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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.

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  • $\begingroup$ Use $\sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \sin \beta \cos \alpha$ to write it as a sum of two sines (times a constant). $\endgroup$
    – Daniel Fischer
    Commented Jul 11, 2013 at 23:28
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    $\begingroup$ Be sure to type in $\LaTeX$ for better readability. Suggestions if you want to type Laplace Transform symbol and curly brackets in $\LaTeX$: Type dollar signs. Between them, type \mathcal{L} and for the brackets, \{ and \}. Also, if you want to type known trig functions, type, for instance, \sin. $\endgroup$
    – Mr Mathster
    Commented Jul 11, 2013 at 23:29
  • $\begingroup$ I'm not exactly sure how'd I use that @DanielFischer since my question doesn't seem to be in the form that you wrote in the comment. $\endgroup$
    – codedude
    Commented Jul 11, 2013 at 23:31
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    $\begingroup$ @codedude The equation Daniel Fischer provided is another trigonometric identity. Be sure to "translate" the given equation with that form. $\endgroup$
    – Mr Mathster
    Commented Jul 11, 2013 at 23:32
  • $\begingroup$ @codedude Your function is one of the terms on the right of the identity. $\endgroup$
    – Daniel Fischer
    Commented Jul 11, 2013 at 23:33

1 Answer 1

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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

$$\mathcal{L}\{\sin(2x)\cos(5x)\} = \frac12 \bigl(\mathcal{L}\{\sin (7x)\} - \mathcal{L}\{\sin (3x)\}\bigr).$$

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