## Algebra 2 (1st Edition)

$\dfrac{3 \sqrt{10}}{10} ; \\ -\dfrac{\sqrt{10}}{10}; \\-3$
We know that $\cos a=-\sqrt{1-\sin^2 a}=-\sqrt{1-\sin (-3/5)^2}=-\dfrac{4}{5}$ Double -angle Theorem can be defined as: $\sin \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{2}}$ Since $\sin a$ is positive in the second quadrant. $\sin \dfrac{a}{2}= \sqrt {\dfrac{1- \cos a}{2}}=\sqrt {\dfrac{1- \cos (-\dfrac{4}{5})}{2}}=\sqrt {\dfrac{9}{10}}= \dfrac{3 \sqrt{10}}{10}$ b) $\cos \dfrac{a}{2}=-\sqrt {\dfrac{1+ \cos a}{2}}=-\sqrt {\dfrac{1- \cos (4/5)}{2}}=-\dfrac{1}{\sqrt {10}}=-\dfrac{\sqrt{10}}{10}$ c) $\tan \dfrac{a}{2}= \dfrac{\sin a/2}{\cos a/2}=\dfrac{ \dfrac{3 \sqrt{10}}{10}}{-\dfrac{\sqrt{10}}{10}}=-3$