Euler's formula: exponential form of a complex number
$$ \forall x \in \hspace{0.05em} \mathbb{R},$$
$$ e^{ix} = cos(x) + i sin(x) \qquad (Euler's \enspace formula)$$
Euler's trigonometric formulas
$$ \forall x \in \hspace{0.05em} \mathbb{R}, \enspace p \in \mathbb{Z}, $$
$$ \Biggl \{ \begin{gather*} 2cos(px) = e^{ipx} + e^{-ipx} \\ 2i.sin(px) = e^{ipx} - e^{-ipx} \end{gather*} \qquad (Euler's \enspace trigonometric \ formulas) $$
Let \( z \in \mathbb{C}\) a complex number having as a module \( |z|= 1\) under its trigononmetric form such as:
$$ z = cos(\theta) + isin(\theta)$$
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:
$$ cos(x) = 1 -\frac{x^2 }{2!}+ \frac{x^4}{4!} - \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (-1)^{n} \frac{x^{2n}}{(2n)!} + o \bigl(x^{2n+1}\bigr) $$
$$ sin(x) = x -\frac{x^3 }{3!}+ \frac{x^5}{5!} - \frac{x^7}{7!} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o \bigl(x^{2n+2}\bigr) $$
$$ e^{ix} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\hspace{0.1em} \hspace{0.1em} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n}}{(2n)!} + o \bigl(x^{2n+1}\bigr) \hspace{0.1em} + ix - i \frac{x^3}{3!} +i \frac{x^5}{5!} - \hspace{0.1em} ... \hspace{0.1em} + i(-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o \bigl(x^{2n+2}\bigr) $$
$$ e^{ix} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\hspace{0.1em} \hspace{0.1em} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n}}{(2n)!} +o \bigl(x^{2n+1}\bigr) \hspace{0.1em} + i \Biggl(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + o \bigl(x^{2n+2}\bigr)\Biggr) $$
Assuming that all the remainder are tending towards \( 0 \) when \( n \to \infty \):
$$ 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)!}} _\text{ \(cos(x) \)} \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)} _\text{\(sin(x)\)} $$
$$ 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,
$$ \forall x \in \hspace{0.05em} \mathbb{R},$$
$$ e^{ix} = cos(x) + i sin(x) \qquad (Euler's \enspace formula) $$
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.
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:
$$ e^{ip\theta} + e^{-ip\theta} = 2cos(p\theta) $$
Moreover, performing the operation \( (1) - (2) \):
$$ e^{ip\theta} - e^{-ip\theta} = 2isin(p\theta) $$
As a result we obtain that,
$$ \forall x \in \hspace{0.05em} \mathbb{R}, \enspace p \in \mathbb{Z}, $$
$$ \Biggl \{ \begin{gather*} 2cos(px) = e^{ipx} + e^{-ipx} \\ 2i.sin(px) = e^{ipx} - e^{-ipx} \end{gather*} \qquad (Euler's \enspace trigonometric \ formulas) $$
Mainly in the frame of integration, it can be useful to work with simplified forms of trigonometric functions.
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,
$$ cos^3(x) = \frac{1}{4}\Bigl( cos(3x) + 3cos(x) \Bigr)$$
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,
$$cos(px)sin(qx) = \frac{1}{2} \Biggl( sin\Bigl[(p+q)x\Bigr] + sin\Bigl[(-p+q)x\Bigr] \Biggr) $$
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