I'm working on the following Laplace transform problem at the moment, and I'm a little stuck.
$$\mathcal{L} \{\sin(2x)\cos(5x) \}$$
I don't recall any trig identity that would apply here. I know that
$$\sin(2x) = 2\sin(x)\cos(x)$$
But I'm not sure if that applies in this situation. If you guys could point in the right direction I'd be most appreciative.
Best Answer
By the addition theorem for the sine, we can write
$$\begin{align}\sin (7x) &= \sin (5x + 2x) = \sin (5x)\cos (2x) + \sin(2x)\cos (5x)\\ \sin (3x) &= \sin (5x - 2x) = \sin (5x) \cos (2x) - \sin (2x)\cos (5x), \end{align}$$
and hence
$$\sin(2x)\cos(5x) = \frac12 \bigl(\sin (7x) - \sin (3x)\bigr).$$
Therefore
$$\mathcal{L}\{\sin(2x)\cos(5x)\} = \frac12 \bigl(\mathcal{L}\{\sin (7x)\} - \mathcal{L}\{\sin (3x)\}\bigr).$$