The trigonometric duplication and addition formulas

The duplication formulas

The sines and cosines function

As well, hare are their respective expressions depending on \( tan(\alpha)\) :

The binomial's duplication formulas

The tangent function

$$ tan(2\alpha) = \frac{2tan(\alpha)}{1 -tan^2(\alpha)} $$

The addition formulas

The sines and cosines function

Furthermore,

$$ with \enspace \Biggl \{ \begin{align*} a = \alpha + \beta \\ b = \alpha - \beta \end{align*} $$

The tangent function

Recap table of the trigonometric duplication and addition formulas

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Demonstrations

The duplication formulas

The sines and cosines function

Rewriting the complex number having as argument \( 2\alpha\) under its complex exponential form, we do have:

We develop the above expression:

Let us identify the respective real and imaginary parts of the complex number \(e^{2i\alpha} \) :

As a result we do have,

Using the famous formula \( cos^2(\alpha) + sin^2(\alpha) = 1 \), we can retrieve the two other formulas.

The binomial's duplication formulas

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:

So in our case,

$$ (e^{ix})^n = cos^n(x) + i \binom{n}{1}cos^{n-1}(x) \ sin(x) - \binom{n}{2}cos^{n-2}(x)sin^2(x) -i\binom{n}{3}cos^{n-3}(x) \ sin^3(x) + \binom{n}{4}cos^{n-4}(x) \ sin^4(x) + \hspace{0.1em} ...$$

$$ \hspace{2em} ... \hspace{0.1em} + \binom{n}{n-3}cos^3(x) \ i^{n-3}.sin^{n-3}(x) + \binom{n}{n-2}cos^2(x) \ i^{n-2}.sin^{n-2}(x) + \binom{n}{n-1}cos(x) \ i^{n-1}.sin^{n-1}(x) + i^n \ sin^n(x) $$

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,

$$ (e^{ix})^n = \hspace{0.2em} \underbrace { \sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k} \ cos^{n-2k}(x) \times sin^{2k}(x) } _\text {real part} \hspace{0.2em} + \hspace{0.2em} i. \underbrace { \Biggl[\sum_{k =0}^{n / 2} (-1)^k \ \binom{n}{2k +1} \ cos^{n-(2k+1)}(x) \times sin^{2k+1}(x) \Biggr] } _\text {imaginary part} $$

And as a result,

1. Example

Let us calculate \( cos(nx)\) for \( n = 3 \) :

$$ cos(3x) = \sum_{k =0}^{3} (-1)^k \ \binom{3}{2k} \ cos^{3-2k}(x) \ sin^{2k}(x)$$
$$ cos(3x) = \binom{3}{0}cos^3(x) - \binom{3}{2}cos(x)sin^2(x) $$
$$ cos(3x) = cos^3(x) - 3cos(x)sin^2(x) $$

The tangent function

We know from the definition of the tangent function that:

1. 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,



$$ sin(2\alpha) = \frac{2 sin(\alpha)}{cos(\alpha) (1 + tan^2(\alpha)) }$$
$$ \forall \alpha \in \mathbb{R}, $$

$$ sin(2\alpha) = \frac{2 tan(\alpha)}{ 1 + tan^2(\alpha) }$$

2. Expression of \( cos(2 \alpha) = f(tan( \alpha)) \)

As well,

$$ cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha)$$
$$ cos(2\alpha) = \frac{1}{1 + tan^2(\alpha)} - sin^2(\alpha)$$
$$ cos(2\alpha) = \frac{1}{1 + tan^2(\alpha)} - tan^2(\alpha)cos^2(\alpha)$$


$$ \forall \alpha \in \mathbb{R}, $$

$$ cos(2\alpha) = \frac{1 - tan^2(\alpha)}{1 + tan^2(\alpha)} $$

3. 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,

$$ tan(2\alpha) = \frac{2tan(\alpha)}{1 -tan^2(\alpha)} $$

The addition formulas

The sines and cosines function

Rewriting the complex number having as argument \( (\alpha + \beta)\) under its complex exponential form, we do have:

We rewrite \( (2) \) under its trigonometric form:

And we develop it:

Let us identify the respective real and imaginary parts of the complex number \(e^{i(\alpha + \beta)} \) :

And as a result,

Using the same process again, it is possible to recover the two other missing formulas:

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:

We do have:

$$ sin(a ) + sin(b) = 2 sin\left(\frac{a+b}{2}\right) cos\left(\frac{a-b}{2}\right) $$

Using the same process again, it is possible to recover the three other missing formulas:

The tangent function

With the sines and cosines addition formulas, we do have:

Let us multiply both numerator and denominator by \(cos(\alpha) cos(\beta)\) :

And finally,

$$ tan(\alpha + \beta) = \frac{tan(\alpha) + tan(\beta)}{ 1 - tan(\alpha)tan(\beta) }$$

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 \( tan( \alpha - \beta) \):

$$ tan(\alpha \textcolor{#8E5B5B}{-} \beta) = \frac{tan(\alpha) \textcolor{#8E5B5B}{-} tan(\beta)}{ 1 \textcolor{#8E5B5B}{+} tan(\alpha)tan(\beta) }$$

Recap table of the trigonometric duplication and addition formulas