Answer
(a) $-\frac{7}{25}$
(b) $-\frac{24}{25}$
(c) $\frac{24}{7}$
(d) $\frac{\sqrt 5}{5}$
(e) $2$
Work Step by Step
Given $cos\theta=-\frac{3}{5}$ and $90^\circ\lt\theta\lt180^\circ$, we have $180^\circ\lt 2\theta\lt360^\circ$ and $45^\circ\lt\frac{\theta}{2}\lt90^\circ$
(a) Use the double-angle formula, we have $cos2\theta=2cos^2\theta -1=2(-\frac{3}{5})^2-1=-\frac{7}{25}$
(b) We can find $sin\theta=\frac{4}{5}$ and $sin2\theta=2sin\theta cos\theta=2(\frac{4}{5})(-\frac{3}{5})=-\frac{24}{25}$
(c) We have $tan\theta=-\frac{4}{3}$ and $tan2\theta=\frac{2tan\theta}{1-tan^2\theta}=\frac{2(-\frac{4}{3})}{1-(-\frac{4}{3})^2}=\frac{24}{7}$
(d) Use the half-angle formula, we have $cos(\frac{\theta}{2})=\sqrt {\frac{1+cos\theta}{2}}=\sqrt {\frac{1+(-\frac{3}{5})}{2}}=\frac{\sqrt 5}{5}$
(e) We have $sin(\frac{\theta}{2})=\sqrt {\frac{1-cos\theta}{2}}=\sqrt {\frac{1-(-\frac{3}{5})}{2}}=\frac{2\sqrt 5}{5}$, thus $tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}=2$
