Find the Exact Value sin(arccos( square root of 5))

Draw a triangle in the plane with vertices ${\displaystyle (\sqrt{5},\sqrt{1^2-{\sqrt{5}}^2})}$, ${\displaystyle (\sqrt{5},0)}$, and the origin. Then ${\displaystyle \arccos \left(\sqrt{5}\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (\sqrt{5},\sqrt{1^2-{\sqrt{5}}^2})}$. Therefore, ${\displaystyle \sin \left(\arccos \left(\sqrt{5}\right)\right)}$ is ${\displaystyle \sqrt{-4}}$.

Rewrite ${\displaystyle -4}$ as ${\displaystyle -1\left(4\right)}$.

Rewrite ${\displaystyle \sqrt{-1\left(4\right)}}$ as ${\displaystyle \sqrt{-1}\cdot \sqrt{4}}$.

Rewrite ${\displaystyle \sqrt{-1}}$ as ${\displaystyle i}$.

Rewrite ${\displaystyle 4}$ as ${\displaystyle 2^2}$.

Pull terms out from under the radical, assuming positive real numbers.