Answer
$ \dfrac{3 \sqrt{13}}{13}$; $\dfrac{2 \sqrt{13}}{13}$; $\dfrac{3}{2}$
Work Step by Step
We know that $\cos a=-\sqrt{1-\sin^2 a}=-\sqrt {1-(12/13)^2}=-\dfrac{5}{13}$
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. $\sin \dfrac{a}{2}= \sqrt {\dfrac{1- \cos a}{2}}=\sqrt {\dfrac{1- \cos (-\dfrac{5}{13})}{2}}= \dfrac{3 \sqrt{13}}{13}$
b) $\cos \dfrac{a}{2}=\sqrt {\dfrac{1+ \cos a}{2}}=\sqrt {\dfrac{1+ \cos (-5/13)}{2}}=\sqrt {\dfrac{4}{13}}=\dfrac{2 \sqrt{13}}{13}$
c) $\tan \dfrac{a}{2}= \dfrac{\sin a/2}{\cos a/2}=\dfrac{\dfrac{3 \sqrt{13}}{13}}{\dfrac{2 \sqrt{13}}{13}}=\dfrac{3}{2}$
