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)estdt(using integration by parts)=1sf(0)1slimtf(t)est+1s0f(t)est dt=1ssin(0)1slimtsin(t)est+1s0cos(t)est dt=1s0cos(t)est dt=1sL{cos(t)}

Similarly

L{cos(t)}=0cos(t)estdt(using integration by parts)=1sf(0)1slimtf(t)est+1s0f(t)est dt=1scos(0)1slimtcos(t)est+1s0(sin(t))est dt=1s1s0sin(t)est dt=1s1sL{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(1s1sL{sin(t)})(1+1s2)L{sin(t)}=1s2(s2+1s2)L{sin(t)}=1s2L{sin(t)}=1s2+1

Then, we can use our scaling property to see that

L{sin(bt)}=1b1(s/b)2+1=bs2+b2


Cosine

Similarly

L{cos(t)}=1s1sL{sin(t)}=1s1s2L{cos(t)}(1+1s2)L{cos(t)}=1s(s2+1s2)L{cos(t)}=1sL{cos(t)}=ss2+1

Then, we can use our scaling property to see that

L{cos(bt)}=1b(s/b)(s/b)2+1=ss2+b2


Translation

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=ssin(c)+bcos(c)s2+b2

and

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+b2sin(c)bs2+b2=scos(c)bsin(c)s2+b2

Laplace Transforms Series