Question

$L\left\{\sin tU\left(t-\frac{\pi }{2}\right)\right\}$

Short answer

The Laplace transform is$\frac{s}{s^2+1}{e}^{-s\frac{\pi }{2}}$

Step-by-step solution

Second Translation Theorem

If $F\left(s\right)=L\left\{f\right(t\left)\right\}$and$a>0$, then$L\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Apply Laplace transform

$L\left\{\sin \right(t\left)U\left(t-\frac{\pi }{2}\right)\right\}=L\left\{\sin \left(\left(t-\frac{\pi }{2}\right)+\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)\right\}$

We know that,

$\sin \left(x+\frac{\pi }{2}\right)=\cos x$

So,

$$\begin{gathered} L\left\{\sin \left(\left(t-\frac{\pi }{2}\right)+\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)\right\}=L\left\{\cos \left(t-\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)\right\} \\ Y\left(s\right)=\frac{s}{s^2+1}{e}^{-s\frac{\pi }{2}} \end{gathered}$$

Hence the solution is $Y\left(s\right)=\frac{s}{s^2+1}{e}^{-s\frac{\pi }{2}}$