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).
\[\begin{align} \mathcal{L}\{ \sin(t) \} &= \int_{0}^{\infty} \sin(t) e^{-st} \; dt \qquad \text{(using integration by parts)} \\[10pt] &= \frac{1}{s} f(0) - \frac{1}{s} \lim_{t \rightarrow \infty} f(t)e^{-st} + \frac{1}{s} \int_{0}^{\infty} f'(t)e^{-st} \ dt \\[10pt] &= \frac{1}{s} \sin(0) - \frac{1}{s} \lim_{t \rightarrow \infty} \sin(t) e^{-st} + \frac{1}{s} \int_{0}^{\infty} \cos(t) e^{-st} \ dt \\[10pt] &= \frac{1}{s} \int_{0}^{\infty} \cos(t) e^{-st} \ dt \\[10pt] &= \frac{1}{s} \mathcal{L}\{ \cos(t) \} \end{align}\]Similarly
\[\begin{align} \mathcal{L}\{ \cos(t) \} &= \int_{0}^{\infty} \cos(t) e^{-st} \; dt \qquad \text{(using integration by parts)} \\[10pt] &= \frac{1}{s} f(0) - \frac{1}{s} \lim_{t \rightarrow \infty} f(t)e^{-st} + \frac{1}{s} \int_{0}^{\infty} f'(t)e^{-st} \ dt \\[10pt] &= \frac{1}{s} \cos(0) - \frac{1}{s} \lim_{t \rightarrow \infty} \cos(t) e^{-st} + \frac{1}{s} \int_{0}^{\infty} (-\sin(t)) e^{-st} \ dt \\[10pt] &= \frac{1}{s} - \frac{1}{s} \int_{0}^{\infty} \sin(t) e^{-st} \ dt \\[10pt] &= \frac{1}{s} - \frac{1}{s} \mathcal{L}\{ \sin(t) \} \end{align}\]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
\[\begin{align} &\mathcal{L}\{ \sin(t) \} = \frac{1}{s} \mathcal{L}\{ \cos(t) \} = \frac{1}{s} \left ( \frac{1}{s} - \frac{1}{s} \mathcal{L}\{ \sin(t) \} \right ) \\[10pt] &\left( 1 + \frac{1}{s^2} \right ) \mathcal{L}\{ \sin(t) \} = \frac{1}{s^2} \\[10pt] &\left( \frac{s^2 + 1}{s^2} \right ) \mathcal{L}\{ \sin(t) \} = \frac{1}{s^2} \\[10pt] &\mathcal{L}\{ \sin(t) \} = \frac{1}{s^2+1} \end{align}\]Then, we can use our scaling property to see that
\[\mathcal{L}\{ \sin(bt) \} = \frac{1}{b} \frac{1}{(s/b)^2+1} = \frac{b}{s^2 + b^2}\]Cosine
Similarly
\[\begin{align} &\mathcal{L}\{ \cos(t) \} = \frac{1}{s} - \frac{1}{s} \mathcal{L}\{ \sin(t) \} = \frac{1}{s} - \frac{1}{s^2} \mathcal{L}\{ \cos(t) \} \\[10pt] &\left( 1 + \frac{1}{s^2} \right ) \mathcal{L}\{ \cos(t) \} = \frac{1}{s} \\[10pt] &\left( \frac{s^2 + 1}{s^2} \right ) \mathcal{L}\{ \cos(t) \} = \frac{1}{s} \\[10pt] &\mathcal{L}\{ \cos(t) \} = \frac{s}{s^2+1} \end{align}\]Then, we can use our scaling property to see that
\[\mathcal{L}\{ \cos(bt) \} = \frac{1}{b} \frac{(s/b)}{(s/b)^2+1} = \frac{s}{s^2 + b^2}\]Translation
For translation, recall the trig identities
\[\sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)\] \[\cos(\theta + \phi) = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi)\]Therefore
\[\begin{align} \mathcal{L}\{ \sin(bt+c) \} &= \mathcal{L}\{ \cos(bt)\sin(c) + \sin(bt)\cos(c) \} \\[10pt] &= \sin(c) \mathcal{L}\{ \cos(bt) \} + \cos(c) \mathcal{L}\{ \sin(bt) \} \\[10pt] &= \sin(c) \frac{s}{s^2 + b^2} + \cos(c) \frac{b}{s^2 + b^2} \\[10pt] &= \frac{s \cdot \sin(c) + b \cdot \cos (c)}{s^2 + b^2} \end{align}\]and
\[\begin{align} \mathcal{L}\{ \cos(bt+c) \} &= \mathcal{L}\{ \cos(bt)\cos(c) - \sin(bt)\sin(c) \} \\[10pt] &= \cos(c) \mathcal{L}\{ \cos(bt) \} - \sin(c) \mathcal{L}\{ \sin(bt) \} \\[10pt] &= \cos(c) \frac{s}{s^2 + b^2} - \sin(c) \frac{b}{s^2 + b^2} \\[10pt] &= \frac{s \cdot \cos(c) - b \cdot \sin(c)}{s^2 + b^2} \end{align}\]