Form 4 Trigonometry

sinA = cosB - cosC2 sin(A/2) cos(A/2) = - 2 sin[(B+C)/2] sin[(B-C)/2]sin(A/2) cos(A/2) = - cos[π/2 - (B+C)/2] sin[(B-C)/2]sin(A/2) cos(A/2) = - cos(A/2) sin[(B-C)/2]cos(A/2) {sin(A/2) + sin[(B-C)/2]} = 0cos(A/2) = 0 or sin(A/2) + sin[(B-C)/2] = 0A/2 = π/2 (rejected) or sin(A/2) + sin[(B-C)/2 = 02 sin[A/2 + (B-C)/2] cos[A/2 - (B-C)/2] = 02 sin[(A+B-C)/2] cos[(A-B+C)/2] = 0sin[(A+B-C)/2] = 0 or cos[(A-B+C)/2] = 0(A+B-C)/2 = 0 or (A-B+C)/2 = π/2 (rejected)A + B = C π - C = CC = π/2 △ABC is a right - angled triangle.

2011-05-19 01:36:18 補充：
Sorry for some mistakes :

From line 9 :

sin(A/2) + sin[(B-C)/2 = 0

2 sin{[A/2 + (B-C)/2]/2} cos{[A/2 - (B-C)/2]/2} = 0

2 sin[(A+B-C)/4] cos[(A-B+C)/4] = 0

sin[(A+B-C)/4] = 0 or cos[(A-B+C)/4] = 0

(A+B-C)/4 = 0 or (A-B+C)/4 = π/2 (rejected)

2011-05-19 01:36:23 補充：
A + B = C

π - C = C

C = π/2

△ABC is a right - angled triangle.