PlanetMath: Laplace transform of cosine and sine

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Laplace transform of cosine and sine

(Derivation)

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

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