Find the Exact Value sin(arccos(0))

Draw a triangle in the plane with vertices ${\displaystyle (0,\sqrt{1^2-0^2})}$, ${\displaystyle (0,0)}$, and the origin. Then ${\displaystyle \arccos \left(0\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (0,\sqrt{1^2-0^2})}$. Therefore, ${\displaystyle \sin \left(\arccos \left(0\right)\right)}$ is ${\displaystyle \sqrt{1}}$.