回覆 3# peter0210 的帖子
第 2 題
三角形\(ABC\)中,三頂點\(A,B,C\)對面的三邊長分別為\(a,b,c\)。若\(a+c=2b\),且角\(\displaystyle A-C=\frac{\pi}{3}\),試求\(cosB=\)
[解答]
a + c = 2b
sinA + sinC = 2sinB
2sin[(A + C)/2]cos[(A - C)/2] = 4sin(B/2)cos(B/2)
2cos[(A - C)/2] = 4sin(B/2)
√3 = 4sin(B/2)
sin(B/2) = √3 / 4
cosB = 1 - 2[sin(B/2)]^2 = 5/8