Please Help! Pre-Calculus and Trigonometry.

Question

cos a = - 12/13, quadrant II, sin B = 1/2, quadrant II. Find cos (a-B).

Answer 1

In quadrant 2, cosine is negative and sine is positive

sina = √[1-cos²a] = 5/13 and cosB = -√[1-sin²B] = -√3/2

So, cos(a-B) = (cosa)(cosB) + (sina)(sinB) = (-12/13)(-√3/2) + (5/13)(1/2) = (12√3 + 5)/26