ansaritaqi99
ansaritaqi99
22-03-2017
Mathematics
contestada
laplace of sin^3(2t)
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LammettHash
LammettHash
23-03-2017
[tex]\sin^32t=\sin2t\sin^22t[/tex]
Recall that [tex]\sin^2x=\dfrac{1-\cos2x}2[/tex]:
[tex]\sin^32t=\sin2t\dfrac{1-\cos4t}2[/tex]
[tex]\sin^32t=\dfrac12\sin2t-\dfrac12\sin2t\cos4t[/tex]
Now recall that [tex]\sin x\cos y=\dfrac{\sin(x+y)+\sin(x-y)}2[/tex], which gives
[tex]\sin^32t=\dfrac12\sin2t-\dfrac14(\sin6t+\sin(-2t))[/tex]
[tex]\sin^32t=\dfrac12\sin2t-\dfrac14\sin6t-\dfrac14\sin(-2t)[/tex]
[tex]\sin^32t=\dfrac34\sin2t-\dfrac14\sin6t[/tex]
So,
[tex]\mathcal L_s\{\sin^32t\}=\dfrac34\dfrac2{s^2+4}-\dfrac14\dfrac6{s^2+36}=\dfrac{48}{s^4+40s^2+144}[/tex]
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