Consider the following for the two (02) items that follow:
Let $p = |\sin\alpha - \sin(\alpha - 90°)|$.

What is the minimum value of $p$?

$p = |\sin\alpha - \sin(\alpha - 90^\circ)|$

Using the trigonometric identity $\sin(\alpha - 90^\circ) = -\sin(90^\circ - \alpha) = -\cos\alpha$, we get:

$p = |\sin\alpha - (-\cos\alpha)| = |\sin\alpha + \cos\alpha|$

Multiplying and dividing the expression inside the modulus by $\sqrt{2}$:

$p = \sqrt{2} \left| \dfrac{1}{\sqrt{2}}\sin\alpha + \dfrac{1}{\sqrt{2}}\cos\alpha \right|$

$p = \sqrt{2} |\sin(\alpha + 45^\circ)|$

Since the absolute value of any real expression is non-negative, the minimum possible value of $p$ is $0$. This minimum is achieved when $\sin(\alpha + 45^\circ) = 0$, which occurs at $\alpha = -45^\circ$.

Therefore, the minimum value of $p$ is $0$.