Sketch on the same axes graphs of$\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{t}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(t-\pi /2\right)$, and$\mathit{s}\mathit{i}\mathit{n}\left(t+\pi /2\right)$, and observe which way the graph shifts. Hint: You can, of course, have your calculator or computer plot these for you, but it's simpler and much more useful to do it in your head. Hint: What values of $\mathit{t}$make the sines equal to zero? For an even simpler example, sketch on the same axes$\mathit{y}\mathbf{=}\mathit{t}\mathbf{,}\mathit{y}\mathbf{=}\mathit{t}\mathbf{-}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{\pi }\mathbf{/}\mathbf{2}$.

The given function is

$$\begin{gathered} y=\sin t...\left(1\right) \\ y=\sin \left(t-\frac{\mathrm{\pi }}{2}\right)...\left(2\right) \\ y=\sin \left(t+\frac{\mathrm{\pi }}{2}\right)...\left(3\right) \end{gathered}$$

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

All the given functions are periodic function of same period$2\mathrm{\pi }$. So, the graph will be quite similar in nature.

Next observe that the first function attains the value 0 at$t=0,\pm \mathrm{\pi },\pm 2\mathrm{\pi }$.

Whereas the second attains the values 0 at $t=+\mathrm{\pi }/2,+3\mathrm{\pi }/2,+5\mathrm{\pi }/2...$and the third function at $t=+\mathrm{\pi }/2,+3\mathrm{\pi }/2,+5\mathrm{\pi }/2...$However the second function attains the value $-1$for $t=0$Whereas third function attains the value 1 at$t=0$.

The graph is given below.

The graph shifts to right $by\mathrm{\pi }/2$for $\mathit{t}$changed to role="math" localid="1654154141557" $t-\mathrm{\pi }/2$and to the left by role="math" localid="1654154128983" $\mathrm{\pi }/2$when $\mathit{t}$is changed$t+\mathrm{\pi }/2$. The shape and size of three graph remains same.

Thus, the graph for $y=\sin t,y=\sin \left(t-\frac{\mathrm{\pi }}{2}\right),y=\sin \left(t+\frac{x}{2}\right)$will be same for all three functions.