Answer
(a) $ -\frac{2\sqrt 2}{3}$
(b) $ \frac{1}{3}$
(c) $ \sqrt {\frac{3-\sqrt 6}{6}}$
(d) $ -\sqrt {\frac{3+\sqrt 6}{6}}$
Work Step by Step
Given $sin\theta=-\frac{\sqrt 3}{3}$ and $\frac{3\pi}{2}\lt \theta\lt 2\pi$, we have $cos\theta=\frac{\sqrt 6}{3}$ and $\frac{3\pi}{4}\lt \frac{\theta}{2}\lt\pi$.
(a) $sin(2\theta)=2sin\theta cos\theta=2(\frac{-\sqrt 3}{3})(\frac{\sqrt 6}{3})=-\frac{2\sqrt 2}{3}$
(b) $cos(2\theta)=1-2sin^2(\theta)=1-2(-\frac{\sqrt 3}{3})^2=\frac{1}{3}$
(c) $sin(\frac{\theta}{2})=\sqrt {\frac{1-cos\theta}{2}}=\sqrt {\frac{1-\frac{\sqrt 6}{3}}{2}}=\sqrt {\frac{3-\sqrt 6}{6}}$
(d) $cos(\frac{\theta}{2})=-\sqrt {\frac{1+cos\theta}{2}}=-\sqrt {\frac{1+\frac{\sqrt 6}{3}}{2}}=-\sqrt {\frac{3+\sqrt 6}{6}}$
