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Express $\cos 5t$ and $\sin 6t$ in terms of exponential functions

The answers are:

$\cos 5t = \frac{e^{i5t} + e^{-i5t}}{2}$

$\sin 6t = \frac{e^{i6t} - e^{-i6t}}{2i}$

I converted these purely using the formula, for example: $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$

But how do we define or come to that formula from the complex number exponential to trigonometric representation:

$ e^{i\theta} = \cos \theta + i \sin \theta$

$e^{-i\theta} = \cos \theta - i \sin \theta$

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  • $\begingroup$ $e^{i5\theta} = \cos 5\theta + i \sin 5\theta$ $\endgroup$
    – Jean Marie
    Commented Jan 27, 2023 at 3:54
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    $\begingroup$ Add the last two equations together, then divide by two. What happens? $\endgroup$
    – Théophile
    Commented Jan 27, 2023 at 3:55
  • $\begingroup$ Also related: de Moivre's formulas $(\cos\theta+i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ $\endgroup$
    – GEdgar
    Commented Feb 2, 2023 at 1:19

1 Answer 1

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You can easily derive the exponential representation of the trigonometric functions by solving one of the two equations for one of the trigonometric functions, substituting it into the other equation, and then just rearranging for the other trigonometric function:

$$ \begin{align*} e^{-\theta \cdot i} = \cos\left( \theta \right) - {\color{red} \sin\left( \theta \right)} \cdot i &\quad\mid\quad +\sin\left( \theta \right)\\ e^{-\theta \cdot i} + {\color{red} \sin\left( \theta \right)} \cdot i = \cos\left( \theta \right) &\quad\\ \cos\left( \theta \right) = e^{-\theta \cdot i} + {\color{red} \sin\left( \theta \right)} \cdot i &\quad\\ \\ e^{\theta \cdot i} = \cos\left( \theta \right) + {\color{red} \sin\left( \theta \right)} \cdot i &\quad\mid\quad \cos\left( \theta \right) = e^{-\theta \cdot i} + {\color{red} \sin\left( \theta \right)} \cdot i\\ e^{\theta \cdot i} = e^{-\theta \cdot i} + {\color{red} \sin\left( \theta \right)} \cdot i + {\color{red} \sin\left( \theta \right)} \cdot i &\quad\mid\quad -e^{-\theta \cdot i} \\ e^{\theta \cdot i} - e^{-\theta \cdot i} = {\color{red} \sin\left( \theta \right)} \cdot i + {\color{red} \sin\left( \theta \right)} \cdot i &\quad\\ e^{\theta \cdot i} - e^{-\theta \cdot i} = 2 \cdot {\color{red} \sin\left( \theta \right)} \cdot i &\quad\mid\quad \div \left( 2 \cdot i \right)\\ \frac{e^{\theta \cdot i} - e^{-\theta \cdot i}}{2 \cdot i} = {\color{red} \sin\left( \theta \right)}\\ \\ \Rightarrow {\color{red} \sin\left( \theta \right)} = \frac{e^{\theta \cdot i} - e^{-\theta \cdot i}}{2 \cdot i} &\quad\\ \\ \\ e^{\theta \cdot i} = {\color{blue} \cos\left( \theta \right)} + \sin\left( \theta \right) \cdot i &\quad\mid\quad -{\color{blue} \cos\left( \theta \right)}\\ e^{\theta \cdot i} - {\color{blue} \cos\left( \theta \right)} = \sin\left( \theta \right) \cdot i &\quad\\ \sin\left( \theta \right) \cdot i = e^{\theta \cdot i} - {\color{blue} \cos\left( \theta \right)} &\quad\\ \\ e^{-\theta \cdot i} = {\color{blue} \cos\left( \theta \right)} - \sin\left( \theta \right) \cdot i &\quad\mid\quad \sin\left( \theta \right) \cdot i = e^{\theta \cdot i} - {\color{blue} \cos\left( \theta \right)}\\ e^{-\theta \cdot i} = {\color{blue} \cos\left( \theta \right)} - \left( e^{\theta \cdot i} - {\color{blue} \cos\left( \theta \right)} \right) &\quad\\ e^{-\theta \cdot i} = {\color{blue} \cos\left( \theta \right)} - e^{\theta \cdot i} + {\color{blue} \cos\left( \theta \right)} &\quad\mid\quad +e^{\theta \cdot i}\\ e^{\theta \cdot i} + e^{-\theta \cdot i} = {\color{blue} \cos\left( \theta \right)} + {\color{blue} \cos\left( \theta \right)} &\quad\\ e^{\theta \cdot i} + e^{-\theta \cdot i} = 2 \cdot {\color{blue} \cos\left( \theta \right)} &\quad\mid\quad \div 2\\ \frac{e^{\theta \cdot i} + e^{-\theta \cdot i}}{2} = {\color{blue} \cos\left( \theta \right)} &\quad\\ \\ \Rightarrow {\color{blue} \cos\left( \theta \right)} = \frac{e^{\theta \cdot i} + e^{-\theta \cdot i}}{2} &\quad\\ \end{align*} $$

These formulas actually have really useful applications too, so you can use them to quickly and easily derive the addition theorems $^{[1]}$ for sine and cosine, to derive the relationship between arc functions (the inverse functions of the trigonometric functions) and complex logarithms $^{[2]}$, but also how to find the sine and cosine of complex numbers can thus be derived, with which one can e.g. understand the close connection between the trigonometric functions and the hyperbolic functions.

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