## Algebra 2 (1st Edition)

a) $\dfrac{\sqrt{3}}{3}$ b) $-\dfrac{\sqrt{6}}{3}$ c) $-\dfrac{\sqrt 2}{2}$
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 second quadrant. $\sin \dfrac{a}{2}= \sqrt {\dfrac{1- \cos a}{2}}=\sqrt {\dfrac{1- \cos (1/3)}{2}}= \dfrac{\sqrt{3}}{3}$ b) Since $\cos a$ is negative in the second quadrant. $\cos \dfrac{a}{2}=- \sqrt {\dfrac{1+ \cos a}{2}}=-\sqrt {\dfrac{1+ \cos (1/3)}{2}}=-\sqrt {\dfrac{4/3}{2}}=-\dfrac{\sqrt{6}}{3}$ c) $\tan \dfrac{a}{2}= \dfrac{\sin a/2}{\cos a/2}=\dfrac{\dfrac{\sqrt{3}}{3}}{-\dfrac{\sqrt{6}}{3}}=-\dfrac{\sqrt 2}{2}$