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Find the Laplace Transform of $\cos^2(\omega t)$, where $\omega$ is a constant.

From a cosine identity: $cos^2(\omega t) = \frac{1}{2}(1+\cos(2\omega t))$.

So then I get:

\begin{align} \mathcal{L}(\cos^2(\omega t)) & = \frac{1}{2}\mathcal{L}(1+\cos(2\omega t))\\ &=\frac{1}{2}(\mathcal{L}(1) + \mathcal{L}(\cos(2\omega t)))\\ & =\frac{1}{2}\left(\frac{1}{s} + \mathcal{L}(\cos(2\omega t))\right) \end{align}

This is the part where I'm unsure how to proceed. Can I use the fact that $\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2+\omega^2}$ to get $\mathcal{L}(\cos(2\omega t)) = \frac{s}{s^2 + (2 \omega)^2}$ or do I have to use the definition of Laplace transformation and do the integration to solve this?

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1 Answer

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What you did is OK, you get $$ \mathcal{L}(\cos^2(\omega t))=\frac{1}{2}\left(\frac{1}{s} + \mathcal{L}(\cos(2\omega t))\right)=\frac{1}{2 s}+\frac{s}{2 \left(s^2+4 w^2\right)} $$ where we have used

$$\mathcal{L}(\cos\omega t) = \frac{s}{s^2 + \omega^2}.$$

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