Evaluate sin(arccos(-4/( square root of 25)))

Draw a triangle in the plane with vertices ${\displaystyle (-\frac{4}{\sqrt{25}},\sqrt{1^2-{(-\frac{4}{\sqrt{25}})}^2})}$, ${\displaystyle (-\frac{4}{\sqrt{25}},0)}$, and the origin. Then ${\displaystyle \arccos (-\frac{4}{\sqrt{25}})}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (-\frac{4}{\sqrt{25}},\sqrt{1^2-{(-\frac{4}{\sqrt{25}})}^2})}$. Therefore, ${\displaystyle \sin \left(\arccos (-\frac{4}{\sqrt{25}})\right)}$ is ${\displaystyle \sqrt{\frac{9}{25}}}$.

Rewrite ${\displaystyle \sqrt{\frac{9}{25}}}$ as ${\displaystyle \frac{\sqrt{9}}{\sqrt{25}}}$.