## Algebra 2 (1st Edition)

a) $\sin \dfrac{a}{2}==\dfrac{\sqrt{10}}{10}$ b)$\cos \dfrac{a}{2}= \dfrac{3 \sqrt{10}}{10}$ c) $\tan \dfrac{a}{2}=\dfrac{1}{3}$
a) 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 first quadrant. $\sin \dfrac{a}{2}= \sqrt {\dfrac{1- \cos a}{2}}=\sqrt {\dfrac{1- \cos 4/5}{2}}= \dfrac{\sqrt{10}}{10}$ b) Since $\cos a$ is positive in the first quadrant. $\cos \dfrac{a}{2}= \sqrt {\dfrac{1+ \cos a}{2}}=\sqrt {\dfrac{1+ \cos 4/5}{2}}=\sqrt {\dfrac{9}{10}}=\dfrac{3 \sqrt{10}}{10}$ c) $\tan \dfrac{a}{2}= \dfrac{\sin a/2}{\cos a/2}=\dfrac{\dfrac{\sqrt{10}}{10}}{\dfrac{3 \sqrt{10}}{10}}=\dfrac{1}{3}$