Question

Have a question dealing with Laplace Transformations. Find the L{(1+cos2t)^2} I know the double angle theorem can translate this into cos^2(t) but we're missing a 1/2 from the equation. I know we can do a change of scale to get it to cos^2t but that does us no good.

Answer 1

expand to \(1 +2 \cos 2 t+ \cos^2 2t\)

then double angle formula \(\cos 4t = 2 \cos^2 2t - 1\) so \(\ \frac{1}{2} \cos 4t + \frac{1}{2} = \cos^2 2t \)

to get \(2 \cos 2t+ \frac{1}{2}\cos 4t+\frac{3}{2}\)

then \(\mathcal{L}\{\cos (at) \}(p) =\frac{p^2}{p^2 + a^2}\)

Answer 2

Second problem you've solved for me irishboy thank you! This one really got me stumped.