \hspace{0.2em} \sin(\theta) \end{gather*} $$

With the exponential form of the complex numbers seen above, we can rewrite this two expressions as:

$$ \Biggl \{ \begin{gather*} z = e^{i\theta} \\ \overline{z} = e^{-i\theta} \end{gather*} \qquad \Bigl(\text{with } \overline{z} = \cos(\theta) - i \hspace{0.2em} \sin(\theta) \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} \overline{z} = \cos(-\theta) + i \sin(-\theta) \Bigr) $$

Raising those two expressions at the power of \(p\):

$$ \Biggl \{ \begin{gather*} z^p = \bigl(e^{i\theta}\bigr)^p = e^{ip\theta} = \cos(p\theta) + i \sin(p\theta) \\ (\overline{z})^p = \bigl(e^{-i\theta}\bigr)^p = e^{-ip\theta} = \cos(p\theta) - i \sin(p\theta) \end{gather*} $$
$$ \Biggl \{ \begin{gather*} e^{ip\theta}= \cos(p\theta) + i \sin(p\theta) \qquad (1) \\ e^{-ip\theta} = \cos(p\theta) - i \sin(p\theta) \qquad (2) \end{gather*} $$

Now, performing the operation \( (1) + (2) \), we do have:

$$ e^{ip\theta} + e^{-ip\theta} = 2 \cos(p\theta) $$

Moreover, performing the operation \( (1) - (2) \):

$$ e^{ip\theta} - e^{-ip\theta} = 2i \sin(p\theta) $$

As a result we obtain that,

$$ \forall x \in \hspace{0.04em} \mathbb{R}, \enspace p \in \mathbb{Z}, $$
$$ \Biggl \{ \begin{gather*} 2 \cos(px) = e^{ipx} + e^{-ipx} \\ 2i \sin(px) = e^{ipx} - e^{-ipx} \end{gather*} \qquad \bigl(\text{Euler's trigonometric formulas}\bigr) $$

Examples

Mainly in the frame of integration , it can be useful to work with simplified forms of trigonometric functions.

  1. Linearization of a trigonometric function

    If we want to linearize \(\cos^3(x) \), we do have:

    $$ \cos^3(x) = \Biggl( \frac{e^{ix} + e^{-ix}}{2} \Biggr)^3 $$

    But, we know from the Newton's binomial that:

    $$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{R}^2,$$
    $$ (a + b)^n = \sum_{p = 0}^n \binom{n}{p} a^{n-p}b^p $$

    Hence,

    $$ \left(e^{ix} + e^{-ix}\right) ^3 = e^{3ix} + 3e^{ix} + 3e^{-ix} + e^{-3ix} $$

    Injecting now our initial expression:

    $$ \cos^3(x) = \frac{e^{3ix} + e^{-3ix} + 3e^{ix} + 3e^{-ix}}{8} $$
    $$ \cos^3(x) = \frac{1}{4}\Biggl( \frac{e^{3ix} + e^{-3ix}}{2} \Biggr) + \frac{3}{4}\Biggl( \frac{e^{ix} + e^{-ix}}{2} \Biggr) $$

    And finally,

    $$ \cos^3(x) = \frac{1}{4}\Bigl( \cos(3x) + 3\cos(x) \Bigr)$$

  2. To transform a trigonometric product into a sum

    If we want to transform the product \(\cos(px)\sin(qx) \) into a sum, we do have:

    $$\cos(px)\sin(qx) = \Biggl( \frac{e^{ipx} + e^{-ipx}}{2} \Biggr) \Biggl( \frac{e^{iqx} - e^{-iqx}}{2i} \Biggr) $$
    $$\cos(px)\sin(qx) = \frac{e^{i(p+q)x} - e^{i(p-q)x} + e^{-i(p-q)x} - e^{-i(p+q)x}}{4i} $$
    $$\cos(px)\sin(qx) = \frac{1}{2} \left( \frac{e^{i(p+q)x} - e^{-i(p+q)x}}{2i} - \frac{e^{i(p-q)x} - e^{-i(p-q)x}}{2i} \right) $$

    And as a result,

    $$\cos(px)\sin(qx) = \frac{1}{2} \Biggl( \sin\Bigl[(p+q)x\Bigr] + \sin\Bigl[(-p+q)x\Bigr] \Biggr) $$
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