Question

Help guys with the question

Answer 1

Answer:

$\sin\Big(\dfrac{\theta }{2}\Big) = \dfrac{2\sqrt 5}{5}$

Step-by-step explanation:

$\cos(2x) = \cos^2 x-\sin^2 x = 1-2\sin^2 x \\ \\ \cos(x) = 1-2\sin^2 (\frac{x}{2}) \\ \\ \Rightarrow \sin^2 (\frac{x}{2}) = \dfrac{1-\cos(x)}{2}\\ \\ \sin(\frac{x}{2}) = \pm \sqrt{\dfrac{1-\cos(x)}{2}},\quad x\in [\frac{3\pi }{2},\pi] \Rightarrow \frac{x}{2}\in [\frac{3\pi}{4},\frac{\pi}{2}]\\ \\ \Rightarrow \sin(\frac{x}{2}) > 0 \Rightarrow \sin(\frac{x}{2}) = \sqrt{\dfrac{1-(-\frac{3}{5})}{2}} \Rightarrow \sin(\frac{x}{2}) = \sqrt{\dfrac{8}{10}}=\dfrac{2\sqrt 2}{\sqrt{10}} = \\ \\ =\dfrac{2\sqrt 5}{5}$