# Laplace transform of cosine and sine

We start from the easily formula

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

where the curved 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  $s>0$.

Title Laplace transform of cosine and sine LaplaceTransformOfCosineAndSine 2013-03-22 18:18:27 2013-03-22 18:18:27 pahio (2872) pahio (2872) 8 pahio (2872) Derivation msc 44A10 Laplace transform of sine and cosine