Originally Posted by

**muddywaters** the answer given is:

since $\displaystyle \frac{\pi}{2}<\theta<\pi, 0<|cos\theta|<1$

as $\displaystyle \theta$ goes from $\displaystyle \frac{\pi}{2}$ to $\displaystyle \pi$, $\displaystyle \cos{\theta}$ goes from 0 to -1, so the absolute value goes from 0 to 1.

since $\displaystyle \pi<2\theta<2\pi \and \2\theta\neq\frac{3}{2}\pi, 0<|sin2\theta|<1$

as $\displaystyle 2\theta$ goes from $\displaystyle \pi$ to $\displaystyle 2\pi$, $\displaystyle \sin{\theta}$ goes from 0 to -1 and back to zero, but the point where it is -1, $\displaystyle 2\theta=\frac{3\pi}{2}$, is excluded so the absolute value is between 0 to 1.