Answer
(a) $-\frac{2\sqrt 5}{5}$
(b) $ \frac{\sqrt 5}{5}$.
(c) $ -\frac{4}{5}$
(d) $ -\frac{3}{5}$
(e) $ \sqrt {\frac{5-\sqrt 5}{10}}$
(f) $ \sqrt {\frac{5+\sqrt 5}{10}}$
Work Step by Step
(a) Given $tan\theta=-2$ and $\frac{3\pi}{2}\lt\theta\lt2\pi$, we have $sin\theta=-\frac{2}{\sqrt {2^2+1^2}}=-\frac{2\sqrt 5}{5}$
(b) $cos\theta=\frac{1}{\sqrt {2^2+1^2}}=\frac{\sqrt 5}{5}$.
(c) $sin2\theta=2sin\theta cos\theta=2(-\frac{2\sqrt 5}{5})(\frac{\sqrt 5}{5})=-\frac{4}{5}$
(d) $cos(2\theta)=2cos^2\theta-1=2(\frac{\sqrt 5}{5})^2-1=-\frac{3}{5}$
(e) $\frac{3\pi}{4}\lt\frac{\theta}{2}\lt\pi$ and $sin(\frac{\theta}{2})=\sqrt {\frac{1-cos\theta}{2}}=\sqrt {\frac{1-\frac{\sqrt 5}{5}}{2}}=\sqrt {\frac{5-\sqrt 5}{10}}$
(f) $\frac{3\pi}{4}\lt\frac{\theta}{2}\lt\pi$ and $cos(\frac{\theta}{2})=-\sqrt {\frac{1+cos\theta}{2}}=\sqrt {\frac{1+\frac{\sqrt 5}{5}}{2}}=\sqrt {\frac{5+\sqrt 5}{10}}$
