Question

Find the Laplace Transform of (cos2t)^3



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Answer 1

Answer:

I've tried to look up some relevant formulae in my book, but I can't find anything that looks useful. I suspect there is something there, that I'm just not seeing.

I've tried to look up some relevant formulae in my book, but I can't find anything that looks useful. I suspect there is something there, that I'm just not seeing.Wolfram Alpha suggests the answer is s+23–√2s2+8s+232s2+8, but I can't "reverse engineer" it either.

I've tried to look up some relevant formulae in my book, but I can't find anything that looks useful. I suspect there is something there, that I'm just not seeing.Wolfram Alpha suggests the answer is s+23–√2s2+8s+232s2+8, but I can't "reverse engineer" it either.Any help appreciated!

Step-by-step explanation:

Case 1: f(t)=cos(2t−π3)u(t)f(t)=cos⁡(2t−π3)u(t)

Case 1: f(t)=cos(2t−π3)u(t)f(t)=cos⁡(2t−π3)u(t)Observing that cos(ωt−ϕ)=cos(ωt)cosϕ+sin(ωt)sinϕcos⁡(ωt−ϕ)=cos⁡(ωt)cos⁡ϕ+sin⁡(ωt)sin⁡ϕ and that L{cos(ωt)}=ss2+ω2L{cos⁡(ωt)}=ss2+ω2 and L{sin(ωt)}=ωs2+ω2L{sin⁡(ωt)}=ωs2+ω2, we have

Case 1: f(t)=cos(2t−π3)u(t)f(t)=cos⁡(2t−π3)u(t)Observing that cos(ωt−ϕ)=cos(ωt)cosϕ+sin(ωt)sinϕcos⁡(ωt−ϕ)=cos⁡(ωt)cos⁡ϕ+sin⁡(ωt)sin⁡ϕ and that L{cos(ωt)}=ss2+ω2L{cos⁡(ωt)}=ss2+ω2 and L{sin(ωt)}=ωs2+ω2L{sin⁡(ωt)}=ωs2+ω2, we haveL{cos(ωt−ϕ)}=L{cos(ωt)}cosϕ+L{sin(ωt)}sinϕ=ss2+ω2cosϕ+ω