三角恒等式證明

1)tan A/2 = tan( 180˚/2 - (B/2 + C/2) )tan A/2 = 1 / tan(B/2 + C/2)tan A/2 = (1 - tan B/2 tan C/2) / (tan B/2 + tan C/2)tan A/2 tan B/2 + tan B/2 tan C/2 + tan C/2 tan A/2 = 1
2)sinA + sinB + sinC= 2 sin (A+B)/2 cos (A-B)/2 + 2sin C/2 cos C/2= 2 sin(180˚/2 - C/2) cos (A-B)/2 + 2 cosC/2 sin(180˚/2 - (A+B)/2)= 2 cos C/2 cos (A-B)/2 + 2 cos C/2 cos (A+B)/2= 2 cos C/2 (cos (A-B)/2 + cos (A+B)/2)= 2 cos C/2 {2 cos [(A-B)/2 + (A+B)/2]/2} cos [(A+B)/2 - (A-B)/2]/2}= 4 cos A/2 cos B/2 cos C/2
3) sin (A-B) + sin (A-C) + sin (B-C)= 2 sin [(A-B)/2 + (A-C)/2] cos [(A -B)/2 - (A-C)/2] + sin (B-C)= 2 sin (2A-B-C)/2 cos (C-B)/2 + sin (B-C) = 2 sin (2A-B-C)/2 cos (B-C)/2 + 2 sin (B-C)/2 cos (B-C)/2 = 2 cos (B-C)/2 (sin (2A-B-C)/2 + sin (B-C)/2)= 2 cos (B-C)/2 {2 sin [(2A-B-C)/4 + (B-C)/4] cos [(2A-B-C)/4 - (B-C)/4]}= 4 cos (B-C)/2 sin (A-C)/2 cos (A-B)/2