The Euler's trigonometric formulas

Euler's formula: exponential form of a complex number

$$ e^{ix} = cos(x) + i sin(x) \qquad (Euler's \enspace formula)$$

Euler's trigonometric formulas

$$ \Biggl \{ \begin{gather*} 2cos(px) = e^{ipx} + e^{-ipx} \\ 2i.sin(px) = e^{ipx} - e^{-ipx} \end{gather*} \qquad (Euler's \enspace trigonometric \ formulas) $$

Demonstration

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:

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

$$ e^{ix} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\hspace{0.1em} \hspace{0.1em} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n}}{(2n)!} + o \bigl(x^{2n+1}\bigr) \hspace{0.1em} + ix - i \frac{x^3}{3!} +i \frac{x^5}{5!} - \hspace{0.1em} ... \hspace{0.1em} + i(-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o \bigl(x^{2n+2}\bigr) $$

$$ e^{ix} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\hspace{0.1em} \hspace{0.1em} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n}}{(2n)!} +o \bigl(x^{2n+1}\bigr) \hspace{0.1em} + i \Biggl(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o \bigl(x^{2n+2}\bigr)\Biggr) $$

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

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

And finally,

$$ e^{ix} = cos(x) + i sin(x) \qquad (Euler's \enspace formula) $$

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

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:

$$ \Biggl \{ \begin{gather*} z = cos(\theta) + isin(\theta) \\ \overline{z} = cos(\theta) - isin(\theta) \end{gather*} $$

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

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

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

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

$$ \Biggl \{ \begin{gather*} e^{ip\theta}= cos(p\theta) + isin(p\theta) \qquad (1) \\ e^{-ip\theta} = cos(p\theta) - isin(p\theta) \qquad (2) \end{gather*} $$

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

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

As a result we obtain that,

$$ \Biggl \{ \begin{gather*} 2cos(px) = e^{ipx} + e^{-ipx} \\ 2i.sin(px) = e^{ipx} - e^{-ipx} \end{gather*} \qquad (Euler's \enspace trigonometric \ formulas) $$

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:

$$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ (a + b)^n = \sum_{p = 0}^n \binom{n}{p} a^{n-p}b^p $$

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,

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



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:

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

Rearraging the whole expression, we obtain:

$$cos(px)sin(qx) = \Biggl( \frac{e^{ix(p+q)} - e^{-ix(p+q)} + e^{ix(-p+q)} - e^{-ix(-p+q)} }{4i} \Biggr) $$

And as a result,

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