The sines and cosines function
$$\forall \alpha \in \mathbb{R}, $$
$$ sin(2\alpha) = 2 sin(\alpha) cos(\alpha) $$
$$ cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha) $$
$$ cos(2\alpha) = 2cos^2(\alpha) - 1 $$
$$ cos(2\alpha) = 1 - 2sin^2(\alpha) $$
As well, hare are their respective expressions depending on \( tan(\alpha)\) :
$$ \forall n \in \mathbb{N}, \ \forall x \in \mathbb{R}, $$
$$\forall \alpha \in \mathbb{R}, $$
$$ sin(2\alpha) = \frac{2 tan(\alpha)}{ 1 + tan^2(\alpha) }$$
$$ cos(2\alpha) = \frac{1 - tan^2(\alpha)}{1 + tan^2(\alpha)} $$
The binomial's duplication formulas
$$ \forall n \in \mathbb{N}, \ \forall x \in \mathbb{R}, $$
$$ cos(nx) = \sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k} \ cos^{n-2k}(x) \times sin^{2k}(x) $$
$$ sin(nx) = \sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k +1} \ cos^{n-(2k+1)}(x) \times sin^{2k+1}(x) $$
$$ \forall k \in \mathbb{Z}, \enspace \forall \alpha \in \Biggl[ \mathbb{R} \hspace{0.2em} \backslash \left\{\frac{\pi}{4} + \frac{k\pi}{2} \right\} \Biggr], $$
$$ tan(2\alpha) = \frac{2tan(\alpha)}{1 -tan^2(\alpha)} $$
The sines and cosines function
$$ \forall (\alpha, \beta) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(\alpha + \beta) = sin(\alpha) cos(\beta) + sin(\beta) cos(\alpha) $$
$$ cos(\alpha + \beta) = cos(\alpha) cos(\beta) - sin(\alpha) sin(\beta) $$
$$ sin(\alpha \textcolor{#A65757}{-} \beta) = sin(\alpha) cos(\beta) \textcolor{#A65757}{-} sin(\beta) cos(\alpha) $$
$$ cos(\alpha \textcolor{#A65757}{-} \beta) = cos(\alpha) cos(\beta) \textcolor{#A65757}{+} sin(\alpha) sin(\beta) $$
Furthermore,
$$ with \enspace \Biggl \{ \begin{align*} a = \alpha + \beta \\ b = \alpha - \beta \end{align*} $$
$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(a ) + sin(b) = 2 sin\left(\frac{a+b}{2}\right) cos\left(\frac{a-b}{2}\right) $$
$$ sin(a ) - sin(b) = 2 cos\left(\frac{a+b}{2}\right) sin\left(\frac{a-b}{2}\right) $$
$$ cos(a ) + cos(b) = 2 cos\left(\frac{a+b}{2}\right) cos\left(\frac{a-b}{2}\right) $$
$$ cos(b ) - cos(a) = 2 sin\left(\frac{a+b}{2}\right) sin\left(\frac{a-b}{2}\right) $$
$$\forall k \in \mathbb{Z}, \enspace \forall (\alpha, \beta) \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \left\{ \frac{\pi}{2} + k\pi \right\} \Bigr]^2,$$
$$ tan(\alpha + \beta) = \frac{tan(\alpha) + tan(\beta)}{ 1 - tan(\alpha)tan(\beta) }$$
$$ tan(\alpha \textcolor{#A65757}{-} \beta) = \frac{tan(\alpha) \textcolor{#A65757}{-} tan(\beta)}{ 1 \textcolor{#A65757}{+} tan(\alpha)tan(\beta) }$$
Recap table of the trigonometric duplication and addition formulas
Click on the title to access to the recap table.
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,
$$ \forall \alpha \in \mathbb{R}, $$
$$ sin(2\alpha) = 2 sin(\alpha) cos(\alpha) $$
$$ cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha) $$
Using the famous formula \( cos^2(\alpha) + sin^2(\alpha) = 1 \), we can retrieve the two other formulas.
$$ \forall \alpha \in \mathbb{R}, $$
$$ cos(2\alpha) = 2cos^2(\alpha) - 1 $$
$$ cos(2\alpha) = 1 - 2sin^2(\alpha) $$
The idea is to find a general expression for the following couple:
$$ \Biggl \{ \begin{align*} cos(nx) \\ sin(nx) \end{align*} $$
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{0.05em} \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 \((1)\):
$$ \Biggl \{ \begin{align*} \mathcal{Re}\Bigl[(e^{ix})^n \Bigr] = cos(nx) \\ \mathcal{Im}\Bigl[(e^{ix})^n \Bigr] = sin(nx) \end{align*} $$
And as a result,
$$ \forall n \in \mathbb{N}, \ \forall x \in \mathbb{R}, $$
$$ cos(nx) = \sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k} \ cos^{n-2k}(x) \times sin^{2k}(x) $$
$$ sin(nx) = \sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k +1} \ cos^{n-(2k+1)}(x) \times sin^{2k+1}(x) $$
Let us calculate \( cos(nx)\) for \( n = 3 \) :
$$ cos(3x) = \sum_{k =0}^{3} (-1)^k \ \binom{3}{2k} \ cos^{3-2k}(x) \ sin^{2k}(x)$$
$$ cos(3x) = \binom{3}{0}cos^3(x) - \binom{3}{2}cos(x)sin^2(x) $$
$$ cos(3x) = cos^3(x) - 3cos(x)sin^2(x) $$
We know from the definition of the tangent function that:
$$ tan(2\alpha) = \frac{sin(2\alpha) }{cos(2\alpha) }$$
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,
$$ sin(2\alpha) = \frac{2 sin(\alpha)}{cos(\alpha) (1 + tan^2(\alpha)) }$$
$$ \forall \alpha \in \mathbb{R}, $$
$$ sin(2\alpha) = \frac{2 tan(\alpha)}{ 1 + tan^2(\alpha) }$$
As well,
$$ cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha)$$
$$ cos(2\alpha) = \frac{1}{1 + tan^2(\alpha)} - sin^2(\alpha)$$
$$ cos(2\alpha) = \frac{1}{1 + tan^2(\alpha)} - tan^2(\alpha)cos^2(\alpha)$$
$$ \forall \alpha \in \mathbb{R}, $$
$$ cos(2\alpha) = \frac{1 - tan^2(\alpha)}{1 + tan^2(\alpha)} $$
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:
$$ \Biggl \{ \begin{align*} \mathcal{Re}\left[e^{i(\alpha + \beta)}\right] = cos(\alpha + \beta) \\ \mathcal{Im}\left[e^{i(\alpha + \beta)}\right] = sin(\alpha + \beta) \end{align*} $$
And as a result,
$$ \forall (\alpha, \beta) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(\alpha + \beta) = sin(\alpha) cos(\beta) + sin(\beta) cos(\alpha) \qquad (3) $$
$$ cos(\alpha + \beta) = cos(\alpha) cos(\beta) - sin(\alpha) sin(\beta) $$
Using the same process again, it is possible to recover the two other missing formulas:
$$ \forall (\alpha, \beta) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(\alpha \textcolor{#A65757}{-} \beta) = sin(\alpha) cos(\beta) \textcolor{#A65757}{-} sin(\beta) cos(\alpha) \qquad (4) $$
$$ cos(\alpha \textcolor{#A65757}{-} \beta) = cos(\alpha) cos(\beta) \textcolor{#A65757}{+} sin(\alpha) sin(\beta) $$
Moreover, with the following formulas \( (3) \) and \( (4) \):
$$ \Biggl \{ \begin{align*} sin(\alpha + \beta) = sin(\alpha) cos(\beta) + sin(\beta) cos(\alpha) \qquad (3) \\ sin(\alpha - \beta) = sin(\alpha) cos(\beta) - sin(\beta) cos(\alpha) \qquad (4) \end{align*} $$
Performing the operation \( (3) +(4) \), we do obtain:
$$ sin(\alpha + \beta) + sin(\alpha - \beta) = 2 sin(\alpha) cos(\beta) $$
Now, setting down two new variables:
$$ \Biggl \{ \begin{align*} a = \alpha + \beta \\ b = \alpha - \beta \end{align*} $$
We do have:
$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(a ) + sin(b) = 2 sin\left(\frac{a+b}{2}\right) cos\left(\frac{a-b}{2}\right) $$
Using the same process again, it is possible to recover the three other missing formulas:
$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ sin(a ) - sin(b) = 2 cos\left(\frac{a+b}{2}\right) sin\left(\frac{a-b}{2}\right) $$
$$ cos(a ) + cos(b) = 2 cos\left(\frac{a+b}{2}\right) cos\left(\frac{a-b}{2}\right) $$
$$ cos(b ) - cos(a) = 2 sin\left(\frac{a+b}{2}\right) sin\left(\frac{a-b}{2}\right) $$
$$ tan(\alpha + \beta ) = \frac{sin(\alpha + \beta )}{cos(\alpha + \beta )}$$
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,
$$\forall k \in \mathbb{Z}, \enspace \forall (\alpha, \beta) \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \left\{ \frac{\pi}{2} + k\pi \right\} \Bigr]^2,$$
$$ tan(\alpha + \beta) = \frac{tan(\alpha) + tan(\beta)}{ 1 - tan(\alpha)tan(\beta) }$$
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 \( tan( \alpha - \beta) \):
$$\forall k \in \mathbb{Z}, \enspace \forall (\alpha, \beta) \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \left\{ \frac{\pi}{2} + k\pi \right\} \Bigr]^2,$$
$$ tan(\alpha \textcolor{#A65757}{-} \beta) = \frac{tan(\alpha) \textcolor{#A65757}{-} tan(\beta)}{ 1 \textcolor{#A65757}{+} tan(\alpha)tan(\beta) }$$
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