\(\sin^32t=\sin2t\sin^22t\)
Recall that \(\sin^2x=\dfrac{1-\cos2x}2\):
\(\sin^32t=\sin2t\dfrac{1-\cos4t}2\)
\(\sin^32t=\dfrac12\sin2t-\dfrac12\sin2t\cos4t\)
Now recall that \(\sin x\cos y=\dfrac{\sin(x+y)+\sin(x-y)}2\), which gives
\(\sin^32t=\dfrac12\sin2t-\dfrac14(\sin6t+\sin(-2t))\)
\(\sin^32t=\dfrac12\sin2t-\dfrac14\sin6t-\dfrac14\sin(-2t)\)
\(\sin^32t=\dfrac34\sin2t-\dfrac14\sin6t\)
So,
\(\mathcal L_s\{\sin^32t\}=\dfrac34\dfrac2{s^2+4}-\dfrac14\dfrac6{s^2+36}=\dfrac{48}{s^4+40s^2+144}\)
