求证sinA+sinC=2*sin[A+C)/2]*cos[(A-C)/2]
求证sinA+sinC=2*sin[A+C)/2]*cos[(A-C)/2]
日期：2021-05-01 02:40:25 人气：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[(
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[(