Proofs
Rewriting the
complex number
having as
argument
\( 2\alpha\) under its
complex exponential form
, we do have:
$$ e^{i2\alpha} = (e^{i\alpha})^2 = (\cos(\alpha) + i.\sin(\alpha))^2 $$
$$ e^{i2\alpha} = \Bigl[\cos(\alpha) + i.\sin(\alpha) \Bigr] \Bigl[\cos(\alpha) + i.\sin(\alpha) \Bigr] $$
We develop the above expression:
$$ e^{i2\alpha} = \cos^2(\alpha) + 2 i.\sin(\alpha)\cos(\alpha) - \sin^2(\alpha) $$
$$ e^{i(\alpha + \beta)} = \enspace \underbrace{\cos^2(\alpha) - \sin^2(\alpha)} _\text{real part} \enspace + \enspace i.\underbrace{2
\sin(\alpha)\cos(\alpha) } _\text{imaginary part} $$
Let us identify the respective real and imaginary parts of the
complex number
\(e^{2i\alpha} \) :
$$ \mathcal{Re}\left(e^{2i\alpha}\right) = \cos^2(\alpha) - \sin^2(\alpha) $$
$$\mathcal{Im}\left(e^{2i\alpha}\right) = 2sin(\alpha)\cos(\alpha) $$
But it is known that:
$$ \mathcal{Re}\left(e^{2i\alpha}\right) = \cos(2\alpha) $$
$$ \mathcal{Im}\left(e^{2i\alpha}\right) = \sin(2\alpha) $$
As a result we do have,
$$ \forall \alpha \in \mathbb{R}, $$
Using the famous formula \( \cos^2(\alpha) + \sin^2(\alpha) = 1 \), we can retrieve the two other formulas.
$$ \forall \alpha \in \mathbb{R}, $$
We know from the definition of the tangent function that:
$$ \tan(2\alpha) = \frac{\sin(2\alpha) }{\cos(2\alpha) }$$
-
Expression of \( \sin(2 \alpha) = f(\tan( \alpha)) \)
Now,
$$ \sin(2\alpha) = 2 \sin(\alpha) \cos(\alpha)$$
$$ \sin(2\alpha) = \frac{2 \sin(\alpha) \cos^2(\alpha)}{\cos(\alpha)}$$
But, this is also
the derivative of \( \tan(x) \)
, which is worth:
$$ \tan(x)' = 1 + \tan^2(\alpha) = \frac{1}{\cos^2(\alpha)}$$
And,
$$ \cos^2(\alpha) = \frac{1}{1 + \tan^2(\alpha)}$$
Consequently we do have now,
$$ \sin(2\alpha) = \frac{2 \sin(\alpha)}{\cos(\alpha) (1 + \tan^2(\alpha)) }$$
$$ \forall \alpha \in \mathbb{R}, $$
$$ \sin(2\alpha) = \frac{2 \tan(\alpha)}{ 1 + \tan^2(\alpha) }$$
-
Expression of \( \cos(2 \alpha) = f(\tan( \alpha)) \)
As well,
$$ \cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha)$$
$$ \cos(2\alpha) = \frac{1}{1 + \tan^2(\alpha)} - \sin^2(\alpha)$$
$$ \cos(2\alpha) = \frac{1}{1 + \tan^2(\alpha)} - \tan^2(\alpha)\cos^2(\alpha)$$
$$ \forall \alpha \in \mathbb{R}, $$
$$ \cos(2\alpha) = \frac{1 - \tan^2(\alpha)}{1 + \tan^2(\alpha)} $$
-
Expression of \( \tan(2 \alpha) \)
Finally,
$$ \tan(2\alpha) = \frac{\sin(2\alpha) }{\cos(2\alpha) }$$
$$ \tan(2\alpha) = \frac{2 \tan(\alpha)}{ 1 + \tan^2(\alpha) } . \frac{1 + \tan^2(\alpha)}{ 1 - \tan^2(\alpha) }$$
$$ \tan(2\alpha) = \frac{2 \tan(\alpha)}{1 - \tan^2(\alpha)} $$
And as a result,
$$ \forall k \in \mathbb{Z}, \enspace \forall \alpha \in \Biggl[ \mathbb{R} \hspace{0.2em} \backslash \left \{\frac{\pi}{4} + \frac{k\pi}{2} \right \}
\Biggr], $$
$$ \tan(2\alpha) = \frac{2tan(\alpha)}{1 -\tan^2(\alpha)} $$
Rewriting the
complex number
having as
argument
\( (\alpha + \beta)\) under its
complex exponential form
, we do have:
$$ e^{i(\alpha + \beta)} = \cos(\alpha + \beta) + i.\sin(\alpha + \beta) $$
Now, factorizing the power we do have:
$$ e^{i(\alpha + \beta)} = e^{i\alpha + i\beta } $$
And,
$$ e^{i(\alpha + \beta)} = e^{i\alpha} . e^{i\beta} \qquad (2) $$
We rewrite \( (2) \) under its trigonometric form:
$$ e^{i(\alpha + \beta)} = (\cos(\alpha) + i.\sin(\alpha)) (\cos(\beta) + i.\sin(\beta)) $$
And we develop it:
$$ e^{i(\alpha + \beta)} = \cos(\alpha)\cos(\beta) + i.\cos(\alpha)\sin(\beta) + i.\sin(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) $$
$$ e^{i(\alpha + \beta)} = \enspace \underbrace{\cos(\alpha)\cos(\beta)- \sin(\alpha)\sin(\beta)} _\text{real part} \enspace + \enspace
i.\underbrace{\bigl[\cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta)\bigr]} _\text{imaginary part} $$
Let us identify the respective real and imaginary parts of the
complex number
\(e^{i(\alpha + \beta)} \) :
$$ \mathcal{Re}\left[e^{i(\alpha + \beta)}\right] = \cos(\alpha)\cos(\beta)- \sin(\alpha)\sin(\beta)$$
$$\mathcal{Im}\left[e^{i(\alpha + \beta)}\right] = \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta)$$
But, we know that:
$$ \Biggl \{ \begin{gather*} \mathcal{Re}\left[e^{i(\alpha + \beta)}\right] = \cos(\alpha + \beta) \\ \mathcal{Im}\left[e^{i(\alpha + \beta)}\right]
= \sin(\alpha + \beta) \end{gather*} $$
And as a result,
$$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$
Using the same process again, it is possible to recover the two other missing formulas:
$$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$
$$ \tan(\alpha + \beta ) = \frac{\sin(\alpha + \beta )}{\cos(\alpha + \beta )}$$
With
the sines and cosines addition formulas
, we do have:
$$ \tan(\alpha + \beta ) = \frac{\sin(\alpha) \cos(\beta) + \sin(\beta) \cos(\alpha)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)}$$
Let us multiply both numerator and denominator by \(\cos(\alpha) \cos(\beta)\) :
$$ \tan(\alpha + \beta ) = \frac{\sin(\alpha) \cos(\beta) + \sin(\beta)\cos(\alpha) }{\cos(\alpha) \cos(\beta)} . \frac{\cos(\alpha)
\cos(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)}$$
$$ \tan(\alpha + \beta ) = (\tan(\alpha) + \tan(\beta)). \frac{1}{ \frac{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)}{\cos(\alpha)
\cos(\beta)}}$$
$$ \tan(\alpha + \beta ) = (\tan(\alpha) + \tan(\beta)). \frac{1}{ 1 - \tan(\alpha)\tan(\beta) }$$
And finally,
$$\forall k \in \mathbb{Z}, \enspace \forall (\alpha, \beta) \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \left \{ \frac{\pi}{2} + k\pi \right \}
\biggr]^2, \enspace \forall m \in \mathbb{Z}, \Bigl[(\alpha + \beta) \neq \pi + 2m\pi \Bigr],$$
$$ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{ 1 - \tan(\alpha)\tan(\beta) }$$
We can also notice that there is a simple sign change from \( \sin( \alpha + \beta) \) to \( \sin( \alpha - \beta) \), as well as the bridge from \(
\cos( \alpha + \beta) \) to \( \cos( \alpha - \beta) \).
Thus, we easily find the same sign change from \( \tan( \alpha + \beta) \) to \( \tan( \alpha - \beta) \):
$$\forall k \in \mathbb{Z}, \enspace \forall (\alpha, \beta) \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \left \{ \frac{\pi}{2} + k\pi \right \}
\biggr]^2, \enspace \forall m \in \mathbb{Z}, \Bigl[(\alpha - \beta) \neq \pi + 2m\pi \Bigr],$$
$$ \tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{ 1 + \tan(\alpha)\tan(\beta) }$$
The idea is to find a general expression for the following couple:
$$ \Biggl \{ \begin{gather*} \cos(nx) \\ \sin(nx) \end{gather*} $$
by determining both real and imaginary parts of the
complex number
\( e^{inx} \).
Rewriting the
complex number
having as
argument
\( nx\) under its
complex exponential form
, we do have:
$$ e^{i(nx)} = \hspace{0.1em} \underbrace { \cos(nx) } _\text { \(\mathcal{Re}(e^{i(nx)} ) \) } \hspace{0.1em} + \hspace{0.1em} i. \underbrace {
\sin(nx) } _\text {\(\mathcal{Im}(e^{i(nx)} ) \)} = (e^{ix})^n \qquad (1) $$
Now,
$$ (e^{ix})^n = \Bigl[\cos(x) + i.\sin(x)\Bigr]^n $$
So, we can apply
the Newton's binomial
which says:
$$\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 \qquad (Newton ) $$
So in our case,
$$ (e^{ix})^n = \sum_{p = 0}^n \binom{n}{p} \cos^{n-p}(x) \ i^p.\sin^p(x) $$
$$ (e^{ix})^n = \cos^n(x) + i \binom{n}{1}\cos^{n-1}(x) \ \sin(x) - \binom{n}{2}\cos^{n-2}(x)\sin^2(x) -i\binom{n}{3}\cos^{n-3}(x) \ \sin^3(x) +
\binom{n}{4}\cos^{n-4}(x) \ \sin^4(x) + \hspace{0.1em} ...$$
$$ \hspace{2em} ... \hspace{0.1em} + \binom{n}{n-3}\cos^3(x) \ i^{n-3}.\sin^{n-3}(x) + \binom{n}{n-2}\cos^2(x) \ i^{n-2}.\sin^{n-2}(x) +
\binom{n}{n-1}\cos(x) \ i^{n-1}.\sin^{n-1}(x) + i^n \ \sin^n(x) $$
It is impossible to assume the sign of terms starting from the end, because all the terms containing \(i^n\) depend of \(n\).
On the other hand, starting from the beginning, we can notice an alternance of even and odd numbers (fistly with their
power
and secondly with their
binom
) respectively corresponding to the \(\cos(x)\) and \(\sin(x)\) functions. By the way, we also notice an alternance of\((+)\) and \((-)\) signs.
$$ (e^{ix})^n = \cos^n(x) + i \binom{n}{1}\cos^{n-1}(x) \ \sin(x) - \binom{n}{2}\cos^{n-2}(x)\sin^2(x) -i\binom{n}{3}\cos^{n-3}(x) \ \sin^3(x) +
\hspace{0.1em} ...$$
$$ \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \binom{n}{4}\cos^{n-4}(x) \ \sin^4(x) + i\binom{n}{5}\cos^{n-5}(x) \ \sin^5(x)
-i\binom{n}{6}\cos^{n-6}(x) \ \sin^6(x) + \binom{n}{7}\cos^{n-7}(x) \ \sin^7(x) + \hspace{0.1em} ... $$
Rearranging all this, we identify both real and imaginary parts of \((e^{ix})^n\).
$$ (e^{ix})^n = \hspace{0.2em} \underbrace { \cos^n(x) - \binom{n}{2}\cos^{n-2}(x)\sin^2(x) + \binom{n}{4}\cos^{n-4}(x) \ \sin^4(x) -
\binom{n}{6}\cos^{n-6}(x) \ \sin^6(x) + \hspace{0.1em} ... } _\text {real part}$$
$$ \hspace{0.2em} ... \hspace{0.2em} + i \underbrace { \Biggl[ \binom{n}{1}\cos^{n-1}(x) \ \sin(x) - \binom{n}{3}\cos^{n-3}(x) \ \sin^3(x) +
\binom{n}{5}\cos^{n-5}(x) \ \sin^5(x) - \hspace{0.1em} ... \Biggr] } _\text {imaginary part} $$
So,
$$ (e^{ix})^n = \hspace{0.2em} \underbrace { \sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k} \ \cos^{n-2k}(x) \times \sin^{2k}(x) } _\text {real part}
\hspace{0.2em} + \hspace{0.2em} i. \underbrace { \Biggl[\sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k +1} \ \cos^{n-(2k+1)}(x) \times \sin^{2k+1}(x)
\Biggr] } _\text {imaginary part} $$
But, we have seen above that with the expression \((1)\):
$$ \Biggl \{ \begin{gather*} \mathcal{Re}\Bigl[(e^{ix})^n \Bigr] = \cos(nx) \\ \mathcal{Im}\Bigl[(e^{ix})^n \Bigr] = \sin(nx) \end{gather*} $$
And as a result,
$$ \forall n \in \mathbb{N}, \ \forall x \in \mathbb{R}, $$
-
Example
Let us calculate \( \cos(nx)\) for \( n = 3 \) :
$$ \cos(3x) = \sum_{k =0}^{3} (-1)^k \ \binom{3}{2k} \ \cos^{3-2k}(x) \ \sin^{2k}(x)$$
$$ \cos(3x) = \binom{3}{0}\cos^3(x) - \binom{3}{2}\cos(x)\sin^2(x) $$
$$ \cos(3x) = \cos^3(x) - 3cos(x)\sin^2(x) $$
Moreover, with the following formulas \( (3) \) and \( (4) \):
$$ \Biggl \{ \begin{gather*} \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \sin(\beta) \cos(\alpha) \qquad (3) \\ \sin(\alpha - \beta) =
\sin(\alpha) \cos(\beta) - \sin(\beta) \cos(\alpha) \qquad (4) \end{gather*} $$
Performing the operation \( (3) +(4) \), we do obtain:
$$ 2 \sin(\alpha) \cos(\beta) = \sin(\alpha + \beta) + \sin(\alpha - \beta) $$
Now performing \( (3) - (4) \), we do obtain this:
$$ 2 \cos(\alpha) \sin(\beta) = \sin(\alpha + \beta) - \sin(\alpha - \beta) $$
Doing the same reasoning but starting from the cosine formulas, we find this these two others:
$$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$
Now, setting down two new variables:
$$ \Biggl \{ \begin{gather*} p = \alpha + \beta \\ q = \alpha - \beta \end{gather*} $$
We do have:
$$ \forall (p, q) \in \hspace{0.04em} \mathbb{R}^2, $$
$$ \sin(p ) + \sin(q) = 2 \sin\left(\frac{p+q}{2}\right) \cos\left(\frac{p-q}{2}\right) $$
Using the same process again, it is possible to recover the three other missing formulas:
$$ \forall (p, q) \in \hspace{0.04em} \mathbb{R}^2, $$
Finally, still with these same formulas:
$$ \tan(p ) + \tan(q) = \frac{\sin(p)}{\cos(p)} + \frac{\sin(q)}{\cos(q)} $$
$$ \tan(p ) + \tan(q) = \frac{\sin(p)}{\cos(p)}\textcolor{rgb(85, 109, 229)}{\times \frac{\sin(q)}{\cos(q)}} + \frac{\sin(q)}{\cos(q)}
\textcolor{rgb(85, 109, 229)}{\times\frac{\sin(p)}{\cos(p)}} $$
$$ \tan(p ) + \tan(q) = \frac{\sin(p)\cos(q) + \sin(q)\cos(p)}{\cos(p)\cos(q)} $$
$$ \tan(p ) + \tan(q) = \frac{\sin(p + q)}{\cos(p)\cos(q)} $$
Then we get the second one immediately afterward, finally obtaining:
$$ \forall (m, n) \in \mathbb{Z}^2, \ \forall (p, q) \in \mathbb{R}^2, \ \biggl[p \neq \frac{\pi}{2} + 2m\pi \biggr] \lor \biggl[q \neq \frac{\pi}{2}
+ 2n\pi \biggr], $$
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