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# Laplace transform of cosine and sine

We start from the easily derivable formula

$\displaystyle e^{{\alpha t}}\;\curvearrowleft\;\frac{1}{s\!-\!\alpha}\qquad(s>% \alpha),$ | (1) |

where the curved arrow points from the Laplace-transformed function to the original function. Replacing $\alpha$ by $-\alpha$ we can write the second formula

$\displaystyle e^{{-\alpha t}}\;\curvearrowleft\;\frac{1}{s\!+\!\alpha}\qquad(s% >-\alpha).$ | (2) |

Adding (1) and (2) and dividing by 2 we obtain (remembering the linearity of the Laplace transform)

$\frac{e^{{\alpha t}}\!+\!e^{{-\alpha t}}}{2}\;\curvearrowleft\;\frac{1}{2}\!% \left(\frac{1}{s\!-\!\alpha}\!+\!\frac{1}{s\!+\!\alpha}\right),$ |

i.e.

$\displaystyle\mathcal{L}\{\cosh{\alpha t}\}=\frac{s}{s^{2}\!-\!\alpha^{2}}.$ | (3) |

Similarly, subtracting (1) and (2) and dividing by 2 give

$\displaystyle\mathcal{L}\{\sinh{\alpha t}\}=\frac{a}{s^{2}\!-\!\alpha^{2}}.$ | (4) |

The formulae (3) and (4) are valid for $s>|\alpha|$.

There are the hyperbolic identities

$\cosh{it}=\cos{t},\quad\frac{1}{i}\sinh{it}=\sin{t}$ |

which enable the transition from hyperbolic to trigonometric functions. If we choose $\alpha:=ia$ in (3), we may calculate

$\cos{at}=\cosh{iat}\;\curvearrowleft\;\frac{s}{s^{2}\!-\!(ia)^{2}}=\frac{s}{s^% {2}+a^{2}},$ |

the formula (4) analogously gives

$\sin{at}=\frac{1}{i}\sinh{iat}\;\curvearrowleft\;\frac{1}{i}\!\cdot\!\frac{ia}% {s^{2}\!-\!(ia)^{2}}=\frac{a}{s^{2}+a^{2}}.$ |

Accordingly, we have derived the Laplace transforms

$\displaystyle\mathcal{L}\{\cos{at}\}=\frac{s}{s^{2}\!+\!a^{2}},$ | (5) |

$\displaystyle\mathcal{L}\{\sin{at}\}=\frac{a}{s^{2}\!+\!a^{2}},$ | (6) |

which are true for $s>0$.

Keywords:

Laplace transform of cosine, Laplace transform of sine

Synonym:

Laplace transform of sine and cosine

Major Section:

Reference

Type of Math Object:

Derivation

Parent:

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new image: plot W(t) = P(waiting time <= t) by robert_dodier

new question: Prove a formula is part of the Gentzen System by LadyAnne

Mar 30

new question: A problem about Euler's totient function by mbhatia

new problem: Problem: Show that phi(a^n-1), (where phi is the Euler totient function), is divisible by n for any natural number n and any natural number a >1. by mbhatia

new problem: MSC browser just displays "No articles found. Up to ." by jaimeglz

Mar 26

new correction: Misspelled name by DavidSteinsaltz

Mar 21

new correction: underline-typo by Filipe

Mar 19

new correction: cocycle pro cocyle by pahio

Mar 7

new image: plot W(t) = P(waiting time <= t) (2nd attempt) by robert_dodier

new image: expected waiting time by robert_dodier

new image: plot W(t) = P(waiting time <= t) by robert_dodier