if cos a = 40/41, and tan b = 5/12 and neither a nor b are in quadrant 1, find sin 2b, and sec (a-b)

Question

I've been stumped for over a week

Answer 1

The fractions indicate that you are looking at right triangles that are Pythagorean triples x-y-hyp = to 40-9-41 (cos = x/hyp = 40/41) in the first case and 12-5-13 (tan = y/x=5/12) in the second case. Once you know one value for a trig function you can work out the other 5. The questions here ask for some things that require identities to put them in terms of trig functions of a and b:

a) sin(2b) = 2sin(b)cos(b) From the 12-5-13 triangle, this is 2(5/13)(12/13)

b) sec(a-b) =1/cos(a-b) = (sin(a)sin(b) + cos(a)cos(b))^-1 = ((9/41)(5/13) + (40/41)(12/13))^-1