Answer
(a) $ -\frac{4}{5}$
(b) $ \frac{3}{5}$
(c) $ \sqrt {\frac{5+2\sqrt 5}{10}}$
(d) $ \sqrt {\frac{5-2\sqrt 5}{10}}$
Work Step by Step
Given $cot\theta=-2, sec\theta\lt0$, we know $\frac{\pi}{2}\lt \theta\lt \pi$. We have $sin\theta=\frac{\sqrt 5}{5},cos\theta=-\frac{2\sqrt 5}{5}$ and $\frac{\pi}{4}\lt \frac{\theta}{2}\lt\frac{\pi}{2}$.
(a) $sin(2\theta)=2sin\theta cos\theta=2(\frac{\sqrt 5}{5})(-\frac{2\sqrt 5}{5})=-\frac{4}{5}$
(b) $cos(2\theta)=1-2sin^2(\theta)=1-2(-\frac{\sqrt 5}{5})^2=\frac{3}{5}$
(c) $sin(\frac{\theta}{2})=\sqrt {\frac{1-cos\theta}{2}}=\sqrt {\frac{1+\frac{2\sqrt 5}{5}}{2}}=\sqrt {\frac{5+2\sqrt 5}{10}}$
(d) $cos(\frac{\theta}{2})=\sqrt {\frac{1+cos\theta}{2}}=\sqrt {\frac{1-\frac{2\sqrt 5}{5}}{2}}=\sqrt {\frac{5-2\sqrt 5}{10}}$
