Laplace transform of cosine and sine

We start from the easily formula

$e^{\alpha t}\curvearrowleft {\displaystyle \frac{1}{s-\alpha }}\mathit{\hspace{1em}\hspace{1em}}(s>\alpha ),$(1)

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

$e^{-\alpha t}\curvearrowleft {\displaystyle \frac{1}{s+\alpha }}\mathit{\hspace{1em}\hspace{1em}}(s>-\alpha ).$(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.

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

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

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

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

There are the hyperbolic identities

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

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

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

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

which are true for  $s>0$.