Question

求证sinA+sinC=2*sin[A+C)/2]*cos[(A-C)/2]

Answer 1

∵A = (A+C)/2 + (A-C)/2

C = (A+C)/2 - (A-C)/2

∴sinA + sinC = sin[(A+C)/2 + (A-C)/2] + sin[(A+C)/2 - (A-C)/2]

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

+ sin[(A+C)/2]cos[(A-C)/2] - cos[(A+C)/2]sin[(A-C)/2]

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