Recall the half-angle identity,
cos²(x/2) = (1 + cos(x))/2
This means
[tex]\cos^2\left(\dfrac12 \arccos\left(-\dfrac{12}{13}\right)\right) = \dfrac{1 + \cos\left(\arccos\left(-\frac{12}{13}\right)\right)}2 = \dfrac{1 - \frac{12}{13}}2 = \boxed{\frac1{26}}[/tex]
