✔ 最佳答案
(a)
a / sinA = k
= k (sinA + sinB + sinC) / (sinA + sinB + sinC)
= (k sinA + k sinB + ksinC) / (sinA + sinB + sinC)
= (a + b + c) / (sinA + sinB + sinC)
(b)
sin C = sin(π - A - B) , 因為 三角形內角和 = A+B+C = π
= sin(A + B)
= 2 sin[(A + B)/2] cos[(A + B)/2]
sinA + sinB + sinC
= (sinA + sinB) + sinC
= 2 sin[(A + B)/2] cos[(A - B)/2]
+ 2 sin[(A + B)/2] cos[(A + B)/2]
= 2 sin[(A + B)/2] • {cos[(A + B)/2] + cos[(A - B)/2]}
= 2 cos[π/2 - (A + B)/2]
• {2 cos ½[(A + B)/2 + (A - B)/2] cos ½[(A + B)/2 + (A - B)/2] }
= 2 cos[(π - A - B)/2]
• [2 cos(A/2) cos(B/2)]
= 2 cos(C/2) • 2 cos(A/2) cos(B/2)
= 4 cos(A/2) cos(B/2) cos(C/2)