Question

Laplace of sin^3(2t)

Answer 1

`\sin^32t=\sin2t\sin^22t`

Recall that
`\sin^2x=\frac{1-\cos2x}2`:


`\sin^32t=\sin2t\frac{1-\cos4t}2`

`\sin^32t=\frac12\sin2t-\frac12\sin2t\cos4t`

Now recall that
`\sin x\cos y=\frac{\sin(x+y)+\sin(x-y)}2`, which gives


`\sin^32t=\frac12\sin2t-\frac14(\sin6t+\sin(-2t))`

`\sin^32t=\frac12\sin2t-\frac14\sin6t-\frac14\sin(-2t)`

`\sin^32t=\frac34\sin2t-\frac14\sin6t`

So,


`\mathcal L_s\{\sin^32t\}=\frac34\frac2{s^2+4}-\frac14\frac6{s^2+36}=(48)/(s^4+40s^2+144)`