Sine and Cosine
I am also aware that decompose the trig functions into exponentials. I will do that in this post. However, I want to do it without complex numbers for now just to show that you don’t need them. Also, the recursive trick used is a nice technique I want to showcase.
Integration by Parts
We have to prove these identities in tandem with each other (we will see why shortly).
L{sin(t)}=∫∞0sin(t)e−stdt(using integration by parts)=1sf(0)−1slimt→∞f(t)e−st+1s∫∞0f′(t)e−st dt=1ssin(0)−1slimt→∞sin(t)e−st+1s∫∞0cos(t)e−st dt=1s∫∞0cos(t)e−st dt=1sL{cos(t)}Similarly
L{cos(t)}=∫∞0cos(t)e−stdt(using integration by parts)=1sf(0)−1slimt→∞f(t)e−st+1s∫∞0f′(t)e−st dt=1scos(0)−1slimt→∞cos(t)e−st+1s∫∞0(−sin(t))e−st dt=1s−1s∫∞0sin(t)e−st dt=1s−1sL{sin(t)}We’ve now expressed both Laplace transforms in terms of each other. So now it’s like we have a system of two equations and two unknowns.
Sine
Combining the two results from above, we get
L{sin(t)}=1sL{cos(t)}=1s(1s−1sL{sin(t)})(1+1s2)L{sin(t)}=1s2(s2+1s2)L{sin(t)}=1s2L{sin(t)}=1s2+1Then, we can use our scaling property to see that
L{sin(bt)}=1b1(s/b)2+1=bs2+b2Cosine
Similarly
L{cos(t)}=1s−1sL{sin(t)}=1s−1s2L{cos(t)}(1+1s2)L{cos(t)}=1s(s2+1s2)L{cos(t)}=1sL{cos(t)}=ss2+1Then, we can use our scaling property to see that
L{cos(bt)}=1b(s/b)(s/b)2+1=ss2+b2Translation
For translation, recall the trig identities
sin(θ+ϕ)=sin(θ)cos(ϕ)+cos(θ)sin(ϕ)
cos(θ+ϕ)=cos(θ)cos(ϕ)−sin(θ)sin(ϕ)
Therefore
L{sin(bt+c)}=L{cos(bt)sin(c)+sin(bt)cos(c)}=sin(c)L{cos(bt)}+cos(c)L{sin(bt)}=sin(c)ss2+b2+cos(c)bs2+b2=s⋅sin(c)+b⋅cos(c)s2+b2and
L{cos(bt+c)}=L{cos(bt)cos(c)−sin(bt)sin(c)}=cos(c)L{cos(bt)}−sin(c)L{sin(bt)}=cos(c)ss2+b2−sin(c)bs2+b2=s⋅cos(c)−b⋅sin(c)s2+b2