If cos a=-12/13 with 90°<A<180° and if sin b=-4/5 with 270°<B<360°, find the cos(A-B)
cos a = -12/13, and a is in quad II
so sin a = 5/13, (you should recognize the 5-12-13 rightangled triangle)
sin b = -4/5, and b is in quad IV
so cos b = 3/5 (you have the famous 3-4-5 right-angled triangle)
cos(a - b) = cosa cosb + sina sinb
= (-12/13)(3/5) + (5/13)(-4/5)
= -36/65 - 20/65
= -56/65