Question

sin(a)+sin(b)=sqrt(5/3) and cos(a)+cos(b)=1. Compute cos(a-b).
I drew out triangles and modeled the sines and cosines of a and b, but that doesnt help me find cos(a-b). What do I do?

Answer 1

Square both given expressions and add them together. Make use of trig identities.

  (sin(a) +sin(b))^2 = sin(a)^2 +2sin(a)sin(b) +sin(b)^2 = 5/3

  (cos(a) +cos(b))^2 = cos(a)^2 +2cos(a)cos(b) +cos(b)^2 = 1

Adding, we have
  sin(a)^2 +cos(a^2) +2(cos(a)cos(b) +sin(a)sin(b)) +sin(b)^2 +cos(b)^2 = 2 2/3

Of course, the first and last pairs of terms are 1 and 1, and the middle product is 2cos(a-b), so you have
  1 + 2cos(a-b) +1 = 2 2/3
  cos(a-b) = 1/3