Answer
$y=2\sin (2x-\pi)+1$
Work Step by Step
The sine function is:
$$y=a\sin (bx-c)+d$$
Rewriting the equation:
$$y=a\sin b\left(x-\frac{c}{b}\right)+d$$
Finding $b$:
$$period=\frac{2\pi}{b}$$ $$b=\frac{2\pi}{\pi}=2$$
With right phase shift of $\frac{\pi}{2}$:
$$\frac{c}{b}=\frac{\pi}{2}$$ $$c=\frac{\pi}{2}b=\frac{\pi}{2}(1)=\pi$$
Substituting $a=2,~b=2,~c=\pi$ and $d=1$, the function is:
$$y=2\sin (2x-\pi)+1$$
