Plane Trigonometry

*35. Prove that in any triangle (Page 182 of Plane Trigonometry)
(a+b)/c = cos [(A-B)/2] / sin (C/2)

[Solution]

According to the law of sine, we can get

Left = (a+b)/c=(sinA+sinB)/sinC

We can multiply the numerator and the denominator of the right side of the original equation by 2cos (C/2). Notice that cos(C/2) = 2sin [(A+B)/2]. Then we get

Right = 2cos [(A-B)/2] sin [(A+B)/2] / [2sin (C/2) cos (C/2)]
          = [sin (A) - sin (-B)]/ sin (C) = [ sin (A) + sin (B) ] / sin (C)
          = Left

[End of Proof]