## Trigonometry (11th Edition) Clone

$sin~(arccos~u) = \sqrt{1-u^2}$
Let $~~\theta = arccos~u$ Then $~~cos~\theta = u$ $sin~\theta = \sqrt{1-cos^2~\theta}$ $sin~\theta = \sqrt{1-u^2}$ Therefore, $~~sin~(arccos~u) = \sqrt{1-u^2}$