25、在△ABC中求证：sinA+sinB+sinC≤cosA/?
25、在△AB中求证：sinA+sinB+sinC≤cosA/2+cosB/2+cosC/2
sinA+sinB+sinC=1\2[(sinA+sinB)+(sinA+sinC)+(sinB+sinC)] =1\2(2sinA+B\2*cosA-B\2+2sinA+C\2*cosA-C\2+2sinB+C\2cosB-C\2) =cosC\2*cosA-B\2+cosB\2cosA-C\2+cosA\2cosB-C\2 ≤cosA/2+cosB/2+cosC/2