一个沉了的三角函数问题……设r,R分别为三角型ABC的内切圆和外接圆半径,为什么4R*sinA/2*sinB/2*sin
1个回答

均由正弦定理证:

(1)由正弦定理:(R为外接圆半径)

a = 2R * sinA, b = 2R * sinB, c = 2R * sinC,

各式相加求得半周长 p:

p =(a+b+c)/2

= R * (sinA + sinB + sinC)

= R * (sinA + sinB + sin(180° - A - B)

= R * (sinA + sinB + sin(A + B)

= R * (sinA + sinB + sinA * cosB + cosA * sinB)

= R * (sinA * (1 + cosB) + sinB * (1 + cosA))

= R * (2sin(A/2)cos(A/2) * 2cos^2(B/2) + 2sin(B/2)cos(B/2) * 2cos^2(A/2))

= 4R * cos(A/2)cos(B/2)(sin(A/2)cos(B/2) + sin(B/2)cos(A/2))

= 4R * cos(A/2)cos(B/2)sin((A + B)/2)

= 4R * cos(A/2)cos(B/2)sin(90° - C/2)

= 4R * cos(A/2)cos(B/2)cos(C/2).

容易证明的是内切圆半径 r = S / p (这个好证,三个小三角形面积和,自己搞定吧,一般就直接当定理用)

所以三角形面积:

S = r * p = 4Rr * cos(A/2)cos(B/2)cos(C/2) (式1)

(2)同样由正弦定理:

a = 2R * sinA, b = 2R * sinB, c = 2R * sinC,

各式相乘求得三角形面积:

S = abc / (4R) = 2R^2 * sinA * sinB * sinC (式2)

三角形面积 S = abc / (4R) 也很好证,画个图用相似三角形证.(这个也一般当个定理用即可)

联立 式1 和 式2:

4Rr * cos(A/2)cos(B/2)cos(C/2) = 2R^2 * sinA * sinB * sinC

即可得:

r = 4R * sin(A/2) * sin(B/2) * sin(C/2)