求证sinA+sinC=2sin[A+C)/2]cos[(A-C)/2]

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

∵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]