Question

高一数学（正弦余弦）

已知三角形ABC，a+c=2b,A-C=60度。则SinB=?

Answer 1

A - C = pi/3
A + C = pi - B
所以 ：
2A = 4pi/3 - B
即：A = 2pi/3 - B
C = pi - A - B
= pi/3 - B/2

由正弦定理及“a+c=2b”，得：
sinA + sinC = 2sinB

sinA + sinC = 2sin((A+C)/2)cos((A-C)/2)
= 2cos(pi/2 - (A+C)/2) cos(pi/6)
= 2cos(B/2) * √3/2
= √3cos(B/2)
2sinB = 4sin(B/2)cos(B/2)

所以：
√3cos(B/2) = 4sin(B/2)cos(B/2)
所以：
cos(B/2) = 0（B=pi，故舍去）
或sin(B/2) = √3/4
因此：
cos(B/2) = √13/4
sinB = 2sin(B/2)cos(B/2) = √39/8
pi=180度