## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\sin (\dfrac{\theta}{2}) =\sqrt {\dfrac{1-\cos \theta}{2}}$
We know that the half-angle formula for $\sin$ is $\sin (\dfrac{x}{2}) = \pm \sqrt {\dfrac{1-\cos x}{2}}$ Since the angle is acute, we will write it as: $\sin (\dfrac{\theta}{2}) =\sqrt {\dfrac{1-\cos \theta}{2}}$