1. The problem is to find the period of the function $y = -\sin\left(\frac{\pi}{2}(x+1)\right)$.\n\n2. The general form of a sine function is $y = \sin(bx)$, where the period is given by $\frac{2\pi}{|b|}$.\n\n3. In the given function, the argument of sine is $\frac{\pi}{2}(x+1)$, so $b = \frac{\pi}{2}$.\n\n4. Calculate the period using the formula: $$\text{Period} = \frac{2\pi}{\left|\frac{\pi}{2}\right|} = \frac{2\pi}{\frac{\pi}{2}} = 2\pi \times \frac{2}{\pi} = 4.$$\n\n5. The negative sign and the horizontal shift $(x+1)$ do not affect the period.\n\n6. Therefore, the period of $y = -\sin\left(\frac{\pi}{2}(x+1)\right)$ is $4$.