MHB Laplace transform of a function II

[Drain Brain]

$\mathscr{L}\{\sin^{2}4t\}$

$\mathscr{L}\{\sin(3t-\frac{1}{2}\}$for the 2nd prob here's what I have tried
$\mathscr{L}\{\sin(3t)\cos(0.5)-\cos(3t)\sin(0.5)}$

$\cos(0.5)\mathscr{L}\{\sin(3t)\}-\sin(0.5) \mathscr{L}\{\cos(3t)\}$

$\frac{3\cos(0.5)-s\sin(0.5)}{s^2+9}$ ---> is this correct?

for the firt prob can you give me hint on how to deal with it.

 

[Prove It]

$\mathscr{L}\{\sin^{2}4t\}$

$\mathscr{L}\{\sin(3t-\frac{1}{2}\}$for the 2nd prob here's what I have tried
$\mathscr{L}\{\sin(3t)\cos(0.5)-\cos(3t)\sin(0.5)}$

$\cos(0.5)\mathscr{L}\{\sin(3t)\}-\sin(0.5) \mathscr{L}\{\cos(3t)\}$

$\frac{3\cos(0.5)-s\sin(0.5)}{s^2+9}$ ---> is this correct?

for the firt prob can you give me hint on how to deal with it.

Hint: $\displaystyle \begin{align*} \sin^2{(x)} \equiv \frac{1}{2} - \frac{1}{2}\cos{ \left( 2\theta \right) } \end{align*}$

 

[Fantini]

Write $\sin^2(4t)$ using the double angle relation for the first. For the second your approach is correct but remember that you can use time translation: $$\mathcal{L}(f(t-\tau))=e^{- \tau} \mathcal{L}(f(t)).$$

 

[Drain Brain]

here's my attempt using the identity you gave me

$\sin^{2}(4t)=\frac{1}{2}-\frac{\cos(8t)}{2}$

so, $\mathscr{L}\{\sin^{2}(4t)\} = \frac{1}{2}\left(\frac{1}{s}-\frac{s}{s^2+64}\right )=\frac{32}{s^3+64s}$

is my answer corect?

 

[Prove It]

here's my attempt using the identity you gave me

$\sin^{2}(4t)=\frac{1}{2}-\frac{\cos(8t)}{2}$

so, $\mathscr{L}\{\sin^{2}(4t)\} = \frac{1}{2}\left(\frac{1}{s}-\frac{s}{s^2+64}\right )=\frac{32}{s^3+64s}$

is my answer corect?

Yes, well done :)

 

[Drain Brain]

did I get the correct answer for the 2nd problem? And can you tell me another way of solving that.